diff --git a/.gitignore b/.gitignore old mode 100644 new mode 100755 diff --git a/README.md b/README.md old mode 100644 new mode 100755 index 88c8aec..28486a9 --- a/README.md +++ b/README.md @@ -1,165 +1,23 @@ # Overview -1. This is my Python (2.7) Leetcode solution. +1. This is my Python Leetcode solution. As time grows, this also become a guide to prepare for software engineer interview. -2. I really take time tried to make the best solution and collect the best resource that I found. -Because I wanted to help others like me. -If you like my answer, a star on GitHub means a lot to me. -https://github.com/wuduhren/leetcode-python +1. The solution is at `problems/python/` or `problems/python3/`. +For example, `merge-sorted-array.py`'s solution is at `https://leetcode.com/problems/python/merge-sorted-array/`. -3. The solution is at `problems/the-file-name/`. -For example, `merge-sorted-array.py`'s solution is at `https://leetcode.com/problems/merge-sorted-array/`. +2. I really take time tried to make the best solution and collect the best resource that I found. +Because I wanted to help others like me. +Please [BUY ME A COFFEE](https://www.buymeacoffee.com/chriswu) if you want to show support. # Leetcode Problem Lists -* https://leetcode.com/list/?selectedList=535ukjh5 (Only 74 problems) -* https://www.programcreek.com/2013/08/leetcode-problem-classification/ -* https://github.com/wisdompeak/LeetCode -* Below [list](https://docs.google.com/spreadsheets/d/1SbpY-04Cz8EWw3A_LBUmDEXKUMO31DBjfeMoA0dlfIA/edit#gid=126913158) is made by **huahua**, I found this on his [youtube](https://www.youtube.com/user/xxfflower/videos). Please visit his [website](https://zxi.mytechroad.com/blog/leetcode-problem-categories/) for more. - -# Similar Problems I found it makes sense to solve similar problems together, so that we can recognize the problem faster when we encounter a new one. My suggestion is to skip the HARD problems when you first go through these list. -### Two Pointers -| Id | Name | Difficulty | Comments | -| ---: | --- | :---: | --- | -| 11 | [Container With Most Water](https://leetcode.com/problems/container-with-most-water/ "Container With Most Water") | ★★ | | -| 167 | [Two Sum II - Input array is sorted](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted "Two Sum II - Input array is sorted") | ★★ | | -| 977 | [Squares of a Sorted Array](https://leetcode.com/problems/squares-of-a-sorted-array "Squares of a Sorted Array") | ★★ | merge sort | - -### Recursion -| Id | Name | Difficulty | | | -| ---: | --- | :---: | :---: | --- | -| 726 | [Number of Atoms](https://leetcode.com/problems/number-of-atoms "Number of Atoms") | ★★★ | [736](https://leetcode.com/problems/parse-lisp-expression/ "736") | [394](https://leetcode.com/problems/decode-string/ "394") | -| 856 | [Score of Parentheses](https://leetcode.com/problems/score-of-parentheses/ "Score of Parentheses") | ★★★ | | | - -### Divide and Conquer -| Id | Name | Difficulty | Comments | -| ---: | --- | :---: | --- | -| 169 | [Majority Element](https://leetcode.com/problems/majority-element "Majority Element") | ★★ | | -| 315 | [Count of Smaller Numbers After Self](https://leetcode.com/problems/count-of-smaller-numbers-after-self/ "Count of Smaller Numbers After Self") | ★★★★ | merge sort / BIT | - -### Search -| Id | Name | Difficulty | | | | | | | Comments | -| ---: | --- | :---: | :---: | --- | --- | --- | --- | --- | --- | -| 17 | [Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number "Letter Combinations of a Phone Number") | ★★ | [39](https://leetcode.com/problems/combination-sum/ "39") | [40](https://leetcode.com/problems/combination-sum-ii/ "40") | [77](https://leetcode.com/problems/combinations/ "77") | [78](https://leetcode.com/problems/subsets/ "78") | [90](https://leetcode.com/problems/subsets-ii/ "90") | [216](https://leetcode.com/problems/combination-sum-iii/ "216") | Combination | -| 46 | [Permutations](https://leetcode.com/problems/permutations/ "Permutations") | ★★ | [47](https://leetcode.com/problems/permutations-ii/ "47") | [784](https://leetcode.com/problems/letter-case-permutation/ "784") | [943](https://leetcode.com/problems/find-the-shortest-superstring "943") | [996](https://leetcode.com/problems/number-of-squareful-arrays/ "996") | | | Permutation | -| 22 | [Generate Parentheses](https://leetcode.com/problems/generate-parentheses/ "Generate Parentheses") | ★★★ | [301](https://leetcode.com/problems/remove-invalid-parentheses/ "301") | | | | | | DFS | -| 37 | [Sudoku Solver](https://leetcode.com/problems/sudoku-solver "Sudoku Solver") | ★★★ | [51](https://leetcode.com/problems/n-queens "51") | [52](https://leetcode.com/problems/n-queens-ii "52") | | | | | DFS | -| 79 | [Word Search](https://leetcode.com/problems/word-search/ "Word Search") | ★★★ | [212](https://leetcode.com/problems/word-search-ii/ "212") | | | | | | DFS | -| 127 | [Word Ladder](https://leetcode.com/problems/word-ladder/ "Word Ladder") | ★★★★ | [126](https://leetcode.com/problems/word-ladder-ii/ "126") | [752](https://leetcode.com/problems/open-the-lock/ "752") | | | | | BFS | -| 542 | [01 Matrix](https://leetcode.com/problems/01-matrix/ "01 Matrix") | ★★★ | [675](https://leetcode.com/problems/cut-off-trees-for-golf-event/ "675") | [934](https://leetcode.com/problems/shortest-bridge/ "934") | | | | | BFS | -| 698 | [Partition to K Equal Sum Subsets](https://leetcode.com/problems/partition-to-k-equal-sum-subsets "Partition to K Equal Sum Subsets") | ★★★ | [93](https://leetcode.com/problems/restore-ip-addresses/ "93") | [131](https://leetcode.com/problems/palindrome-partitioning/ "131") | [241](https://leetcode.com/problems/different-ways-to-add-parentheses/ "241") | [282](https://leetcode.com/problems/expression-add-operators/ "282") | [842](https://leetcode.com/problems/split-array-into-fibonacci-sequence/ "842") | | Partition | - -### Hash Table -| Id | Name | Difficulty | | -| ---: | --- | :---: | :---: | -| 1 | [Two Sum](https://leetcode.com/problems/two-sum/ "Two Sum") | ★★ | [560](https://leetcode.com/problems/subarray-sum-equals-k/ "560") | - -### List -| Id | Name | Difficulty | | Comments | -| ---: | --- | :---: | :---: | --- | -| 2 | [Add Two Numbers](https://leetcode.com/problems/add-two-numbers/ "Add Two Numbers") | ★★ | [445](https://leetcode.com/problems/add-two-numbers-ii/ "445") | | -| 24 | [Swap Nodes in Pairs](https://leetcode.com/problems/swap-nodes-in-pairs/ "Swap Nodes in Pairs") | ★★ | | | -| 206 | [Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/ "Reverse Linked List") | ★★ | | | -| 141 | [Linked List Cycle](https://leetcode.com/problems/linked-list-cycle/ "Linked List Cycle") | ★★ | [142](https://leetcode.com/problems/linked-list-cycle-ii "142") | fast/slow | -| 23 | [Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/ "Merge k Sorted Lists") | ★★★ | [21](https://leetcode.com/problems/merge-two-sorted-lists/ "21") | priority_queue | -| 147 | [Insertion Sort List](https://leetcode.com/problems/insertion-sort-list/ "Insertion Sort List") | ★★★ | | insertion sort | -| 148 | [Sort List](https://leetcode.com/problems/sort-list/ "Sort List") | ★★★★ | | merge sort O(1) space | -| 707 | [Design Linked List](https://leetcode.com/problems/design-linked-list "Design Linked List") | ★★★★ | | | - -### Tree -| Id | Name | Difficulty | | | | | | | Comments | -| ---: | --- | :---: | :---: | --- | --- | --- | --- | --- | --- | -| 94 | [Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/ "Binary Tree Inorder Traversal") | ★ | [589](https://leetcode.com/problems/n-ary-tree-preorder-traversal "589") | [590](https://leetcode.com/problems/n-ary-tree-postorder-traversal "590") | | | | | traversal | -| 100 | [Same Tree](https://leetcode.com/problems/same-tree/ "Same Tree") | ★★ | [101](https://leetcode.com/problems/symmetric-tree/ "101") | [104](https://leetcode.com/problems/maximum-depth-of-binary-tree/ "104") | [110](https://leetcode.com/problems/balanced-binary-tree/ "110") | [111](https://leetcode.com/problems/minimum-depth-of-binary-tree "111") | [572](https://leetcode.com/problems/subtree-of-another-tree "572") | [965](https://leetcode.com/problems/univalued-binary-tree/ "965") | | -| 102 | [Binary Tree Level Order Traversal](https://leetcode.com/problems/binary-tree-level-order-traversal/ "Binary Tree Level Order Traversal") | ★★ | [107](https://leetcode.com/problems/binary-tree-level-order-traversal-ii "107") | [429](https://leetcode.com/problems/n-ary-tree-level-order-traversal "429") | [872](https://leetcode.com/problems/leaf-similar-trees/ "872") | [987](https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree "987") | | | collecting nodes | -| 814 | [Binary Tree Pruning](https://leetcode.com/problems/binary-tree-pruning/ "Binary Tree Pruning") | ★★ | [669](https://leetcode.com/problems/trim-a-binary-search-tree/ "669") | | | | | | | -| 112 | [Path Sum](https://leetcode.com/problems/path-sum/ "Path Sum") | ★★★ | [113](https://leetcode.com/problems/path-sum-ii "113") | [437](https://leetcode.com/problems/path-sum-iii "437") | | | | | | -| 124 | [Binary Tree Maximum Path Sum](https://leetcode.com/problems/binary-tree-maximum-path-sum/ "Binary Tree Maximum Path Sum") | ★★★ | [543](https://leetcode.com/problems/diameter-of-binary-tree/ "543") | [687](https://leetcode.com/problems/longest-univalue-path/ "687") | | | | | Use both children, return one | -| 129 | [Sum Root to Leaf Numbers](https://leetcode.com/problems/sum-root-to-leaf-numbers/ "Sum Root to Leaf Numbers") | ★★★ | [257](https://leetcode.com/problems/binary-tree-paths/ "257") | | | | | | | -| 236 | [Lowest Common Ancestor of a Binary Tree](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/ "Lowest Common Ancestor of a Binary Tree") | ★★★ | [235](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree "235") | | | | | | | -| 297 | [Serialize and Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/ "Serialize and Deserialize Binary Tree") | ★★★ | [449](https://leetcode.com/problems/serialize-and-deserialize-bst "449") | | | | | | | -| 508 | [Most Frequent Subtree Sum](https://leetcode.com/problems/most-frequent-subtree-sum/ "Most Frequent Subtree Sum") | ★★★ | | | | | | | | -| 968 | [Binary Tree Cameras](https://leetcode.com/problems/binary-tree-cameras/ "Binary Tree Cameras") | ★★★★ | [337](https://leetcode.com/problems/house-robber-iii/ "337") | [979](https://leetcode.com/problems/distribute-coins-in-binary-tree "979") | | | | | | - -### Binary Search -| Id | Name | Difficulty | | | | | | Comments | -| ---: | --- | :---: | :---: | --- | --- | --- | --- | --- | -| 35 | [Search Insert Position](https://leetcode.com/problems/search-insert-position/ "Search Insert Position") | ★★ | [34](https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/ "34") | [704](https://leetcode.com/problems/binary-search/ "704") | [981](https://leetcode.com/problems/time-based-key-value-store "981") | | | upper_bound | -| 33 | [Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array "Search in Rotated Sorted Array") | ★★★ | [81](https://leetcode.com/problems/search-in-rotated-sorted-array-ii/ "81") | [153](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ "153") | [154](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii "154") | [162](https://leetcode.com/problems/find-peak-element "162") | [852](https://leetcode.com/problems/peak-index-in-a-mountain-array/ "852") | rotated / peak | -| 69 | [Sqrt(x)](https://leetcode.com/problems/sqrtx "Sqrt(x)") | ★★★ | | | | | | upper_bound | -| 74 | [Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/ "Search a 2D Matrix") | ★★★ | | | | | | treat 2d as 1d | -| 875 | [Koko Eating Bananas](https://leetcode.com/problems/koko-eating-bananas/ "Koko Eating Bananas") | ★★★ | [1011](https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/ "1011") | | | | | guess ans and check | -| 378 | [Kth Smallest Element in a Sorted Matrix](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/ "Kth Smallest Element in a Sorted Matrix") | ★★★ | [668](https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/ "668") | | | | | kth + matrix | -| 778 | [Swim in Rising Water](https://leetcode.com/problems/swim-in-rising-water/ "Swim in Rising Water") | ★★★ | [174](https://leetcode.com/problems/dungeon-game/ "174") | [875](https://leetcode.com/problems/koko-eating-bananas/ "875") | | | | guess ans and check | -| 4 | [Median of Two Sorted Arrays](https://leetcode.com/problems/median-of-two-sorted-arrays/ "Median of Two Sorted Arrays") | ★★★★ | | | | | | | -| 719 | [Find K-th Smallest Pair Distance](https://leetcode.com/problems/find-k-th-smallest-pair-distance/ "Find K-th Smallest Pair Distance") | ★★★★ | [786](https://leetcode.com/problems/k-th-smallest-prime-fraction/ "786") | | | | | kth + two pointers | - -### Binary Search Tree -| Id | Name | Difficulty | | Comments | -| ---: | --- | :---: | :---: | --- | -| 98 | [Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/ "Validate Binary Search Tree") | ★★ | [530](https://leetcode.com/problems/minimum-absolute-difference-in-bst "530") | inorder | -| 700 | [Search in a Binary Search Tree](https://leetcode.com/problems/search-in-a-binary-search-tree/ "Search in a Binary Search Tree") | ★★ | [701](https://leetcode.com/problems/insert-into-a-binary-search-tree/ "701") | binary search | -| 230 | [Kth Smallest Element in a BST](https://leetcode.com/problems/kth-smallest-element-in-a-bst "Kth Smallest Element in a BST") | ★★★ | | inorder | -| 99 | [Recover Binary Search Tree](https://leetcode.com/problems/recover-binary-search-tree/ "Recover Binary Search Tree") | ★★★ | | inorder | -| 108 | [Convert Sorted Array to Binary Search Tree](https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/ "Convert Sorted Array to Binary Search Tree") | ★★★ | | | -| 501 | [Find Mode in Binary Search Tree](https://leetcode.com/problems/find-mode-in-binary-search-tree/ "Find Mode in Binary Search Tree") | ★★★ | | inorder | -| 450 | [Delete Node in a BST](https://leetcode.com/problems/delete-node-in-a-bst/ "Delete Node in a BST") | ★★★★ | | binary search | - -### Graph -| Id | Name | Difficulty | | | | | Comments | -| ---: | --- | :---: | :---: | --- | --- | --- | --- | -| 133 | [Clone Graph](https://leetcode.com/problems/clone-graph/ "Clone Graph") | ★★ | [138](https://leetcode.com/problems/copy-list-with-random-pointer/ "138") | | | | queue + hashtable | -| 200 | [Number of Islands](https://leetcode.com/problems/number-of-islands/ "Number of Islands") | ★★ | [547](https://leetcode.com/problems/friend-circles/ "547") | [695](https://leetcode.com/problems/max-area-of-island "695") | [733](https://leetcode.com/problems/flood-fill/ "733") | [827](https://leetcode.com/problems/making-a-large-island/ "827") | grid + connected components | -| 841 | [Keys and Rooms](https://leetcode.com/problems/keys-and-rooms/ "Keys and Rooms") | ★★ | | | | | connected components | -| 207 | [Course Schedule](https://leetcode.com/problems/course-schedule/ "Course Schedule") | ★★★ | [210](https://leetcode.com/problems/course-schedule-ii/ "210") | [802](https://leetcode.com/problems/find-eventual-safe-states "802") | | | topology sorting | -| 399 | [Evaluate Division](https://leetcode.com/problems/evaluate-division "Evaluate Division") | ★★★ | [839](https://leetcode.com/problems/similar-string-groups "839") | [952](https://leetcode.com/problems/largest-component-size-by-common-factor/ "952") | [990](https://leetcode.com/problems/satisfiability-of-equality-equations "990") | [721](https://leetcode.com/problems/accounts-merge/ "721") | union find | -| 785 | [Is Graph Bipartite?](https://leetcode.com/problems/is-graph-bipartite "Is Graph Bipartite?") | ★★★ | | | | | bipartition | -| 684 | [Redundant Connection](https://leetcode.com/problems/redundant-connection "Redundant Connection") | ★★★★ | [685](https://leetcode.com/problems/redundant-connection-ii "685") | [787](https://leetcode.com/problems/cheapest-flights-within-k-stops/ "787") | | | cycle, union find | -| 743 | [Network Delay Time](https://leetcode.com/problems/network-delay-time "Network Delay Time") | ★★★★ | [882](https://leetcode.com/problems/reachable-nodes-in-subdivided-graph/ "882") | | | | shortest path | -| 847 | [Shortest Path Visiting All Nodes](https://leetcode.com/problems/shortest-path-visiting-all-nodes/ "Shortest Path Visiting All Nodes") | ★★★★ | [815](https://leetcode.com/problems/bus-routes/ "815") | [864](https://leetcode.com/problems/shortest-path-to-get-all-keys/ "864") | [924](https://leetcode.com/problems/minimize-malware-spread/ "924") | | BFS | -| 943 | [Find the Shortest Superstring](https://leetcode.com/problems/find-the-shortest-superstring/ "Find the Shortest Superstring") | ★★★★ | [980](https://leetcode.com/problems/unique-paths-iii/ "980") | [996](https://leetcode.com/problems/number-of-squareful-arrays/ "996") | | | Hamiltonian path (DFS / DP) | -| 959 | [Regions Cut By Slashes](https://leetcode.com/problems/regions-cut-by-slashes/ "Regions Cut By Slashes") | ★★★★ | | | | | union find / grid + connected component | -| 332 | [Reconstruct Itinerary](https://leetcode.com/problems/reconstruct-itinerary/ "Reconstruct Itinerary") | ★★★★ | | | | | Eulerian path | -| 1192 | [Critical Connections in a Network](https://leetcode.com/problems/critical-connections-in-a-network/ "Critical Connections in a Network") | ★★★★ | | | | | Tarjan | - - -### Dynamic Programming -| Id | Name | Difficulty | | | | | | | Comments | -| ---: | --- | :---: | :---: | --- | --- | --- | --- | --- | --- | -| 70 | [Climbing Stairs](https://leetcode.com/problems/climbing-stairs "Climbing Stairs") | ★ | [746](https://leetcode.com/problems/min-cost-climbing-stairs "746") | | | | | | I: O(n), S = O(n), T = O(n) | -| 303 | [Range Sum Query - Immutable](https://leetcode.com/problems/range-sum-query-immutable "Range Sum Query - Immutable") | ★ | | | | | | | | -| 53 | [Maximum Subarray](https://leetcode.com/problems/maximum-subarray "Maximum Subarray") | ★★ | [121](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/ "121") | | | | | | | -| 198 | [House Robber](https://leetcode.com/problems/house-robber/ "House Robber") | ★★★ | [213](https://leetcode.com/problems/house-robber-ii/ "213") | [309](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/ "309") | [740](https://leetcode.com/problems/delete-and-earn/ "740") | [790](https://leetcode.com/problems/domino-and-tromino-tiling/ "790") | [801](https://leetcode.com/problems/minimum-swaps-to-make-sequences-increasing/ "801") | | I: O(n), S = O(3n), T = O(3n) | -| 139 | [Word Break](https://leetcode.com/problems/word-break "Word Break") | ★★★ | [140](https://leetcode.com/problems/word-break-ii "140") | [818](https://leetcode.com/problems/race-car/ "818") | | | | | I: O(n), S = O(n), T = O(n^2) | -| 300 | [Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/ "Longest Increasing Subsequence") | ★★★ | [673](https://leetcode.com/problems/number-of-longest-increasing-subsequence "673") | | | | | | | -| 72 | [Edit Distance](https://leetcode.com/problems/edit-distance "Edit Distance") | ★★★ | [10](https://leetcode.com/problems/regular-expression-matching "10") | [44](https://leetcode.com/problems/wildcard-matching/ "44") | [97](https://leetcode.com/problems/interleaving-string "97") | [115](https://leetcode.com/problems/distinct-subsequences/ "115") | [583](https://leetcode.com/problems/delete-operation-for-two-strings/ "583") | [712](https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings "712") | I: O(m+n), S = O(mn), T = O(mn) | -| 322 | [Coin Change](https://leetcode.com/problems/coin-change "Coin Change") | ★★★ | [377](https://leetcode.com/problems/combination-sum-iv/ "377") | [416](https://leetcode.com/problems/partition-equal-subset-sum/ "416") | [494](https://leetcode.com/problems/target-sum "494") | | | | I: O(n) + k, S = O(n), T = O(kn) | -| 813 | [Largest Sum of Averages](https://leetcode.com/problems/largest-sum-of-averages/ "Largest Sum of Averages") | ★★★ | | | | | | | I: O(n) + k, S = O(n), T = O(kn^2) | -| 312 | [Burst Balloons](https://leetcode.com/problems/burst-balloons/ "Burst Balloons") | ★★★★ | [664](https://leetcode.com/problems/strange-printer/ "664") | [1024](https://leetcode.com/problems/video-stitching/ "1024") | [1039](https://leetcode.com/problems/minimum-score-triangulation-of-polygon/ "1039") | | | | I: O(n), S = O(n^2), T = O(n^3) | -| 741 | [Cherry Pickup](https://leetcode.com/problems/cherry-pickup/ "Cherry Pickup") | ★★★★ | | | | | | | I: O(n^2), S = O(n^3), T = O(n^3) | -| 546 | [Remove Boxes](https://leetcode.com/problems/remove-boxes "Remove Boxes") | ★★★★★ | | | | | | | I: O(n), S = O(n^3), T = O(n^4) | -| 943 | [Find the Shortest Superstring](https://leetcode.com/problems/find-the-shortest-superstring/ "Find the Shortest Superstring") | ★★★★ | [980](https://leetcode.com/problems/unique-paths-iii/ "980") | [996](https://leetcode.com/problems/number-of-squareful-arrays/ "996") | | | | | I: O(n), S = O(n*2^n), T = (n^2*2^n) | -| 62 | [Unique Paths](https://leetcode.com/problems/unique-paths "Unique Paths") | ★★ | [63](https://leetcode.com/problems/unique-paths-ii "63") | [64](https://leetcode.com/problems/minimum-path-sum "64") | [120](https://leetcode.com/problems/triangle "120") | [174](https://leetcode.com/problems/dungeon-game "174") | [931](https://leetcode.com/problems/minimum-falling-path-sum/ "931") | | I: O(mn), S = O(mn), T = O(mn) | -| 85 | [Maximal Rectangle](https://leetcode.com/problems/delete-operation-for-two-strings/ "Maximal Rectangle") | ★★★ | [221](https://leetcode.com/problems/maximal-square/ "221") | [304](https://leetcode.com/problems/range-sum-query-2d-immutable "304") | | | | | | -| 688 | [Knight Probability in Chessboard](https://leetcode.com/problems/knight-probability-in-chessboard/ "Knight Probability in Chessboard") | ★★★ | [576](https://leetcode.com/problems/out-of-boundary-paths/ "576") | [935](https://leetcode.com/problems/knight-dialer/ "935") | | | | | I: O(mn) + k, S = O(kmn) T = O(kmn) | -| 322 | [Coin Change](https://leetcode.com/problems/reconstruct-itinerary/ "Coin Change") | ★★★ | [377](https://leetcode.com/problems/combination-sum-iv/ "377") | [416](https://leetcode.com/problems/partition-equal-subset-sum/ "416") | [494](https://leetcode.com/problems/target-sum/ "494") | [1043](https://leetcode.com/problems/partition-array-for-maximum-sum/ "1043") | [1049](https://leetcode.com/problems/last-stone-weight-ii/ "1049") | | I: O(n) + k, S = O(n), T = O(kn) | -| | | | [1220](https://leetcode.com/problems/count-vowels-permutation/ "1220") | [1230](https://leetcode.com/problems/toss-strange-coins/ "1230") | [1262](https://leetcode.com/problems/greatest-sum-divisible-by-three/ "1262") | [1269](https://leetcode.com/problems/number-of-ways-to-stay-in-the-same-place-after-some-steps/ "1269") | | | -| 813 | [Largest Sum of Averages](https://leetcode.com/problems/largest-sum-of-averages/ "Largest Sum of Averages") | ★★★★ | [1278](https://leetcode.com/problems/palindrome-partitioning-iii/ "1278") | [1335](https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/ "1335") | [410](https://leetcode.com/problems/split-array-largest-sum/ "410") | | | | I: O(n) + k
S = O(n*k), T = O(kn^2) | -| 1223 | [Dice Roll Simulation](https://leetcode.com/problems/dice-roll-simulation/ "Dice Roll Simulation") | ★★★★ | | | | | | | I: O(n) + k + p
S = O(k*p), T = O(n^2kp) | -| 312 | [Burst Balloons](https://leetcode.com/problems/burst-balloons/ "Burst Balloons") | ★★★★ | [664](https://leetcode.com/problems/strange-printer/ "664") | [1024](https://leetcode.com/problems/video-stitching/ "1024") | [1039](https://leetcode.com/problems/minimum-score-triangulation-of-polygon/ "1039") | [1140](https://leetcode.com/problems/stone-game-ii/ "1140") | [1130](https://leetcode.com/problems/minimum-cost-tree-from-leaf-values/ "1130") | | I: O(n), S = O(n^2), T = O(n^3) | -| 741 | [Cherry Pickup](https://leetcode.com/problems/cherry-pickup/ "Cherry Pickup") | ★★★★ | | | | | | | I: O(n^2), S = O(n^3), T = O(n^3) | -| 546 | [Remove Boxes](https://leetcode.com/problems/remove-boxes/ "Remove Boxes") | ★★★★★ | | | | | | | I: O(n), S = O(n^3), T = O(n^4) | -| 943 | [Find the Shortest Superstring](https://leetcode.com/problems/find-the-shortest-superstring/ "Find the Shortest Superstring") | ★★★★★ | [980](https://leetcode.com/problems/unique-paths-iii/ "980") | [996](https://leetcode.com/problems/number-of-squareful-arrays/ "996") | [1125](https://leetcode.com/problems/smallest-sufficient-team/ "1125") | | | | I: O(n)
S = O(n*2^n), T = (n^2*2^n) | - -### Advanced -| Id | Name | Difficulty | | | | | | Comments | -| ---: | --- | :---: | :---: | --- | --- | --- | --- | --- | -| 208 | [Implement Trie (Prefix Tree)](https://leetcode.com/problems/implement-trie-prefix-tree "Implement Trie (Prefix Tree)") | ★★★ | [648](https://leetcode.com/problems/replace-words/ "648") | [676](https://leetcode.com/problems/implement-magic-dictionary "676") | [677](https://leetcode.com/problems/map-sum-pairs "677") | [720](https://leetcode.com/problems/longest-word-in-dictionary "720") | [745](https://leetcode.com/problems/prefix-and-suffix-search "745") | Trie | -| 307 | [Range Sum Query - Mutable](https://leetcode.com/problems/range-sum-query-mutable "Range Sum Query - Mutable") | ★★★ | | | | | | BIT/Segment Tree | -| 901 | [Online Stock Span](https://leetcode.com/problems/online-stock-span "Online Stock Span") | ★★★ | [907](https://leetcode.com/problems/sum-of-subarray-minimums "907") | [1019](https://leetcode.com/problems/next-greater-node-in-linked-list/ "1019") | | | | Monotonic Stack | -| 239 | [Sliding Window Maximum](https://leetcode.com/problems/sliding-window-maximum/ "Sliding Window Maximum") | ★★★ | | | | | | Monotonic Queue | +* https://neetcode.io/practice (150 problems with video explaination) +* https://www.programcreek.com/2013/08/leetcode-problem-classification/ +* https://github.com/wisdompeak/LeetCode +* https://docs.google.com/spreadsheets/d/1SbpY-04Cz8EWw3A_LBUmDEXKUMO31DBjfeMoA0dlfIA/edit#gid=126913158 ([huahua](https://www.youtube.com/user/xxfflower/videos)). +* https://leetcode.com/list/?selectedList=535ukjh5 (Only 74 problems) # Software Engineer Interview @@ -180,7 +38,7 @@ I found it makes sense to solve similar problems together, so that we can recogn 3. [What should I know from the CLRS 3rd edition book if my aim is to get into Google?](https://www.quora.com/What-should-I-know-from-the-CLRS-3rd-edition-book-if-my-aim-is-to-get-into-Google/answer/Jimmy-Saade) ## Data Structures and Algorithms for beginners -If you are new or know nothing about data structures and algorithms, I recommend [this course](). This course is taught in Python and design to help you find job and do well in the interview. +If you are new or know nothing about data structures and algorithms, I recommend [this course](). This course is taught in Python and design to help you find job and do well in the interview. # System Design diff --git "a/common/bellman\342\200\223ford.py" "b/common/bellman\342\200\223ford.py" old mode 100644 new mode 100755 diff --git a/common/binary-search-tree.py b/common/binary-search-tree.py old mode 100644 new mode 100755 diff --git a/common/bst-in-order.py b/common/bst-in-order.py new file mode 100755 index 0000000..0ffe82e --- /dev/null +++ b/common/bst-in-order.py @@ -0,0 +1,15 @@ +def inOrderTraverse(root): + stack = [] + node = root + + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + + #do something + print node.val + + node = node.right \ No newline at end of file diff --git a/common/dijkstra.py b/common/dijkstra.py old mode 100644 new mode 100755 diff --git a/common/heap-sort.py b/common/heap-sort.py old mode 100644 new mode 100755 index eb2e11c..3d56e39 --- a/common/heap-sort.py +++ b/common/heap-sort.py @@ -1,35 +1,33 @@ +#O(LogN), used when part of the array is already heapified. +def heapify(A, N, i): + largest = i + l = i*2+1 + r = i*2+2 + + if lA[largest]: largest = l + if rA[largest]: largest = r + if largest!=i: + A[largest], A[i] = A[i], A[largest] + heapify(A, N, largest) + +#O(NLogN) def heapSort(A): - #build max heap - def heapify(A, n, i): - if i>=n: return - l, r = i*2+1, i*2+2 - left = A[l] if lA[i] or right>A[i]: - if left>right: - A[i], A[l] = A[l], A[i] - heapify(A, n, l) - else: - A[i], A[r] = A[r], A[i] - heapify(A, n, r) - - n = len(A) - - for i in reversed(xrange(n)): - heapify(A, n, i) - - for i in reversed(xrange(1, n)): - A[i], A[0] = A[0], A[i] + N = len(A) + + #build max heap, O(NLogN). Can be optimized to the O(N). + for i in range(N//2-1, -1, -1): + heapify(A, N, i) + + #keep swapping the largest + for i in range(N-1, -1, -1): + A[0], A[i] = A[i], A[0] heapify(A, i, 0) - -A = [21, 4, 1, 3, 9, 20, 25, 6, 21, 14] +A = [12, 11, 13, 5, 6, 7] heapSort(A) -print A +print(A) -""" -Time Complexity O(NLogN) in best, average, worst case. -Space Complexity O(1) -""" \ No newline at end of file +A = [1, 3, 5, 4, 6, 13, 10, 9, 8, 15, 17] +heapSort(A) +print(A) \ No newline at end of file diff --git a/common/insertion-sort.py b/common/insertion-sort.py old mode 100644 new mode 100755 diff --git a/common/knapsack.py b/common/knapsack.py old mode 100644 new mode 100755 diff --git a/common/merge-sort-in-place.py b/common/merge-sort-in-place.py old mode 100644 new mode 100755 diff --git a/common/merge-sort.py b/common/merge-sort.py old mode 100644 new mode 100755 diff --git a/common/min-heap.py b/common/min-heap.py old mode 100644 new mode 100755 diff --git a/common/quick-sort.py b/common/quick-sort.py old mode 100644 new mode 100755 diff --git a/common/radix-sort.py b/common/radix-sort.py old mode 100644 new mode 100755 diff --git a/common/trie.py b/common/trie.py old mode 100644 new mode 100755 diff --git a/problems/accounts-merge.py b/problems/accounts-merge.py deleted file mode 100644 index 30e4787..0000000 --- a/problems/accounts-merge.py +++ /dev/null @@ -1,52 +0,0 @@ -from collections import defaultdict - -class Solution(object): - def accountsMerge(self, accounts): - graph = defaultdict(list) - merged = set() - ans = [] - - #[0] - for data in accounts: - emails = data[1:] - for i, email in enumerate(emails): - graph[email].extend(emails[:i]) - graph[email].extend(emails[i+1:]) - - for data in accounts: - name = data[0] - visited = set() - stack = [data[1]] #[2] - - if data[1] in merged: continue #[1] - - while stack: - e = stack.pop() - if e in visited: continue - visited.add(e) - stack.extend(graph[e]) - - merged.update(visited) - ans.append([name]+sorted(list(visited))) #[3] - - return ans - -""" -Treat each email as a node. -Build an adjacency graph. [0] - -For every account's data -First, lets check if the first email is already merged. [1] -If the first email is already merged to other groups, then other emails will be in another group as well. -So don't need to check. - -Second, do a DFS starting from the first email. Put all the connected nodes into visited. [2] -And append the sorted(visited) to the ans with name. [3] - -Let N be the total number of emails. M be the total number of final groups. -Build the graph takes O(N). -For each final groups, we make a DFS to all nodes, taking O(MN). -Sort the group takes O((N/M)*Log(N/M)) -> O((N/M)*(logN-LogM)) -Time: O(MN) -Space: O(N) -""" \ No newline at end of file diff --git a/problems/binary-search-tree-iterator.py b/problems/binary-search-tree-iterator.py deleted file mode 100644 index 2d171bd..0000000 --- a/problems/binary-search-tree-iterator.py +++ /dev/null @@ -1,36 +0,0 @@ -# Definition for a binary tree node. -# class TreeNode(object): -# def __init__(self, x): -# self.val = x -# self.left = None -# self.right = None - -class BSTIterator(object): - - def __init__(self, root): - self.stack = [] - while root: - self.stack.append(root) - root = root.left - - - - def next(self): - node = self.stack.pop() - r = node.right - while r: - self.stack.append(r) - r = r.left - return node.val - - - - def hasNext(self): - return len(self.stack)!=0 - - - -# Your BSTIterator object will be instantiated and called as such: -# obj = BSTIterator(root) -# param_1 = obj.next() -# param_2 = obj.hasNext() \ No newline at end of file diff --git a/problems/binary-tree-level-order-traversal-ii.py b/problems/binary-tree-level-order-traversal-ii.py deleted file mode 100644 index db49ccd..0000000 --- a/problems/binary-tree-level-order-traversal-ii.py +++ /dev/null @@ -1,20 +0,0 @@ -from collections import deque - -class Solution(object): - def levelOrderBottom(self, root): - opt = [] - - if not root: return opt - - q = deque() - q.append((root, 0)) - - while q: - node, depth = q.popleft() - if depth=0: return dp[t] - - ans = 0 - for num in nums: - ans += helper(t-num) - - dp[t] = ans - return ans - - dp = [-1]*(target+1) - dp[0] = 1 - - for i in xrange(target+1): - helper(i) - return dp[target] - -""" -Time: O(TN), T is the value of target and N is the count of nums. -Space: O(T) -""" \ No newline at end of file diff --git a/problems/convert-sorted-array-to-binary-search-tree.py b/problems/convert-sorted-array-to-binary-search-tree.py deleted file mode 100644 index d0f2fd8..0000000 --- a/problems/convert-sorted-array-to-binary-search-tree.py +++ /dev/null @@ -1,22 +0,0 @@ -class Solution(object): - def sortedArrayToBST(self, nums): - if not nums: return None - - m = len(nums)/2 - root = TreeNode(nums[m], self.sortedArrayToBST(nums[:m]), self.sortedArrayToBST(nums[m+1:])) - return root - - -""" -`nums` is a sorted array. -To construct the most balanced BST, we must use the median number in `nums` as root. -So the number of elements under left subtree and right subtree will be the same. - -Now we know which element in the `nums` will be root. -For `root.left` we only need to consider `nums[:m]`. -For `root.right` we only need to consider `nums[m+1:]`. -Which needs to be a balanced BST, too. So we apply the same logic. - -Time: O(N). Because there will be N element being construct. -Space: O(LogN). There will be LogN level of recursive call. -""" \ No newline at end of file diff --git a/problems/find-and-replace-in-string.py b/problems/find-and-replace-in-string.py deleted file mode 100644 index ed407d1..0000000 --- a/problems/find-and-replace-in-string.py +++ /dev/null @@ -1,23 +0,0 @@ -""" -`p` is the index on s which already processed. -""" -class Solution(object): - def findReplaceString(self, s, indices, sources, targets): - p = -1 - ans = '' - - memo = {} - for i, index in enumerate(indices): - memo[index] = (sources[i], targets[i]) - - for i in xrange(len(s)): - if i<=p: continue - - if i in memo and (s[i:i+len(memo[i][0])] if i+len(memo[i][0])=0 and nums[i+1]<=nums[i]: - i -= 1 - - if i>=0: - j = len(nums)-1 - while nums[j]<=nums[i]: j -= 1 - swap(i, j) - - reverse(i+1) - return nums \ No newline at end of file diff --git a/problems/partition-to-k-equal-sum-subsets.py b/problems/partition-to-k-equal-sum-subsets.py deleted file mode 100644 index 5d98491..0000000 --- a/problems/partition-to-k-equal-sum-subsets.py +++ /dev/null @@ -1,18 +0,0 @@ -class Solution(object): - def canPartitionKSubsets(self, nums, k): - def search(subs): - if not nums: return True - n = nums.pop() - for i, sub in enumerate(subs): - if sub+n<=target: - subs[i]+=n - if search(subs): return True - subs[i]-=n - if not sub: break - nums.append(n) - return False - - if sum(nums)%k!=0: return False - target = sum(nums)/k - nums.sort() - return search([0]*k) diff --git a/problems/path-sum-ii.py b/problems/path-sum-ii.py deleted file mode 100644 index a900baa..0000000 --- a/problems/path-sum-ii.py +++ /dev/null @@ -1,20 +0,0 @@ -class Solution(object): - def pathSum(self, root, S): - if not root: return False - - opt = [] - stack = [] - stack.append((root, 0, [])) - - while stack: - node, s, path = stack.pop() - s += node.val - path = path + [node.val] - if not node.left and not node.right and s==S: opt.append(path) - if node.left: stack.append((node.left, s, path)) - if node.right: stack.append((node.right, s, path)) - return opt -""" -Time complexity is O(N), because we traverse all the nodes. -Space complexity is O(N^2), because in the worst case, all node could carry all the other nodes in the `path`. -""" \ No newline at end of file diff --git a/problems/path-sum.py b/problems/path-sum.py deleted file mode 100644 index cd2ae5a..0000000 --- a/problems/path-sum.py +++ /dev/null @@ -1,14 +0,0 @@ -class Solution(object): - def hasPathSum(self, root, S): - if not root: return False - - stack = [] - stack.append((root, 0)) - - while stack: - node, s = stack.pop() - s += node.val - if not node.left and not node.right and s==S: return True - if node.left: stack.append((node.left, s)) - if node.right: stack.append((node.right, s)) - return False \ No newline at end of file diff --git a/problems/permutations-ii.py b/problems/permutations-ii.py deleted file mode 100644 index 89ba9ee..0000000 --- a/problems/permutations-ii.py +++ /dev/null @@ -1,26 +0,0 @@ -""" -This is the extended question of the [first one](https://leetcode.com/problems/permutations/submissions/). -The differece is there will be duplicates in the `nums`. - -```python -if i>0 and options[i]==options[i-1]: continue -``` -Above lets us skip exploring the same path. -We can directly skip the `i` if the value is the same with `i-1`, because `dfs()` on `i-1` has already cover up all the possiblities. - -The time complexity is `O(N!)`. -The space complexity is `O(N!)`, too. -""" -class Solution(object): - def permuteUnique(self, nums): - def dfs(path, options): - if len(path)==len(nums): - opt.append(path) - return - for i, n in enumerate(options): - if i>0 and options[i]==options[i-1]: continue - dfs(path+[n], options[:i]+options[i+1:]) - opt = [] - nums.sort() - dfs([], nums) - return opt diff --git a/problems/permutations.py b/problems/permutations.py deleted file mode 100644 index 6659d57..0000000 --- a/problems/permutations.py +++ /dev/null @@ -1,29 +0,0 @@ -""" -`nums = [1, 2, 3, 4, 5]` -The first time you choose you can either choose 1 or 2 or 3 or 4 or 5. Lets say you choose 2. So the path so far is `[2]`. -The second time you choose you can only choose 1 or 3 or 4 or 5. Lets say you choose 3. So the path so far is `[2, 3]`. -The third time you choose you can only choose 1 or 4 or 5. Lets say you choose 1. So the path so far is `[2, 3, 1]`. -The third time you choose you can only choose 4 or 5... -. -. -. - -We put the numbers we can choose in the `options` parameter. And the path so far in the `path` parameter. -In each `dfs()` we check if the path has used up all the numbers in the `nums`. If true. Append it in the output. -If not, we explore all the posible path in the `options` by `dfs()`. - -The time complexity is O(N!). Since in this example our choices is 5 at the beginning, then 4, then 3, then 2, then 1. -The space complexity is O(N!), too. And the recursion takes N level of recursion. -""" -class Solution(object): - def permute(self, nums): - def dfs(path, options): - if len(nums)==len(path): - opt.append(path) - return - for i, nums in enumerate(options): - dfs(path+[nums], options[:i]+options[i+1:]) - - opt = [] - dfs([], nums) - return opt diff --git a/problems/profitable-schemes.py b/problems/profitable-schemes.py deleted file mode 100644 index ead8e92..0000000 --- a/problems/profitable-schemes.py +++ /dev/null @@ -1,19 +0,0 @@ -""" -TLE -dp[i][n][p] := considering profit[:i], what is the number of ways produce profit p with n people. -""" -class Solution(object): - def profitableSchemes(self, maxMember, minProfit, group, profit): - P = sum(profit) - N = sum(group) - dp = [[[0 for _ in xrange(P+1)] for _ in xrange(N+1)] for _ in xrange(len(profit)+1)] - dp[0][0][0] = 1 - - count = 0 - for i in xrange(1, len(profit)+1): - for n in xrange(N+1): - for p in xrange(P+1): - dp[i][n][p] = (dp[i-1][n-group[i-1]][p-profit[i-1]] if p-profit[i-1]>=0 and n-group[i-1]>=0 else 0) + dp[i-1][n][p] - if i==len(profit) and p>=minProfit and n<=maxMember: count += dp[i][n][p] - return count - \ No newline at end of file diff --git a/problems/01-matrix.py b/problems/python/01-matrix.py old mode 100644 new mode 100755 similarity index 100% rename from problems/01-matrix.py rename to problems/python/01-matrix.py diff --git a/problems/3sum-closest.py b/problems/python/3sum-closest.py old mode 100644 new mode 100755 similarity index 100% rename from problems/3sum-closest.py rename to problems/python/3sum-closest.py diff --git a/problems/3sum.py b/problems/python/3sum.py old mode 100644 new mode 100755 similarity index 100% rename from problems/3sum.py rename to problems/python/3sum.py diff --git a/problems/4sum.py b/problems/python/4sum.py old mode 100644 new mode 100755 similarity index 100% rename from problems/4sum.py rename to problems/python/4sum.py diff --git a/problems/python/accounts-merge.py b/problems/python/accounts-merge.py new file mode 100755 index 0000000..91e03ef --- /dev/null +++ b/problems/python/accounts-merge.py @@ -0,0 +1,134 @@ +""" +Treat each email as a node. +Build an adjacency graph. [0] + +For every account's data +First, lets check if the first email is already merged. [1] +If the first email is already merged to other groups, then other emails will be in another group as well. +So don't need to check. + +Second, do a DFS starting from the first email. Put all the connected nodes into visited. [2] +And append the sorted(visited) to the ans with name. [3] + +Let N be the total number of emails. M be the total number of final groups. +Build the graph takes O(N). +For each final groups, we make a DFS to all nodes, taking O(MN). +Sort the group takes O((N/M)*Log(N/M)) -> O((N/M)*(logN-LogM)) +Time: O(MN) +Space: O(N) +""" +from collections import defaultdict + +class Solution(object): + def accountsMerge(self, accounts): + graph = defaultdict(list) + merged = set() + ans = [] + + #[0] + for data in accounts: + emails = data[1:] + for i, email in enumerate(emails): + graph[email].extend(emails[:i]) + graph[email].extend(emails[i+1:]) + + for data in accounts: + name = data[0] + visited = set() + stack = [data[1]] #[2] + + if data[1] in merged: continue #[1] + + while stack: + e = stack.pop() + if e in visited: continue + visited.add(e) + stack.extend(graph[e]) + + merged.update(visited) + ans.append([name]+sorted(list(visited))) #[3] + + return ans + + +#Union Find +class Solution(object): + def accountsMerge(self, accounts): + def find(x): + p = parents[x] + while p!=parents[p]: + p = find(p) + parents[x] = p + return p + + def union(x, y): + p1, p2 = find(x), find(y) + if p1==p2: return + parents[p2] = p1 + + parents = {} + mailToName = {} + + for account in accounts: + name = account[0] + root = account[1] + if root not in parents: parents[root] = root + root = find(root) + mailToName[root] = name + + for i in xrange(2, len(account)): + email = account[i] + if email in parents: + union(parents[email], root) + root = find(root) + parents[email] = root + + rootToMails = collections.defaultdict(list) + for email in parents: + rootToMails[find(email)].append(email) + + ans = [] + for root in rootToMails: + name = mailToName[root] + mails = rootToMails[root] + ans.append([name]+sorted(mails)) + + return ans + +#DFS +class Solution(object): + def accountsMerge(self, accounts): + + #build adjacency list + adj = collections.defaultdict(list) + for account in accounts: + name = account[0] + email0 = account[1] + for i in xrange(2, len(account)): + email = account[i] + adj[email0].append(email) + adj[email].append(email0) + + #iterate accounts and dfs each email group + ans = [] + visited = set() #store all the visited email + for account in accounts: + name = account[0] + email0 = account[1] + if email0 in visited: continue + + #dfs + group = set() #store the email group related to email0 + stack = [email0] + while stack: + email = stack.pop() + if email in group or email in visited: continue + group.add(email) + visited.add(email) + for nei in adj[email]: + stack.append(nei) + + ans.append([name]+sorted(list(group))) + + return ans + \ No newline at end of file diff --git a/problems/add-binary.py b/problems/python/add-binary.py old mode 100644 new mode 100755 similarity index 100% rename from problems/add-binary.py rename to problems/python/add-binary.py diff --git a/problems/add-digits.py b/problems/python/add-digits.py old mode 100644 new mode 100755 similarity index 100% rename from problems/add-digits.py rename to problems/python/add-digits.py diff --git a/problems/python/add-strings.py b/problems/python/add-strings.py new file mode 100755 index 0000000..9cdd7f9 --- /dev/null +++ b/problems/python/add-strings.py @@ -0,0 +1,34 @@ +class Solution(object): + def addStrings(self, nums1, nums2): + ans = '' + i = len(nums1)-1 + j = len(nums2)-1 + + carry = 0 + while 0<=i and 0<=j: + n1 = int(nums1[i]) + n2 = int(nums2[j]) + total = n1+n2+carry + n = total%10 + carry = 1 if total>=10 else 0 + ans = str(n)+ans + i -= 1 + j -= 1 + + while 0<=i: + total = int(nums1[i])+carry + n = total%10 + carry = 1 if total>=10 else 0 + ans = str(n)+ans + i -= 1 + + while 0<=j: + total = int(nums2[j])+carry + n = total%10 + carry = 1 if total>=10 else 0 + ans = str(n)+ans + j -= 1 + + if carry: ans = str(carry)+ans + + return ans \ No newline at end of file diff --git a/problems/add-two-numbers-ii.py b/problems/python/add-two-numbers-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/add-two-numbers-ii.py rename to problems/python/add-two-numbers-ii.py diff --git a/problems/add-two-numbers.py b/problems/python/add-two-numbers.py old mode 100644 new mode 100755 similarity index 100% rename from problems/add-two-numbers.py rename to problems/python/add-two-numbers.py diff --git a/problems/python/alien-dictionary.py b/problems/python/alien-dictionary.py new file mode 100755 index 0000000..5c27918 --- /dev/null +++ b/problems/python/alien-dictionary.py @@ -0,0 +1,37 @@ +""" +Topological Sort +""" +class Solution(object): + def alienOrder(self, words): + #return true if cycles are detected. + def dfs(c): + if c in path: return True + if c in visited: return False + path.add(c) + for nei in adj[c]: + if dfs(nei): return True + res.append(c) + path.remove(c) + visited.add(c) + return False + + #build adjacency list + adj = {c: set() for word in words for c in word} + for i in xrange(len(words)-1): + w1, w2 = words[i], words[i+1] + minLen = min(len(w1), len(w2)) + if w1[:minLen]==w2[:minLen] and len(w1)>len(w2): return "" + + for j in xrange(minLen): + if w1[j]!=w2[j]: + adj[w1[j]].add(w2[j]) + break + + #topological sort + path = set() #path currently being reversed + visited = set() #done processing + res = [] + for c in adj: + if dfs(c): return "" + + return "".join(reversed(res)) \ No newline at end of file diff --git a/problems/python/all-nodes-distance-k-in-binary-tree.py b/problems/python/all-nodes-distance-k-in-binary-tree.py new file mode 100755 index 0000000..77152e3 --- /dev/null +++ b/problems/python/all-nodes-distance-k-in-binary-tree.py @@ -0,0 +1,40 @@ +class Solution(object): + def distanceK(self, root, target, k): + graph = collections.defaultdict(list) + q = collections.deque([root]) #for traverse binary tree + q2 = collections.deque([(target, k)]) #for bfs the graph + visited = set() #for bfs the graph + ans = [] + + #build graph + while q: + node = q.popleft() + + if node.left: + graph[node].append(node.left) + graph[node.left].append(node) + q.append(node.left) + + if node.right: + graph[node].append(node.right) + graph[node.right].append(node) + q.append(node.right) + + + #bfs graph + while q2: + node, distance = q2.popleft() + if node.val in visited: continue + visited.add(node.val) + if distance==0: ans.append(node.val) + if distance<0 or distance>k: continue + + for nei in graph[node]: + q2.append((nei, distance-1)) + + return ans + + + + + \ No newline at end of file diff --git a/problems/python/amount-of-new-area-painted-each-day.py b/problems/python/amount-of-new-area-painted-each-day.py new file mode 100755 index 0000000..a5504d5 --- /dev/null +++ b/problems/python/amount-of-new-area-painted-each-day.py @@ -0,0 +1,40 @@ +""" +1. Build sorted records = [(position, index, isStart)...] +2. Iterate through all positions and maintain a box with all the "index" of the records its position is in start ~ end +3. The smallest index in the box is the actual one that is paiting. + +Time: O(NLogN+P), N is the count of paint. Sorting the records takes NLogN. P is the max position. +Although there is a while loop when iterate through P, each record is only being iterated once. +O(NLogN + P + NLogN) ~= O(NLogN + P) +Space: O(N) +""" +from sortedcontainers import SortedList + +class Solution(object): + def amountPainted(self, paint): + ans = [0]*len(paint) + box = SortedList() + records = [] + maxPos = float('-inf') + + #[1] + for i, (start, end) in enumerate(paint): + records.append((start, i, -1)) + records.append((end, i, 1)) + maxPos = max(maxPos, end) + + records.sort() + + #[2] + i = 0 + for pos in xrange(maxPos+1): + while i3 or i>=len(websites): + return + else: + helper(comb+[websites[i]], i+1) + helper(comb[:], i+1) + + combs = set() + helper([], 0) + return combs + + + + \ No newline at end of file diff --git a/problems/backspace-string-compare.py b/problems/python/backspace-string-compare.py old mode 100644 new mode 100755 similarity index 81% rename from problems/backspace-string-compare.py rename to problems/python/backspace-string-compare.py index f3df6d2..cec9e8a --- a/problems/backspace-string-compare.py +++ b/problems/python/backspace-string-compare.py @@ -82,4 +82,24 @@ def backspaceCompare(self, s1, s2): j -= 1 return True - \ No newline at end of file + + +class Solution(object): + def backspaceCompare(self, s, t): + def helper(S): + ans = '' + i = len(S)-1 + backspaceCount = 0 + + while i>=0: + if S[i]=='#': + backspaceCount += 1 + else: + if backspaceCount>0: + backspaceCount -= 1 + else: + ans += S[i] + i -= 1 + return ans + + return helper(s)==helper(t) \ No newline at end of file diff --git a/problems/python/balance-a-binary-search-tree.py b/problems/python/balance-a-binary-search-tree.py new file mode 100755 index 0000000..015934d --- /dev/null +++ b/problems/python/balance-a-binary-search-tree.py @@ -0,0 +1,36 @@ +""" +Imagine a list of sorted nodes, if we wanted to use the nodes to form a balanced BST, which root should we use? +Yes, the one in the middle, since it can evenly spread two groups of nodes. This is what `getRoot()` does. +And we do the same for the left half and right half. And so on. And so on... + +Time: O(N), N is the number of nodes. +Space: O(N) +""" +class Solution(object): + def balanceBST(self, root): + def getInorderNodes(root): + nodes = [] + stack = [] + node = root + + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + nodes.append(node) + node = node.right + + return nodes + + def getRoot(l, r): + if l>r: return None + m = (l+r)/2 + root = inorderNodes[m] + root.left = getRoot(l, m-1) + root.right = getRoot(m+1, r) + return root + + inorderNodes = getInorderNodes(root) + return getRoot(0, len(inorderNodes)-1) \ No newline at end of file diff --git a/problems/balanced-binary-tree.py b/problems/python/balanced-binary-tree.py old mode 100644 new mode 100755 similarity index 53% rename from problems/balanced-binary-tree.py rename to problems/python/balanced-binary-tree.py index 3061e57..36232a8 --- a/problems/balanced-binary-tree.py +++ b/problems/python/balanced-binary-tree.py @@ -24,4 +24,25 @@ def helper(node, depth): return -1 if not root: return True - return helper(root, 0)!=-1 \ No newline at end of file + return helper(root, 0)!=-1 + + +""" +Time: O(N), since we recursively traverse each node once. +Space: O(LogN), for the recursive call. O(N) if the tree is not balanced. +""" +class Solution(object): + def isBalanced(self, root): + + #return (if the node isBalanced, the height of the node) + def helper(node): + if not node: return True, -1 + + isLeftBalanced, leftHeight = helper(node.left) + isRightBalanced, rightHeight = helper(node.right) + + height = max(leftHeight, rightHeight)+1 + + return isLeftBalanced and isRightBalanced and abs(leftHeight-rightHeight)<=1, height + + return helper(root)[0] \ No newline at end of file diff --git a/problems/python/basic-calculator-ii.py b/problems/python/basic-calculator-ii.py new file mode 100755 index 0000000..90ab539 --- /dev/null +++ b/problems/python/basic-calculator-ii.py @@ -0,0 +1,31 @@ +class Solution(object): + def calculate(self, s): + s += '+' #edge case, for last operation to be executed. + lastOperation = '+' #edge case, for the first currNum + operations = set(['+', '-', '*', '/']) + + stack = [] + currNum = 0 + + for c in s: + if c.isdigit(): + currNum = currNum*10 + int(c) + elif c in operations: + if lastOperation=='+': + stack.append(currNum) + currNum = 0 + elif lastOperation=='-': + stack.append(-currNum) + currNum = 0 + elif lastOperation=='*': + currNum = stack.pop() * currNum + stack.append(currNum) + currNum = 0 + elif lastOperation=='/': + currNum = stack.pop() / currNum + if currNum<0: currNum += 1 + stack.append(currNum) + currNum = 0 + lastOperation = c + + return sum(stack) \ No newline at end of file diff --git a/problems/best-time-to-buy-an-stock.py b/problems/python/best-time-to-buy-an-stock.py old mode 100644 new mode 100755 similarity index 100% rename from problems/best-time-to-buy-an-stock.py rename to problems/python/best-time-to-buy-an-stock.py diff --git a/problems/best-time-to-buy-and-sell-stock-iii.py b/problems/python/best-time-to-buy-and-sell-stock-iii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/best-time-to-buy-and-sell-stock-iii.py rename to problems/python/best-time-to-buy-and-sell-stock-iii.py diff --git a/problems/best-time-to-buy-and-sell-stock-with-cooldown.py b/problems/python/best-time-to-buy-and-sell-stock-with-cooldown.py old mode 100644 new mode 100755 similarity index 100% rename from problems/best-time-to-buy-and-sell-stock-with-cooldown.py rename to problems/python/best-time-to-buy-and-sell-stock-with-cooldown.py diff --git a/problems/best-time-to-buy-and-sell-stock.py b/problems/python/best-time-to-buy-and-sell-stock.py old mode 100644 new mode 100755 similarity index 100% rename from problems/best-time-to-buy-and-sell-stock.py rename to problems/python/best-time-to-buy-and-sell-stock.py diff --git a/problems/big-countries.sql b/problems/python/big-countries.sql old mode 100644 new mode 100755 similarity index 100% rename from problems/big-countries.sql rename to problems/python/big-countries.sql diff --git a/problems/python/binary-search-tree-iterator.py b/problems/python/binary-search-tree-iterator.py new file mode 100755 index 0000000..48d1458 --- /dev/null +++ b/problems/python/binary-search-tree-iterator.py @@ -0,0 +1,88 @@ +class BSTIterator(object): + + def __init__(self, root): + self.stack = [] + while root: + self.stack.append(root) + root = root.left + + + + def next(self): + node = self.stack.pop() + r = node.right + while r: + self.stack.append(r) + r = r.left + return node.val + + + + def hasNext(self): + return len(self.stack)!=0 + +""" +Time: init() O(1). next() O(LogN). hasNext() O(1). +Space: O(N) +""" +class BSTIterator(object): + + def __init__(self, root): + self.node = root + self.stack = [] + + def next(self): + while self.node: + self.stack.append(self.node) + self.node = self.node.left + + node = self.stack.pop() + self.node = node.right + return node.val + + def hasNext(self): + return self.stack or self.node + + +""" +FYI. Template for in-order traverse. +""" +def inOrderTraverse(root): + stack = [] + node = root + + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + + #do something + print node.val + + node = node.right + + + +class BSTIterator(object): + + def __init__(self, root): + self.stack = [] + self.node = root + + + + def next(self): + while self.node: + self.stack.append(self.node) + self.node = self.node.left + self.node = self.stack.pop() + returnVal = self.node.val + self.node = self.node.right + return returnVal + + + + def hasNext(self): + return self.node or self.stack \ No newline at end of file diff --git a/problems/binary-search.py b/problems/python/binary-search.py old mode 100644 new mode 100755 similarity index 100% rename from problems/binary-search.py rename to problems/python/binary-search.py diff --git a/problems/python/binary-subarrays-with-sum.py b/problems/python/binary-subarrays-with-sum.py new file mode 100755 index 0000000..dafb077 --- /dev/null +++ b/problems/python/binary-subarrays-with-sum.py @@ -0,0 +1,19 @@ +class Solution(object): + def numSubarraysWithSum(self, nums, goal): + #number of subarrays with sum at most "goal" + def atMost(goal): + ans = 0 + total = 0 + i = 0 + + for j, num in enumerate(nums): + if num==1: total += 1 + + while igoal: + if nums[i]==1: total -= 1 + i += 1 + + ans += j-i+1 #number of subarrays that can generate from nums[i~j] + return ans + + return atMost(goal)-atMost(goal-1) if goal>0 else atMost(goal) \ No newline at end of file diff --git a/problems/binary-tree-inorder-traversal.py b/problems/python/binary-tree-inorder-traversal.py old mode 100644 new mode 100755 similarity index 100% rename from problems/binary-tree-inorder-traversal.py rename to problems/python/binary-tree-inorder-traversal.py diff --git a/problems/python/binary-tree-level-order-traversal-ii.py b/problems/python/binary-tree-level-order-traversal-ii.py new file mode 100755 index 0000000..07084a3 --- /dev/null +++ b/problems/python/binary-tree-level-order-traversal-ii.py @@ -0,0 +1,67 @@ +from collections import deque + +class Solution(object): + def levelOrderBottom(self, root): + opt = [] + + if not root: return opt + + q = deque() + q.append((root, 0)) + + while q: + node, depth = q.popleft() + if depth'+str(node.val)) if path else str(node.val) + + if node.left: q.append((node.left, path)) + if node.right: q.append((node.right, path)) + + if not node.left and not node.right: ans.append(path) + return ans \ No newline at end of file diff --git a/problems/binary-tree-pruning.py b/problems/python/binary-tree-pruning.py old mode 100644 new mode 100755 similarity index 100% rename from problems/binary-tree-pruning.py rename to problems/python/binary-tree-pruning.py diff --git a/problems/binary-tree-right-side-view.py b/problems/python/binary-tree-right-side-view.py old mode 100644 new mode 100755 similarity index 61% rename from problems/binary-tree-right-side-view.py rename to problems/python/binary-tree-right-side-view.py index ddae3a5..76d807f --- a/problems/binary-tree-right-side-view.py +++ b/problems/python/binary-tree-right-side-view.py @@ -28,4 +28,28 @@ def rightSideView(self, root): view.append(node.val) queue.append((node.right, level+1)) queue.append((node.left, level+1)) - return view \ No newline at end of file + return view + +""" +Time: O(N). +Space: O(N). + +Similar solution using DFS. +""" +class Solution(object): + def rightSideView(self, root): + if not root: return [] + + ans = [] + currLevel = -1 + stack = [(root, 0)] + + while stack: + node, level = stack.pop() + if level>currLevel: + ans.append(node.val) + currLevel = level + if node.left: stack.append((node.left, level+1)) + if node.right: stack.append((node.right, level+1)) + + return ans \ No newline at end of file diff --git a/problems/python/binary-tree-vertical-order-traversal.py b/problems/python/binary-tree-vertical-order-traversal.py new file mode 100755 index 0000000..1510ac0 --- /dev/null +++ b/problems/python/binary-tree-vertical-order-traversal.py @@ -0,0 +1,50 @@ +""" +Time: O(N) +Space: O(N) + +Use BFS to traverse the tree. +x is the x coordinate of the node, the smaller the x is the leftmost node will be. +Since we add the left node first, the nodes "in the same row and column" will automatically sorted from left to right. +""" +class Solution(object): + def verticalOrder(self, root): + if not root: return [] + + q = collections.deque([(root, 0)]) + minX = 0 + maxX = 0 + ans = collections.defaultdict(list) + + while q: + node, x = q.popleft() + ans[x].append(node.val) + minX = min(minX, x) + maxX = max(maxX, x) + + if node.left: q.append((node.left, x-1)) + if node.right: q.append((node.right, x+1)) + + return [ans[x] for x in xrange(minX, maxX+1)] + + +class Solution(object): + def verticalOrder(self, root): + if not root: return [] + ans = [] + minX = float('inf') + maxX = float('-inf') + locations = collections.defaultdict(list) + q = collections.deque([(root, 0)]) + + while q: + node, x = q.popleft() + locations[x].append(node.val) + minX = min(minX, x) + maxX = max(maxX, x) + if node.left: q.append((node.left, x-1)) + if node.right: q.append((node.right, x+1)) + + for x in xrange(minX, maxX+1): + if locations[x]: ans.append(locations[x]) + + return ans \ No newline at end of file diff --git a/problems/binary-watch.py b/problems/python/binary-watch.py old mode 100644 new mode 100755 similarity index 100% rename from problems/binary-watch.py rename to problems/python/binary-watch.py diff --git a/problems/python/buildings-with-an-ocean-view.py b/problems/python/buildings-with-an-ocean-view.py new file mode 100755 index 0000000..0025b13 --- /dev/null +++ b/problems/python/buildings-with-an-ocean-view.py @@ -0,0 +1,16 @@ +""" +Traverse from the right and keep track of the highest building. + +Time: O(N) +Space: O(1) +""" +class Solution(object): + def findBuildings(self, heights): + ans = [] + currMaxHeight = 0 + for i in xrange(len(heights)-1, -1, -1): + h = heights[i] + if h>currMaxHeight: ans.append(i) + currMaxHeight = max(currMaxHeight, h) + + return reversed(ans) \ No newline at end of file diff --git a/problems/burst-balloons.py b/problems/python/burst-balloons.py old mode 100644 new mode 100755 similarity index 100% rename from problems/burst-balloons.py rename to problems/python/burst-balloons.py diff --git a/problems/python/campus-bikes-ii.py b/problems/python/campus-bikes-ii.py new file mode 100755 index 0000000..a292d91 --- /dev/null +++ b/problems/python/campus-bikes-ii.py @@ -0,0 +1,35 @@ +""" +state[i] := ith bike has been selected. +For example, there are M bikes, 0000, 0100, 0110, 1111..... + +Time: O(ELogE) +Space: O(E) +""" +class Solution(object): + def assignBikes(self, workers, bikes): + M = len(bikes) + N = len(workers) + costs = [[0]*N for _ in xrange(M)] + + for i in xrange(M): + for j in xrange(N): + costs[i][j] = abs(workers[j][0]-bikes[i][0])+abs(workers[j][1]-bikes[i][1]) + + pq = [(0, '0'*M)] + visited = set() + while pq: + cost, state = heapq.heappop(pq) + if state in visited: continue + visited.add(state) + + j = state.count('1') #j users have selected bikes + if j>=N: return cost + + for i in xrange(M): + if state[i]=='1': continue + + nextState = state[:i] + '1' + state[i+1:] + if nextState in visited: continue + heapq.heappush(pq, (cost+costs[i][j], nextState)) + + return float('inf') \ No newline at end of file diff --git a/problems/candy.py b/problems/python/candy.py old mode 100644 new mode 100755 similarity index 100% rename from problems/candy.py rename to problems/python/candy.py diff --git a/problems/capacity-to-ship-packages-within-d-days.py b/problems/python/capacity-to-ship-packages-within-d-days.py old mode 100644 new mode 100755 similarity index 100% rename from problems/capacity-to-ship-packages-within-d-days.py rename to problems/python/capacity-to-ship-packages-within-d-days.py diff --git a/problems/cheapest-flights-within-k-stops.py b/problems/python/cheapest-flights-within-k-stops.py old mode 100644 new mode 100755 similarity index 62% rename from problems/cheapest-flights-within-k-stops.py rename to problems/python/cheapest-flights-within-k-stops.py index f5e8f0c..eb785b3 --- a/problems/cheapest-flights-within-k-stops.py +++ b/problems/python/cheapest-flights-within-k-stops.py @@ -23,16 +23,19 @@ class Solution1(object): def findCheapestPrice(self, n, flights, src, dst, K): graph = collections.defaultdict(list) pq = [] + visited = set() for u, v, w in flights: graph[u].append((w, v)) heapq.heappush(pq, (0, K+1, src)) while pq: price, stops, city = heapq.heappop(pq) + visited.add(city) if city is dst: return price if stops>0: for price_to_nei, nei in graph[city]: + if nei in visited: continue heapq.heappush(pq, (price+price_to_nei, stops-1, nei)) return -1 @@ -66,3 +69,37 @@ def findCheapestPrice(self, n, flights, src, dst, K): return min_price if min_price!=float('inf') else -1 + +""" +Standard Dijkstra, except this time instead of only explore the ones with least price +We also need to explore the ones with less steps. So add stepFromSrc to check. + +Time: O(ELogE), since there will be at most E edges that get pushed into the heap. +Space: O(E) +""" +class Solution(object): + def findCheapestPrice(self, n, flights, src, dst, K): + priceFromSrc = {} + stepFromSrc = {} + h = [(0, 0, src)] + G = collections.defaultdict(list) + + #build graph + for s, d, p in flights: + G[s].append((d, p)) + + #dijkstra + while h: + price, k, node = heapq.heappop(h) + + if node==dst: return price + if k>K: continue + + for nei, price2 in G[node]: + #explore next destination with less price or less steps + if nei not in priceFromSrc or price+price2<=priceFromSrc[nei] or kabs(node.val-target): + ans = node.val + + if target>node.val: + node = node.right + else: + node = node.left + + return ans \ No newline at end of file diff --git a/problems/coin-change.py b/problems/python/coin-change.py old mode 100644 new mode 100755 similarity index 100% rename from problems/coin-change.py rename to problems/python/coin-change.py diff --git a/problems/python/coloring-a-border.py b/problems/python/coloring-a-border.py new file mode 100755 index 0000000..0b1ecc9 --- /dev/null +++ b/problems/python/coloring-a-border.py @@ -0,0 +1,43 @@ +""" +A node is on the "border" if +1. It is on the actual border of the grid. +Or +2. Any of its neighbor has different color. + +BFS the grid starting from (row, col), if the node "isBorder" add it to nodesToColor. +Color the nodes in the nodesToColor. + +Time: O(N) +Space: O(N) +""" +class Solution(object): + def colorBorder(self, grid, row, col, color): + def isValid(i, j): + return i>=0 and j>=0 and istart and candidates[i]==candidates[i-1]: continue + candidate = candidates[i] + comb.append(candidate) + helper(remain-candidate, comb, i+1) + comb.pop() + + ans = [] + candidates.sort() + helper(target, [], 0) + return ans + + + \ No newline at end of file diff --git a/problems/combination-sum-iii.py b/problems/python/combination-sum-iii.py old mode 100644 new mode 100755 similarity index 54% rename from problems/combination-sum-iii.py rename to problems/python/combination-sum-iii.py index c5fa010..d5fccc1 --- a/problems/combination-sum-iii.py +++ b/problems/python/combination-sum-iii.py @@ -38,3 +38,40 @@ def dfs(path, min_num): dfs([], 1) return opt + + +""" +Use a helper to check the remain. +comb is the combination to sum up to remain. +k is the max length of comb. +start is the starting index of "nums", since we already explore the all the possible comb from previous index. + +Time: O(9!/(9-k)! * k), 9*8*7... for k times. And each ans takes O(k) to copy the comb list. +Space: O(k) +""" +class Solution(object): + def combinationSum3(self, k, n): + def helper(remain, comb, k, start): + if remain==0 and len(comb)==k: + ans.append(comb[:]) + elif remain<0 or len(comb)>k: + return + else: + for i in xrange(start, len(nums)): + used = nums[i] + num = i+1 + + if used: continue + + comb.append(num) + nums[i] = True + + helper(remain-num, comb, k, i+1) + + comb.pop() + nums[i] = False + + nums = [False]*9 + ans = [] + helper(n, [], k, 0) + return ans \ No newline at end of file diff --git a/problems/python/combination-sum-iv.py b/problems/python/combination-sum-iv.py new file mode 100755 index 0000000..88bc853 --- /dev/null +++ b/problems/python/combination-sum-iv.py @@ -0,0 +1,66 @@ +class Solution(object): + def combinationSum4(self, nums, target): + def helper(t): + if t<0: return 0 + if dp[t]>=0: return dp[t] + + ans = 0 + for num in nums: + ans += helper(t-num) + + dp[t] = ans + return ans + + dp = [-1]*(target+1) + dp[0] = 1 + + for i in xrange(target+1): + helper(i) + return dp[target] + +""" +Time: O(TN), T is the value of target and N is the count of nums. +Space: O(T) +""" + + +""" +dp[n] := number of combs sums up to n + +For example, lets say target is 32 and nums is [4,2,1]. +dp[32] = dp[28]+dp[30]+dp[31] +Since the combs of dp[28] adds 4 will all equals to 32. +Since the combs of dp[30] adds 2 will all equals to 32. +Since the combs of dp[31] adds 1 will all equals to 32. + +... + +dp[28] = dp[24]+dp[26]+dp[27] +Since the combs of dp[28] adds 4 will all equals to 24. +Since the combs of dp[28] adds 2 will all equals to 26. +Since the combs of dp[28] adds 1 will all equals to 27. + +... + + + +Time: O(TN), T is the value of target and N is the count of nums. +Space: O(T) +""" +class Solution(object): + def combinationSum4(self, nums, target): + dp = [0]*(target+1) + dp[0] = 1 + + t = 1 + + while t<=target: + combs = 0 + for num in nums: + if t-num<0: continue + combs += dp[t-num] + + dp[t] = combs + t += 1 + + return dp[target] \ No newline at end of file diff --git a/problems/combination-sum.py b/problems/python/combination-sum.py old mode 100644 new mode 100755 similarity index 84% rename from problems/combination-sum.py rename to problems/python/combination-sum.py index 423f3c1..c76cc6a --- a/problems/combination-sum.py +++ b/problems/python/combination-sum.py @@ -118,3 +118,26 @@ def dfs(index, target, path): candidates.sort() dfs(0, T, []) return opt + + +""" +Time: O(N^(T/M)), N is the number of candidates. T is target. M is min(candidates). +Space: O(T/M) +""" +class Solution(object): + def combinationSum(self, candidates, target): + def helper(remain, comb, start=0): + if remain==0: + ans.append(comb[:]) + elif remain<0: + return + elif remain>0: + for i in xrange(start, len(candidates)): + candidate = candidates[i] + comb.append(candidate) + helper(remain-candidate, comb, i) + comb.pop() + + ans = [] + helper(target, []) + return ans \ No newline at end of file diff --git a/problems/combinations.py b/problems/python/combinations.py old mode 100644 new mode 100755 similarity index 87% rename from problems/combinations.py rename to problems/python/combinations.py index 49f7ebf..f3a0937 --- a/problems/combinations.py +++ b/problems/python/combinations.py @@ -83,6 +83,21 @@ def dfs(n_min, path): +class Solution(object): + def combine(self, N, K): + def dfs(comb, start, N, K): + if len(comb)==K: ans.append(comb[:]) + + for n in xrange(start, N+1): + comb.append(n) + dfs(comb, n+1, N, K) + comb.pop() + + ans = [] + dfs([], 1, N, K) + return ans + + diff --git a/problems/combine-two-tables.sql b/problems/python/combine-two-tables.sql old mode 100644 new mode 100755 similarity index 100% rename from problems/combine-two-tables.sql rename to problems/python/combine-two-tables.sql diff --git a/problems/compare-version-numbers.py b/problems/python/compare-version-numbers.py old mode 100644 new mode 100755 similarity index 100% rename from problems/compare-version-numbers.py rename to problems/python/compare-version-numbers.py diff --git a/problems/python/connecting-cities-with-minimum-cost.py b/problems/python/connecting-cities-with-minimum-cost.py new file mode 100755 index 0000000..50a04e5 --- /dev/null +++ b/problems/python/connecting-cities-with-minimum-cost.py @@ -0,0 +1,38 @@ +class Solution(object): + def minimumCost(self, N, connections): + def union(n1, n2): + p1 = find(n1) + p2 = find(n2) + + if p1==p2: return False + + if ranks[p1]>ranks[p2]: + parents[p2] = p1 + ranks[p1] += 1 + else: + parents[p1] = p2 + ranks[p2] += 1 + return True + + def find(n): + p = parents[n] + while p!=parents[p]: p = find(p) + parents[n] = p + return p + + if not connections: return 0 + + parents = [n for n in xrange(N+1)] + ranks = [0]*(N+1) + totalCost = 0 + count = N # the count of nodes not yet union + + # sort by cost, union the ones with less cost first + sortedConnections = sorted([(cost, x, y) for x, y, cost in connections]) + + for cost, x, y in sortedConnections: + if union(x, y): + totalCost += cost + count -= 1 + + return totalCost if count==1 else -1 \ No newline at end of file diff --git a/problems/python/consecutive-numbers-sum.py b/problems/python/consecutive-numbers-sum.py new file mode 100755 index 0000000..127faa6 --- /dev/null +++ b/problems/python/consecutive-numbers-sum.py @@ -0,0 +1,43 @@ + +# Sliding window +class Solution(object): + def consecutiveNumbersSum(self, N): + i = 1 + j = 2 + s = 1 #sum(range(i, j)) + ans = 0 + + while i<=j and j<=N+1: + if s==N: + ans += 1 + s += j + j += 1 + elif s>N: + s -= i + i += 1 + else: + s += j + j += 1 + return ans + + +#Math +class Solution(object): + def consecutiveNumbersSum(self, N): + ans = 0 + upperLimit = int((2 * N + 0.25)**0.5 - 0.5) + 2 + + for x in xrange(1, upperLimit): + if x%2==0: + if float(N)/x-N/x==0.5: + upper = N/x + x/2 + lower = N/x - x/2 + 1 + if 1<=lower and upper<=N: + ans += 1 + else: + if N%x==0: + upper = N/x + (x-1)/2 + lower = N/x - (x-1)/2 + if 1<=lower and upper<=N: + ans += 1 + return ans \ No newline at end of file diff --git a/problems/python/construct-binary-tree-from-inorder-and-postorder-traversal.py b/problems/python/construct-binary-tree-from-inorder-and-postorder-traversal.py new file mode 100755 index 0000000..f8304fc --- /dev/null +++ b/problems/python/construct-binary-tree-from-inorder-and-postorder-traversal.py @@ -0,0 +1,102 @@ +""" +inorder: [LEFT]root[RIGHT] +postorder: [LEFT][RIGHT]root + +First thing we know is the value of root, which is the last element of `postorder`. +Find the index of the root in `inorder`. So find out the interval of [LEFT] and [RIGHT] in `inorder`. + +The length of the [LEFT] and [RIGHT] in `inorder` are the same with the length of the [LEFT] and [RIGHT] in `postorder`. +""" +class Solution(object): + def buildTree(self, inorder, postorder): + if not inorder or not postorder: return None + + root = TreeNode(postorder[-1]) + if len(inorder)==1: return root + + r = inorder.index(root.val) + + leftInOrder = inorder[:r] + leftPostOrder = postorder[:r] + rightInOrder = inorder[r+1:] + rightPostOrder = postorder[r:len(postorder)-1] + + root.left = self.buildTree(leftInOrder, leftPostOrder) + root.right = self.buildTree(rightInOrder, rightPostOrder) + + return root +""" +Time: O(NLogN). For each node, we need to do an iteration to its children. To be precise.. +O(N) for constructing root. + +O(N/2) for constructing root.left +O(N/2) for constructing root.right + +O(N/4) for constructing root.left.left +O(N/4) for constructing root.left.right +O(N/4) for constructing root.right.left +O(N/4) for constructing root.right.right +... + +To improve this, we can use a hash table to get the index of `i` below + + +Space: O(NLogN). +For each node, we need to construct inorder/postorder arrays of its children. +We can improve this by using pointers. +""" + +""" +Improved version. +Time: O(N). +Space: O(N). For `index`. +""" +class Solution(object): + def buildTree(self, inorder, postorder): + def helper(i, j, k, l): + if j-i<=0: return None + if l-k<=0: return None + + root = TreeNode(postorder[l-1]) + if j-i==1: return root + + r = index[root.val] + + root.left = helper(i, r, k, k+r-i) + root.right = helper(r+1, j, k+r-i, l-1) + return root + + index = {} #the index of inorder + for i, n in enumerate(inorder): index[n] = i + return helper(0, len(inorder), 0, len(postorder)) + + + + + +""" +Inorder: [LEFT]root[RIGHT] +Postorder: [LEFT][RIGHT]root +As you can see, rootVal is the last val in postorder. + +i and j is the range of the inorder +k and l is the range of the postorder +Get the root +Also recursively use the same method to get the left and right node, but this time we change the inorder and postorder. +""" +class Solution(object): + def buildTree(self, inorder, postorder): + def helper(i, j, k, l): + if i>j or k>l: return None + rootVal = postorder[l] + root = TreeNode(rootVal) + r = inorderIndex[rootVal] + leftLength = r-i + rightLength = j-r + root.left = helper(i, r-1, k, k+leftLength-1) + root.right = helper(r+1, j, k+leftLength, k+leftLength+rightLength-1) + return root + + inorderIndex = {} + for i, n in enumerate(inorder): inorderIndex[n] = i + return helper(0, len(inorder)-1, 0, len(postorder)-1) diff --git a/problems/python/construct-binary-tree-from-preorder-and-inorder-traversal.py b/problems/python/construct-binary-tree-from-preorder-and-inorder-traversal.py new file mode 100755 index 0000000..3f021d7 --- /dev/null +++ b/problems/python/construct-binary-tree-from-preorder-and-inorder-traversal.py @@ -0,0 +1,30 @@ +""" +Time: O(N) +Space: O(N) + +preorder: root[LEFT][RIGHT] +inorder: [LEFT]root[RIGHT] + +The `helper()` will return the root node from `preorder` and `inorder`. +If they have left or right child, call `helper()` again to get the left or right node. + +Use pointers to represent array, avoiding keep copying `preorder` and `inorder`. +i and j represent the preorder[i:j] +k and l epresent the inorder[k:l] +""" +class Solution(object): + def buildTree(self, preorder, inorder): + def helper(i, j, k, l): + root = TreeNode(preorder[i]) + + r = inorderIndex[root.val] #the index of the root val in inorder + leftLength = r-k + rightLength = l-(r+1) + + if leftLength>0: root.left = helper(i+1, i+1+leftLength, k, k+leftLength) + if rightLength>0: root.right = helper(i+1+leftLength, i+1+leftLength+rightLength, r+1, r+1+rightLength) + return root + + inorderIndex = {} + for i, n in enumerate(inorder): inorderIndex[n] = i + return helper(0, len(preorder), 0, len(inorder)) \ No newline at end of file diff --git a/problems/container-with-most-water.py b/problems/python/container-with-most-water.py old mode 100644 new mode 100755 similarity index 100% rename from problems/container-with-most-water.py rename to problems/python/container-with-most-water.py diff --git a/problems/contains-duplicate-ii.py b/problems/python/contains-duplicate-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/contains-duplicate-ii.py rename to problems/python/contains-duplicate-ii.py diff --git a/problems/contains-duplicate-iii.py b/problems/python/contains-duplicate-iii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/contains-duplicate-iii.py rename to problems/python/contains-duplicate-iii.py diff --git a/problems/contains-duplicate.py b/problems/python/contains-duplicate.py old mode 100644 new mode 100755 similarity index 100% rename from problems/contains-duplicate.py rename to problems/python/contains-duplicate.py diff --git a/problems/python/continuous-subarray-sum.py b/problems/python/continuous-subarray-sum.py new file mode 100755 index 0000000..75dd08b --- /dev/null +++ b/problems/python/continuous-subarray-sum.py @@ -0,0 +1,35 @@ +class Solution(object): + def checkSubarraySum(self, nums, k): + prefixSumEndings = collections.defaultdict(list) + prefixSumEndings[0].append(-1) + + prefixSum = 0 + for i, n in enumerate(nums): + prefixSum += n + + x = prefixSum/k + while x>=0: + p2 = prefixSum-k*x + if len(prefixSumEndings[p2])>0 and prefixSumEndings[p2][0]=1: return True + if len(prefixSumKRemain[remain])>0 and prefixSumKRemain[remain][0]0: node.left = helper(i, m) + if rightLength>0: node.right = helper(m+1, j) + return node + + return helper(0, len(nums)) + + + + +class Solution(object): + def sortedArrayToBST(self, nums): + def helper(s, e): + if s>e: return None + m = (s+e)/2 + node = TreeNode(nums[m]) + node.left = helper(s, m-1) + node.right = helper(m+1, e) + return node + return helper(0, len(nums)-1) \ No newline at end of file diff --git a/problems/python/convert-sorted-list-to-binary-search-tree.py b/problems/python/convert-sorted-list-to-binary-search-tree.py new file mode 100755 index 0000000..5b127b7 --- /dev/null +++ b/problems/python/convert-sorted-list-to-binary-search-tree.py @@ -0,0 +1,45 @@ +""" +Time: O(NLogN), +The root need to traverse N node to get its midNode. + +The root.left need to traverse N/2 node to get its midNode. +The root.right need to traverse N/2 node to get its midNode. + +The root.left.left need to traverse N/4 node to get its midNode. +The root.left.right need to traverse N/4 node to get its midNode. +The root.right.left need to traverse N/4 node to get its midNode. +The root.right.right need to traverse N/4 node to get its midNode. + +... +So total: +N + (N/2+N/2) + (N/4+N/4+N/4+N/4) + ... => N + N + N + ... +There are LogN levels in the tree, so LogN x N => NLogN + +Space: O(LogN) for recursion depth. +""" +class Solution(object): + def sortedListToBST(self, head): + if not head: return None + + midNode = self.getMiddle(head) + root = TreeNode(midNode.val) + + if head.val==midNode.val: return root + + root.left = self.sortedListToBST(head) + root.right = self.sortedListToBST(midNode.next) + + return root + + def getMiddle(self, node): + pre = None + slow = node + fast = node + + while fast and fast.next: + fast = fast.next.next + pre = slow + slow = slow.next + + if pre: pre.next = None + return slow \ No newline at end of file diff --git a/problems/copy-list-with-random-pointer.py b/problems/python/copy-list-with-random-pointer.py old mode 100644 new mode 100755 similarity index 67% rename from problems/copy-list-with-random-pointer.py rename to problems/python/copy-list-with-random-pointer.py index 0790e26..ddf901e --- a/problems/copy-list-with-random-pointer.py +++ b/problems/python/copy-list-with-random-pointer.py @@ -55,5 +55,38 @@ def copyRandomList(self, head): """ Time: O(N). Two iteration. First iteration make a the clones. Second iteration setup the links. -Space: O(1). No extra space except the cloned node. -""" \ No newline at end of file +Space: O(N). +""" + + +""" +Time: O(N) +Space: O(1) +""" +class Solution(object): + def copyRandomList(self, head): + if not head: return head + curr = head + while curr: + nextCurr = curr.next + copy = Node(curr.val) + curr.next = copy + copy.next = nextCurr + curr = nextCurr + + copyHead = head.next + + curr = head + while curr: + if curr.random: curr.next.random = curr.random.next + curr = curr.next.next + + curr = head + while curr: + nextCurr = curr.next.next + if nextCurr: + curr.next.next = nextCurr.next + curr.next = nextCurr + curr = nextCurr + + return copyHead \ No newline at end of file diff --git a/problems/python/count-all-valid-pickup-and-delivery-opti.py b/problems/python/count-all-valid-pickup-and-delivery-opti.py new file mode 100755 index 0000000..8bfe82a --- /dev/null +++ b/problems/python/count-all-valid-pickup-and-delivery-opti.py @@ -0,0 +1,5 @@ +class Solution(object): + def countOrders(self, n): + d = 1 + for i in xrange(1, 2*n, 2): d*=i + return math.factorial(n)*d % (10**9 + 7) \ No newline at end of file diff --git a/problems/python/count-binary-substrings.py b/problems/python/count-binary-substrings.py new file mode 100755 index 0000000..5a2520e --- /dev/null +++ b/problems/python/count-binary-substrings.py @@ -0,0 +1,21 @@ +class Solution(object): + def countBinarySubstrings(self, s): + groups = [] + count = 0 + last = s[0] + ans = 0 + + for bit in s: + if bit==last: + count += 1 + else: + groups.append(count) + count = 1 + last = bit + + groups.append(count) + + for i in xrange(1, len(groups)): + ans += min(groups[i], groups[i-1]) + + return ans \ No newline at end of file diff --git a/problems/python/count-complete-tree-nodes.py b/problems/python/count-complete-tree-nodes.py new file mode 100755 index 0000000..ab9c7bc --- /dev/null +++ b/problems/python/count-complete-tree-nodes.py @@ -0,0 +1,27 @@ +""" +Time: O(H^2). The time complexity will countNodes() will be LogH if the tree is perfect. +If not, we will need Log(H-1) + +Space: O(H) +""" +class Solution(object): + def countNodes(self, root): + if not root: return 0 + + node = root + l = 1 #level on the right side + while node.left: + node = node.left + l += 1 + + node = root + r = 1 #level on the left side + while node.right: + node = node.right + r += 1 + + if l==r: + #perfect tree + return (2**r)-1 + else: + return 1+self.countNodes(root.left)+self.countNodes(root.right) \ No newline at end of file diff --git a/problems/python/count-number-of-nice-subarrays.py b/problems/python/count-number-of-nice-subarrays.py new file mode 100755 index 0000000..33df804 --- /dev/null +++ b/problems/python/count-number-of-nice-subarrays.py @@ -0,0 +1,19 @@ +class Solution(object): + def numberOfSubarrays(self, nums, K): + def atMost(k): + oddCount = 0 + ans = 0 + i = 0 + + for j in xrange(len(nums)): + if nums[j]%2!=0: oddCount += 1 + + while oddCount>k: + if nums[i]%2!=0: oddCount -= 1 + i += 1 + + ans += j-i+1 + + return ans + + return atMost(K)-atMost(K-1) \ No newline at end of file diff --git a/problems/python/count-unique-characters-of-all-substrings-of-a-given-string.py b/problems/python/count-unique-characters-of-all-substrings-of-a-given-string.py new file mode 100755 index 0000000..fe58877 --- /dev/null +++ b/problems/python/count-unique-characters-of-all-substrings-of-a-given-string.py @@ -0,0 +1,18 @@ +class Solution(object): + def uniqueLetterString(self, s): + index = {} + ans = 0 + for c in 'abcdefghijklmnopqrstuvwxyz': + index[c.upper()] = [-1, -1] + + # count the substring that s[j] is the unique letter + for k, c in enumerate(s): + i, j = index[c] + ans += (j-i) * (k-j) + index[c] = [j, k] + + # count the substring that s[j] is the unique letter, because last iteration did not count the last letter + for c in index: + i, j = index[c] + ans += (j-i) * (len(s)-j) + return ans \ No newline at end of file diff --git a/problems/course-schedule-ii.py b/problems/python/course-schedule-ii.py old mode 100644 new mode 100755 similarity index 61% rename from problems/course-schedule-ii.py rename to problems/python/course-schedule-ii.py index 9715692..a7c139f --- a/problems/course-schedule-ii.py +++ b/problems/python/course-schedule-ii.py @@ -35,16 +35,42 @@ def findOrder(self, numCourses, prerequisites): We can use it to know which node comes after which or detect cycles. The algorithm is easy to understand. First, we build the adjacent list (`graph`) and count all the inbound of the node. -Then we start from the node whose inbound count is 0, adding it in to the `q`. -For every node we pop out from `q` +Then we start from the node whose inbound count is 0, adding it in to the `pq` (priority queue). +For every node we pop out from `pq` * We remove the node's outbound by decrease 1 on all its neighbor's inbound. - * Put the node's neighbor to `q` if it has no inbound - * Put the node into the `order` + * Put the node's neighbor to `pq` if it has no inbound + * Put the node into the `sortedNodes` Repeat the process until there is no more node. -The order in the `order` is the order we are going to encounter when we run through the directed graph. +The order in the `sortedNodes` is the order we are going to encounter when we run through the directed graph. If we cannot sort all the nodes in the graph, it means that there are some nodes we couldn't find its starting point, in other words, there are cycles in the graph. Time: O(E+2V) ~= O(E+V) we used O(E) to build the graph #[1], O(V) to find the starting point #[2], then traverse all the nodes again #[3]. Space: O(E+3V) ~= O(E+V), O(E+V) for the adjacent list. O(V) for the `q`, O(V) for the `q_next`. -""" \ No newline at end of file +""" +class Solution(object): + def findOrder(self, numCourses, prerequisites): + graph = collections.defaultdict(list) + inbounds = collections.defaultdict(int) + pq = collections.deque() + sortedNodes = [] + + for c1, c2 in prerequisites: + graph[c2].append(c1) + inbounds[c1]+=1 + + for node in xrange(numCourses): + if inbounds[node]==0: + pq.append(node) + + while pq: + node = pq.popleft() + + for nei in graph[node]: + inbounds[nei] -= 1 + if inbounds[nei]==0: + pq.append(nei) + sortedNodes.append(node) + + return sortedNodes if len(sortedNodes)==numCourses else [] + \ No newline at end of file diff --git a/problems/course-schedule.py b/problems/python/course-schedule.py old mode 100644 new mode 100755 similarity index 100% rename from problems/course-schedule.py rename to problems/python/course-schedule.py diff --git a/problems/python/custom-sort-string.py b/problems/python/custom-sort-string.py new file mode 100755 index 0000000..2e63523 --- /dev/null +++ b/problems/python/custom-sort-string.py @@ -0,0 +1,20 @@ +""" +Take a look at the char in s. +For the char that is in the order, rearrange them to sortedChars with respect to the "order". +For the char that is in not the order, put them in postString. +""" +class Solution(object): + def customSortString(self, order, s): + sortedChars = '' + counter = collections.Counter(s) + for c in order: + if c in order: + sortedChars += c*counter[c] + + orderSet = set(order) + postString = '' + for c in s: + if c not in orderSet: + postString += c + + return sortedChars+postString \ No newline at end of file diff --git a/problems/python/cutting-ribbons.py b/problems/python/cutting-ribbons.py new file mode 100755 index 0000000..175e336 --- /dev/null +++ b/problems/python/cutting-ribbons.py @@ -0,0 +1,13 @@ +class Solution(object): + def maxLength(self, ribbons, k): + maxLen = max(ribbons) + minLen = 0 + + while minLen=k: + minLen = l + else: + maxLen = l-1 + return maxLen \ No newline at end of file diff --git a/problems/data-stream-as-disjoint-intervals.py b/problems/python/data-stream-as-disjoint-intervals.py old mode 100644 new mode 100755 similarity index 100% rename from problems/data-stream-as-disjoint-intervals.py rename to problems/python/data-stream-as-disjoint-intervals.py diff --git a/problems/decode-string.py b/problems/python/decode-string.py old mode 100644 new mode 100755 similarity index 100% rename from problems/decode-string.py rename to problems/python/decode-string.py diff --git a/problems/python/decode-ways.py b/problems/python/decode-ways.py new file mode 100755 index 0000000..82f4682 --- /dev/null +++ b/problems/python/decode-ways.py @@ -0,0 +1,19 @@ +""" +Time: O(N), due to the "memo", the time can reduce from O(2^N) to O(N). +Space: O(N) +""" +class Solution(object): + def numDecodings(self, s): + def helper(s, i): + if (s, i) in memo: return memo[(s, i)] + if i>=len(s): return 1 + if s[i]=='0': return 0 + + count = 0 + count += helper(s, i+1) + if len(s)-i>=2 and int(s[i:i+2])<=26: count += helper(s, i+2) + memo[(s, i)] = count + return count + + memo = {} + return helper(s, 0) \ No newline at end of file diff --git a/problems/delete-and-earn.py b/problems/python/delete-and-earn.py old mode 100644 new mode 100755 similarity index 100% rename from problems/delete-and-earn.py rename to problems/python/delete-and-earn.py diff --git a/problems/python/delete-duplicate-folders-in-system.py b/problems/python/delete-duplicate-folders-in-system.py new file mode 100755 index 0000000..512cd1a --- /dev/null +++ b/problems/python/delete-duplicate-folders-in-system.py @@ -0,0 +1,42 @@ +class Node(object): + def __init__(self, val): + self.val = val + self.key = None + self.children = {} + +class Solution(object): + def deleteDuplicateFolder(self, paths): + def setKey(node): + node.key = '' + for c in sorted(node.children.keys()): #need to be sorted. so when child structs are the same, we won't generate different key from different iteration order. + setKey(node.children[c]) + node.key += node.children[c].val + '|' + node.children[c].key + '|' #generate a key for each node. only considering its children structure. (see the "identical" definition, it does not consider the val of the node itself.) + + keyCount[node.key] += 1 + + def addPath(node, path): + if node.children and keyCount[node.key]>1: return #leaf node does not apply to this rule + ans.append(path+[node.val]) + for c in node.children: + addPath(node.children[c], path+[node.val]) + + + ans = [] + root = Node('/') + keyCount = collections.Counter() + + #build the tree + for path in paths: + node = root + for c in path: + if c not in node.children: node.children[c] = Node(c) + node = node.children[c] + + #set all nodes key recursively + setKey(root) + + #build ans + for c in root.children: + addPath(root.children[c], []) + + return ans \ No newline at end of file diff --git a/problems/delete-node-in-a-bst.py b/problems/python/delete-node-in-a-bst.py old mode 100644 new mode 100755 similarity index 69% rename from problems/delete-node-in-a-bst.py rename to problems/python/delete-node-in-a-bst.py index de1101a..f1e0fe2 --- a/problems/delete-node-in-a-bst.py +++ b/problems/python/delete-node-in-a-bst.py @@ -1,3 +1,23 @@ +""" +Find the parant and the node to be deleted. [0] +Deleting the node means replacing its reference by something else. [1] + +For the node to be deleted, if it only has no child, just remove it from the parent. Return None. [2] + +If it has one child, return the child. So its parent will directly connect to its child. [3] + +If it has both child. Update the node's val to the minimum value in the right subtree. Remove the minimum value node in the right subtree. [4] +This is equivalent to replacing the node by the minimum value node in the right subtree. +Another option is to replace the node by the maximum value node in the left subtree. + +Find the minimum value in the left subtree is easy. The leftest node value in the tree is the smallest. [5] + +Time Complexity: O(LogN). O(LogN) for finding the node to be deleted. +The recursive call in `remove()` will be apply to a much smaller subtree. And much smaller subtree... +So can be ignored. +Space complexity is O(LogN). Because the recursive call will at most be called LogN times. +N is the number of nodes. And LogN can be consider the height of the tree. +""" class Solution(object): def deleteNode(self, root, key): def find_min(root): @@ -39,23 +59,41 @@ def remove(node): return root -""" -Find the parant and the node to be deleted. [0] -Deleting the node means replacing its reference by something else. [1] -For the node to be deleted, if it only has no child, just remove it from the parent. Return None. [2] -If it has one child, return the child. So its parent will directly connect to its child. [3] -If it has both child. Update the node's val to the minimum value in the right subtree. Remove the minimum value node in the right subtree. [4] -This is equivalent to replacing the node by the minimum value node in the right subtree. -Another option is to replace the node by the maximum value node in the left subtree. +class Solution(object): + def deleteNode(self, node, key): + if not node: + return node + elif keynode.val: + node.right = self.deleteNode(node.right, key) + return node + elif key==node.val: + if not node.left and not node.right: + return None + elif not node.left: + return node.right + elif not node.right: + return node.left + else: + node.val = self.findMin(node.right) #min value in the right subtree + node.right = self.deleteNode(node.right, node.val) + return node + + def findMin(self, root): + ans = float('inf') + stack = [root] + + while stack: + node = stack.pop() + ans = min(ans, node.val) + if node.left: stack.append(node.left) + if node.right: stack.append(node.right) + return ans + -Find the minimum value in the left subtree is easy. The leftest node value in the tree is the smallest. [5] -Time Complexity: O(LogN). O(LogN) for finding the node to be deleted. -The recursive call in `remove()` will be apply to a much smaller subtree. And much smaller subtree... -So can be ignored. -Space complexity is O(LogN). Because the recursive call will at most be called LogN times. -N is the number of nodes. And LogN can be consider the height of the tree. -""" \ No newline at end of file diff --git a/problems/delete-operation-for-two-strings.py b/problems/python/delete-operation-for-two-strings.py old mode 100644 new mode 100755 similarity index 100% rename from problems/delete-operation-for-two-strings.py rename to problems/python/delete-operation-for-two-strings.py diff --git a/problems/python/design-add-and-search-words-data-structure.py b/problems/python/design-add-and-search-words-data-structure.py new file mode 100755 index 0000000..4b0d71f --- /dev/null +++ b/problems/python/design-add-and-search-words-data-structure.py @@ -0,0 +1,35 @@ +class Node(object): + def __init__(self, char): + self.char = char + self.children = {} + +class WordDictionary(object): + + def __init__(self): + self.root = Node('') + self.endSign = ';' + + def addWord(self, word): + word = word + self.endSign + node = self.root + for c in word: + if c not in node.children: + node.children[c] = Node(c) + node = node.children[c] + + def search(self, word): + word = word + self.endSign + return self.searchFromNode(self.root, word) + + def searchFromNode(self, node, word): + if not word: return True + + char = word[0] + if char in node.children: + return self.searchFromNode(node.children[char], word[1:]) + elif char=='.': + for c in node.children: + if self.searchFromNode(node.children[c], word[1:]): + return True + return False + \ No newline at end of file diff --git a/problems/python/design-in-memory-file-system.py b/problems/python/design-in-memory-file-system.py new file mode 100755 index 0000000..3e9d808 --- /dev/null +++ b/problems/python/design-in-memory-file-system.py @@ -0,0 +1,46 @@ +""" +self.dirs is a nested hashmap to store the file structure. +self.files is a hashmap, mainly used to determine if the path is a "file", also used to store the content of the file. + +Time: +ls O(K) +mkdir O(K) +addContentToFile O(K) +readContentFromFile O(1) +K is the path count, for example /a/b/c/d, K=4. + +Space: +O(N), N is the dir counts. +""" +class FileSystem(object): + + def __init__(self): + self.files = {} + self.dirs = {} + + def ls(self, path): + if path in self.files: + return [path.split('/')[-1]] + else: + if path=='/': return sorted(self.dirs.keys()) + curr = self.dirs + for d in path.split('/')[1:]: + curr = curr[d] + return sorted(curr.keys()) + + def mkdir(self, path): + curr = self.dirs + for d in path.split('/')[1:]: + if d not in curr: + curr[d] = {} + curr = curr[d] + + def addContentToFile(self, filePath, content): + if filePath not in self.files: + self.mkdir(filePath) + self.files[filePath] = content + else: + self.files[filePath] += content + + def readContentFromFile(self, filePath): + return self.files[filePath] \ No newline at end of file diff --git a/problems/design-linked-list.py b/problems/python/design-linked-list.py old mode 100644 new mode 100755 similarity index 100% rename from problems/design-linked-list.py rename to problems/python/design-linked-list.py diff --git a/problems/python/design-tic-tac-toe.py b/problems/python/design-tic-tac-toe.py new file mode 100755 index 0000000..5e72cda --- /dev/null +++ b/problems/python/design-tic-tac-toe.py @@ -0,0 +1,52 @@ +""" +Time: O(1) for move(). +Space: O(N). + +We need to find a way to quikly check if the player wins after the move. +Checking the whole map row by row or column by column will take a lots of time. +What we do instead is to record the count of the placement for each row and colomn. Also the count of topright-bottomleft, topleft-bottomright placement. +So if any of the count adds up to n, the player wins. + +self.records[0] stores player1's record. +self.records[1] stores player2's record. + +For player1's record (self.records[0]), +record[0][row] stores the placement count of the row for player1. +record[1][col] stores the placement count of the col for player1. +record[2] stores the topright-bottomleft placement count for player1. +record[3] stores the topleft-bottomright placement count for player1. + +For player2's record (self.records[1]), +record[0][row] stores the placement count of the row for player2. +record[1][col] stores the placement count of the col for player2. +record[2] stores the topright-bottomleft placement count for player2. +record[3] stores the topleft-bottomright placement count for player2. + +Note that, when checkRecord() we only need to check the (row, col) we just placed. +So we can achieve O(1) in time. +""" +class TicTacToe(object): + + def __init__(self, n): + self.records = [[[0]*n, [0]*n, 0, 0], [[0]*n, [0]*n, 0, 0]] + self.n = n + + + def move(self, row, col, player): + record = self.records[player-1] + record[0][row] += 1 + record[1][col] += 1 + if row==col: record[2] += 1 + if row+col==self.n-1: record[3] += 1 + + if self.checkRecord(record, row, col): return player + + return 0 + + def checkRecord(self, record, row, col): + if record[0][row]==self.n: return True + if record[1][col]==self.n: return True + if record[2]==self.n: return True + if record[3]==self.n: return True + + return False \ No newline at end of file diff --git a/problems/python/diagonal-traverse.py b/problems/python/diagonal-traverse.py new file mode 100755 index 0000000..0510201 --- /dev/null +++ b/problems/python/diagonal-traverse.py @@ -0,0 +1,25 @@ +class Solution(object): + def findDiagonalOrder(self, mat): + def helper(i, j, reverse): + output = [] + while 0<=i=2 as lca. + def findCount(node): + if not node: return 0 + count = 0 + if node.val==p or node.val==q: count += 1 + count += findPQCount(node.left) + count += findPQCount(node.right) + if count>=2 and not self.lca: self.lca = node + return count + + def findHeight(node, h): + if not node: return + if node.val==p: self.pHeight = h + if node.val==q: self.qHeight = h + if self.pHeight and self.qHeight: return + findHeight(node.left, h+1) + findHeight(node.right, h+1) + + findCount(root) + findHeight(self.lca, 0) + return self.qHeight + self.pHeight \ No newline at end of file diff --git a/problems/python/find-duplicate-subtrees.py b/problems/python/find-duplicate-subtrees.py new file mode 100755 index 0000000..5729e8e --- /dev/null +++ b/problems/python/find-duplicate-subtrees.py @@ -0,0 +1,16 @@ +class Solution(object): + def findDuplicateSubtrees(self, root): + def dfs(node): + if not node: return '#' + string = str(node.val) + ',' + dfs(node.left) + ',' + dfs(node.right) + data[string].append(node) + return string + + data = collections.defaultdict(list) + ans = [] + + dfs(root) + + for s in data: + if len(data[s])>=2: ans.append(data[s][0]) + return ans \ No newline at end of file diff --git a/problems/find-eventual-safe-states.py b/problems/python/find-eventual-safe-states.py old mode 100644 new mode 100755 similarity index 100% rename from problems/find-eventual-safe-states.py rename to problems/python/find-eventual-safe-states.py diff --git a/problems/find-first-and-last-position-of-element-in-sorted-array.py b/problems/python/find-first-and-last-position-of-element-in-sorted-array.py old mode 100644 new mode 100755 similarity index 100% rename from problems/find-first-and-last-position-of-element-in-sorted-array.py rename to problems/python/find-first-and-last-position-of-element-in-sorted-array.py diff --git a/problems/python/find-k-closest-elements.py b/problems/python/find-k-closest-elements.py new file mode 100755 index 0000000..802279f --- /dev/null +++ b/problems/python/find-k-closest-elements.py @@ -0,0 +1,20 @@ +class Solution(object): + def findClosestElements(self, arr, K, X): + def isCloserThan(n1, n2, x): + return abs(x-n1)=len(arr) or (l>=0 and isCloserThan(arr[l], arr[r], X)): + output1.append(arr[l]) + l -= 1 + else: + output2.append(arr[r]) + r += 1 + + return output1[::-1]+output2 diff --git a/problems/find-k-pairs-with-smallest-sums.py b/problems/python/find-k-pairs-with-smallest-sums.py old mode 100644 new mode 100755 similarity index 100% rename from problems/find-k-pairs-with-smallest-sums.py rename to problems/python/find-k-pairs-with-smallest-sums.py diff --git a/problems/python/find-leaves-of-binary-tree.py b/problems/python/find-leaves-of-binary-tree.py new file mode 100755 index 0000000..8b08b11 --- /dev/null +++ b/problems/python/find-leaves-of-binary-tree.py @@ -0,0 +1,57 @@ +""" +Time: O(N), each nodes is traversed 2 times. +Space: O(N) for `parents` and `q`, `ans`, `ans2` + +[1] +Through BFS +Store the parent of each node in the `parents`. +Also store the leaf node in the `temp`. + +[2] +Store the `temp` to the ans. +For each node stored to the ans, we need to "detach" it from its parent. +Also check if the parent become a leaf node, if so, store it in the new `temp`. + +[3] +Store the val from node reference in ans to ans2. +""" +class Solution(object): + def findLeaves(self, root): + ans = [] + q = collections.deque([root]) + parents = {} + + #[1] + temp = [] + while q: + node = q.popleft() + + if not node.left and not node.right: + temp.append(node) + + if node.left: + q.append(node.left) + parents[node.left] = node + + if node.right: + q.append(node.right) + parents[node.right] = node + + #[2] + while temp: + ans.append(temp) + temp = [] + + for node in ans[-1]: + if node not in parents: break + p = parents[node] + if p.left==node: p.left = None + if p.right==node: p.right = None + if not p.left and not p.right: temp.append(p) + + #[3] + ans2 = [] + for temp in ans: + ans2.append([node.val for node in temp]) + + return ans2 \ No newline at end of file diff --git a/problems/find-median-from-data-stream.py b/problems/python/find-median-from-data-stream.py old mode 100644 new mode 100755 similarity index 100% rename from problems/find-median-from-data-stream.py rename to problems/python/find-median-from-data-stream.py diff --git a/problems/find-minimum-in-rotated-sorted-array-ii.py b/problems/python/find-minimum-in-rotated-sorted-array-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/find-minimum-in-rotated-sorted-array-ii.py rename to problems/python/find-minimum-in-rotated-sorted-array-ii.py diff --git a/problems/find-minimum-in-rotated-sorted-array.py b/problems/python/find-minimum-in-rotated-sorted-array.py old mode 100644 new mode 100755 similarity index 100% rename from problems/find-minimum-in-rotated-sorted-array.py rename to problems/python/find-minimum-in-rotated-sorted-array.py diff --git a/problems/find-mode-in-binary-search-tree.py b/problems/python/find-mode-in-binary-search-tree.py old mode 100644 new mode 100755 similarity index 81% rename from problems/find-mode-in-binary-search-tree.py rename to problems/python/find-mode-in-binary-search-tree.py index 4705a77..b4d3aff --- a/problems/find-mode-in-binary-search-tree.py +++ b/problems/python/find-mode-in-binary-search-tree.py @@ -1,3 +1,8 @@ +""" +Time: O(N). For we traverse every node recursively. +Space: O(N). Becuase we store all element's val and count in `counter`. +Note that, we do not use the feature of BST. +""" from collections import Counter class Solution(object): def findMode(self, root): @@ -14,11 +19,21 @@ def count(node): return [val for val, count in counter.items() if count==max_count] """ -Time: O(N). For we traverse every node recursively. -Space: O(N). Becuase we store all element's val and count in `counter`. -Note that, we do not use the feature of BST. -""" +To use the feature of BST, we are going to inorder traverse the BST. +So it will be like we are iterating a sorted array. + +[1] +While iterating, we can put only the element count that is greater or equal than `max_count` to `ans`. +If we encounter a new element with larger `curr_count`, we reset the `ans`. +[0] +With the help of `prev_val` we can know that `curr_node` is the same to the previous or not. +If not, its a new element, we need to reset the `curr_count`. + +Time: O(N). Space: O(LogN) + +For better understanding, below is a template for inorder traverse. +""" class Solution(object): def findMode(self, root): if not root: return [] @@ -54,23 +69,6 @@ def findMode(self, root): return ans -""" -To use the feature of BST, we are going to inorder traverse the BST. -So it will be like we are iterating a sorted array. - -[1] -While iterating, we can put only the element count that is greater or equal than `max_count` to `ans`. -If we encounter a new element with larger `curr_count`, we reset the `ans`. - -[0] -With the help of `prev_val` we can know that `curr_node` is the same to the previous or not. -If not, its a new element, we need to reset the `curr_count`. - -Time: O(N). Space: O(LogN) - -For better understanding, below is a template for inorder traverse. -""" - #inorder traversal of BST def inorder_traverse(root): curr = root @@ -85,3 +83,28 @@ def inorder_traverse(root): print curr.val curr = curr.right + + + + +""" +Time: O(N) +Space: O(N) +""" +class Solution(object): + def findMode(self, root): + def helper(node): + if not node: return + counter[node.val] += 1 + counter['maxCount'] = max(counter['maxCount'], counter[node.val]) + helper(node.left) + helper(node.right) + + ans = [] + counter = collections.Counter() + helper(root) + for v in counter: + if v!='maxCount' and counter[v]==counter['maxCount']: + ans.append(v) + + return ans \ No newline at end of file diff --git a/problems/python/find-original-array-from-doubled-array,py b/problems/python/find-original-array-from-doubled-array,py new file mode 100755 index 0000000..e9e2726 --- /dev/null +++ b/problems/python/find-original-array-from-doubled-array,py @@ -0,0 +1,26 @@ +""" +Time: O(NLogN) +Space: O(N) +""" +class Solution(object): + def findOriginalArray(self, changed): + if len(changed)%2!=0: return [] + + ans = [] + counter = collections.Counter() #store the count of the doubled number + count = 0 #sum of count in counter + + changed.sort() #need to be sorted, otherwise we cannot identify which number is orginal or it is doubled. + + for num in changed: + if counter[num]>0: + #num is a doubled num + counter[num] -= 1 + count -= 1 + ans.append(num/2) + else: + #num is an original num + counter[num*2] += 1 + count += 1 + + return ans if count==0 else [] \ No newline at end of file diff --git a/problems/find-peak-element.py b/problems/python/find-peak-element.py old mode 100644 new mode 100755 similarity index 78% rename from problems/find-peak-element.py rename to problems/python/find-peak-element.py index 5b9a8b3..aeb4726 --- a/problems/find-peak-element.py +++ b/problems/python/find-peak-element.py @@ -54,5 +54,24 @@ def findPeakElement(self, nums): return l - + +""" +[l, r] is the possible range. +Keep decreasing the range unsing binary search until l==r. +Pay attention to +m = l+(r-l+1)/2 +Sometimes you need m = l+(r-l)/2 to avoid infinite loop. +""" +class Solution(object): + def findPeakElement(self, nums): + l = 0 + r = len(nums)-1 + + while l2: + counter[fruits[i]] -= 1 + if counter[fruits[i]]==0: uniqueFruits -= 1 + i += 1 + + ans = max(ans, j-i+1) + return ans \ No newline at end of file diff --git a/problems/game-of-life.py b/problems/python/game-of-life.py old mode 100644 new mode 100755 similarity index 100% rename from problems/game-of-life.py rename to problems/python/game-of-life.py diff --git a/problems/generate-parentheses.py b/problems/python/generate-parentheses.py old mode 100644 new mode 100755 similarity index 76% rename from problems/generate-parentheses.py rename to problems/python/generate-parentheses.py index b2d5b70..cadbe96 --- a/problems/generate-parentheses.py +++ b/problems/python/generate-parentheses.py @@ -34,3 +34,20 @@ def helper(open_remain, close_remain, s): ans = [] helper(N, N, '') return ans + +""" +Time: O(2^N) +Space: O(N) +""" +class Solution(object): + def generateParenthesis(self, n): + def helper(curr, openCount, left, right): + if left==0 and right==0: ans.append(curr) + if left>0: + helper(curr+'(', openCount+1, left-1, right) + if right>0 and openCount>0: + helper(curr+')', openCount-1, left, right-1) + + ans = [] + helper('', 0, n, n) + return ans \ No newline at end of file diff --git a/problems/graph-valid-tree.py b/problems/python/graph-valid-tree.py old mode 100644 new mode 100755 similarity index 100% rename from problems/graph-valid-tree.py rename to problems/python/graph-valid-tree.py diff --git a/problems/greatest-sum-divisible-by-three.py b/problems/python/greatest-sum-divisible-by-three.py old mode 100644 new mode 100755 similarity index 100% rename from problems/greatest-sum-divisible-by-three.py rename to problems/python/greatest-sum-divisible-by-three.py diff --git a/problems/group-anagrams.py b/problems/python/group-anagrams.py old mode 100644 new mode 100755 similarity index 100% rename from problems/group-anagrams.py rename to problems/python/group-anagrams.py diff --git a/problems/python/group-shifted-strings.py b/problems/python/group-shifted-strings.py new file mode 100755 index 0000000..d5ac664 --- /dev/null +++ b/problems/python/group-shifted-strings.py @@ -0,0 +1,20 @@ +class Solution(object): + def groupStrings(self, strings): + def getHash(string): + h = '' + offset = getNumByChar(string[0])*-1 + for c in string: + h += getCharByNum((getNumByChar(c)+offset) if (getNumByChar(c)+offset)>=0 else 26+(getNumByChar(c)+offset)) + return h + + def getNumByChar(letter): + return ord(letter) - 97 + + def getCharByNum(pos): + return chr(pos + 97) + + groups = collections.defaultdict(list) + for string in strings: + h = getHash(string) + groups[h].append(string) + return groups.values() \ No newline at end of file diff --git a/problems/guess-number-higher-or-lower-ii.py b/problems/python/guess-number-higher-or-lower-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/guess-number-higher-or-lower-ii.py rename to problems/python/guess-number-higher-or-lower-ii.py diff --git a/problems/guess-number-higher-or-lower.py b/problems/python/guess-number-higher-or-lower.py old mode 100644 new mode 100755 similarity index 100% rename from problems/guess-number-higher-or-lower.py rename to problems/python/guess-number-higher-or-lower.py diff --git a/problems/python/guess-the-word.py b/problems/python/guess-the-word.py new file mode 100755 index 0000000..278a659 --- /dev/null +++ b/problems/python/guess-the-word.py @@ -0,0 +1,32 @@ +# """ +# This is Master's API interface. +# You should not implement it, or speculate about its implementation +# """ +#class Master(object): +# def guess(self, word): +# """ +# :type word: str +# :rtype int +# """ + +class Solution(object): + def findSecretWord(self, wordlist, master): + def similarity(w1, w2): + count = 0 + for i in xrange(6): + if w1[i]==w2[i]: count += 1 + return count + + for _ in xrange(10): + + temp = [] + + word = random.choice(wordlist) + k = master.guess(word) + if k==6: return + + for otherWord in wordlist: + if otherWord==word: continue + if similarity(word, otherWord)==k: temp.append(otherWord) + + wordlist = temp \ No newline at end of file diff --git a/problems/h-index-ii.py b/problems/python/h-index-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/h-index-ii.py rename to problems/python/h-index-ii.py diff --git a/problems/h-index.py b/problems/python/h-index.py old mode 100644 new mode 100755 similarity index 100% rename from problems/h-index.py rename to problems/python/h-index.py diff --git a/problems/hamming-distance.py b/problems/python/hamming-distance.py old mode 100644 new mode 100755 similarity index 100% rename from problems/hamming-distance.py rename to problems/python/hamming-distance.py diff --git a/problems/house-robber-ii.py b/problems/python/house-robber-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/house-robber-ii.py rename to problems/python/house-robber-ii.py diff --git a/problems/house-robber-iii.py b/problems/python/house-robber-iii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/house-robber-iii.py rename to problems/python/house-robber-iii.py diff --git a/problems/house-robber.py b/problems/python/house-robber.py old mode 100644 new mode 100755 similarity index 100% rename from problems/house-robber.py rename to problems/python/house-robber.py diff --git a/problems/python/implement-trie-prefix-tree.py b/problems/python/implement-trie-prefix-tree.py new file mode 100755 index 0000000..fd64dd7 --- /dev/null +++ b/problems/python/implement-trie-prefix-tree.py @@ -0,0 +1,46 @@ +""" +Time for insert() O(N), search() O(N), startsWith() O(N). N is the number of characters. +Space for insert() O(N), search() O(1), startsWith() O(1). N is the number of characters. + +Use . to represent the end of a string. +""" +class Node(object): + def __init__(self, char): + self.char = char + self.children = {} + +class Trie(object): + + def __init__(self): + self.period = '.' + self.root = Node('') + + + def insert(self, word): + word = word + self.period + node = self.root + + for c in word: + if c not in node.children: + node.children[c] = Node(c) + node = node.children[c] + + + def search(self, word): + word = word + self.period + node = self.root + + for c in word: + if c not in node.children: + return False + node = node.children[c] + return True + + + def startsWith(self, prefix): + node = self.root + for c in prefix: + if c not in node.children: + return False + node = node.children[c] + return True \ No newline at end of file diff --git a/problems/increasing-triplet-subsequence.py b/problems/python/increasing-triplet-subsequence.py old mode 100644 new mode 100755 similarity index 100% rename from problems/increasing-triplet-subsequence.py rename to problems/python/increasing-triplet-subsequence.py diff --git a/problems/python/inorder-successor-in-bst-ii.py b/problems/python/inorder-successor-in-bst-ii.py new file mode 100755 index 0000000..67440d7 --- /dev/null +++ b/problems/python/inorder-successor-in-bst-ii.py @@ -0,0 +1,21 @@ +""" +Time: O(H), since we are just moving vertically in the tree. +Space: O(1) + +First, check if the node has right child, if so, continue the traversal. +If the node doesn't have right child, this mean that the node and all its children are visited. +We need to move up. +While moving up a node, if a node itself is a right child, we need to keep moving up, since the node's parent is already vistied. +Keep moving up until the node itself is a left child. Return its parent. +""" +class Solution(object): + def inorderSuccessor(self, node): + if node.right: + node = node.right + while node.left: + node = node.left + return node + + while node.parent and node.parent.right==node: + node = node.parent + return node.parent \ No newline at end of file diff --git a/problems/python/inorder-successor-in-bst.py b/problems/python/inorder-successor-in-bst.py new file mode 100755 index 0000000..83ae4d2 --- /dev/null +++ b/problems/python/inorder-successor-in-bst.py @@ -0,0 +1,52 @@ +""" +Time: O(N) +Space: O(N) + +First, check if the p has right child, if it has a right child, we can continue the inorder traversal from it. +Second, if p does not have right child, we need to do the inorder traversal from the beginning. +During the traversal, do check if the prev is p. +If so, return the current node. +""" +class Solution(object): + def inorderSuccessor(self, root, p): + if p.right: + p = p.right + while p.left: + p = p.left + return p + else: + stack = [] + node = root + prev = None + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + + if prev==p: return node + + prev = node + node = node.right + return None + + +""" +Inorder Traversal Template. +""" +def inOrderTraverse(root): + stack = [] + node = root + + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + + #do something + print node.val + + node = node.right \ No newline at end of file diff --git a/problems/insert-interval.py b/problems/python/insert-interval.py old mode 100644 new mode 100755 similarity index 100% rename from problems/insert-interval.py rename to problems/python/insert-interval.py diff --git a/problems/insert-into-a-binary-search-tree.py b/problems/python/insert-into-a-binary-search-tree.py old mode 100644 new mode 100755 similarity index 50% rename from problems/insert-into-a-binary-search-tree.py rename to problems/python/insert-into-a-binary-search-tree.py index 2670175..9b3da5a --- a/problems/insert-into-a-binary-search-tree.py +++ b/problems/python/insert-into-a-binary-search-tree.py @@ -1,3 +1,7 @@ +""" +Time complexity: O(LogN) +Space complexity: O(1) +""" class Solution(object): def insertIntoBST(self, root, val): if not root: return TreeNode(val) @@ -19,8 +23,26 @@ def insertIntoBST(self, root, val): return None - """ Time complexity: O(LogN) -Space complexity: O(1) -""" \ No newline at end of file +Space complexity: O(LogN) +""" +class Solution(object): + def insertIntoBST(self, root, val): + def helper(node, val): + if not node: return + if val=tail.val or insertVal<=head.val: + tail.next = newNode + newNode.next = head + else: + curr = head + while not curr.val<=insertVal<=curr.next.val: curr = curr.next + temp = curr.next + curr.next = newNode + newNode.next = temp + + return random \ No newline at end of file diff --git a/problems/insertion-sort-list.py b/problems/python/insertion-sort-list.py old mode 100644 new mode 100755 similarity index 100% rename from problems/insertion-sort-list.py rename to problems/python/insertion-sort-list.py diff --git a/problems/python/integer-to-english-words.py b/problems/python/integer-to-english-words.py new file mode 100755 index 0000000..a4e8d88 --- /dev/null +++ b/problems/python/integer-to-english-words.py @@ -0,0 +1,28 @@ +class Solution(object): + def numberToWords(self, num): + underTwenty = ['', 'One', 'Two', 'Three', 'Four', 'Five', 'Six', 'Seven', 'Eight', 'Nine', 'Ten', 'Eleven', 'Twelve', 'Thirteen', 'Fourteen', 'Fifteen', 'Sixteen', 'Seventeen', 'Eighteen', 'Nineteen', ] + tens = ['', '', 'Twenty', 'Thirty', 'Forty', 'Fifty', 'Sixty', 'Seventy', 'Eighty', 'Ninety'] + + def toWords(n): + if n==0: + return [] + elif n<20: + return [underTwenty[n]] + elif n<100: + c, r = n/10, n%10 + return [tens[c]] + toWords(r) + elif n<1000: + c, r = n/100, n%100 + return toWords(c) + ['Hundred'] + toWords(r) + else: + if n>=1000000000: + c, r = n/1000000000, n%1000000000 + return toWords(c) + ['Billion'] + toWords(r) + elif n>=1000000: + c, r = n/1000000, n%1000000 + return toWords(c) + ['Million'] + toWords(r) + elif n>=1000: + c, r = n/1000, n%1000 + return toWords(c) + ['Thousand'] + toWords(r) + + return ' '.join(toWords(num)) or 'Zero' \ No newline at end of file diff --git a/problems/interleaving-string.py b/problems/python/interleaving-string.py old mode 100644 new mode 100755 similarity index 100% rename from problems/interleaving-string.py rename to problems/python/interleaving-string.py diff --git a/problems/intersection-of-two-arrays-ii.py b/problems/python/intersection-of-two-arrays-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/intersection-of-two-arrays-ii.py rename to problems/python/intersection-of-two-arrays-ii.py diff --git a/problems/intersection-of-two-arrays.py b/problems/python/intersection-of-two-arrays.py old mode 100644 new mode 100755 similarity index 100% rename from problems/intersection-of-two-arrays.py rename to problems/python/intersection-of-two-arrays.py diff --git a/problems/python/invert-binary-tree.py b/problems/python/invert-binary-tree.py new file mode 100755 index 0000000..5dc3ec9 --- /dev/null +++ b/problems/python/invert-binary-tree.py @@ -0,0 +1,16 @@ +""" +Time: O(N) +Time: O(N) +""" +class Solution(object): + def invertTree(self, root): + if not root: return root + q = collections.deque([root]) + + while q: + node = q.popleft() + node.left, node.right = node.right, node.left + if node.left: q.append(node.left) + if node.right: q.append(node.right) + + return root \ No newline at end of file diff --git a/problems/is-graph-bipartite.py b/problems/python/is-graph-bipartite.py old mode 100644 new mode 100755 similarity index 100% rename from problems/is-graph-bipartite.py rename to problems/python/is-graph-bipartite.py diff --git a/problems/is-subsequence.py b/problems/python/is-subsequence.py old mode 100644 new mode 100755 similarity index 100% rename from problems/is-subsequence.py rename to problems/python/is-subsequence.py diff --git a/problems/isomorphic-strings.py b/problems/python/isomorphic-strings.py old mode 100644 new mode 100755 similarity index 100% rename from problems/isomorphic-strings.py rename to problems/python/isomorphic-strings.py diff --git a/problems/jewels-and-stones.py b/problems/python/jewels-and-stones.py old mode 100644 new mode 100755 similarity index 100% rename from problems/jewels-and-stones.py rename to problems/python/jewels-and-stones.py diff --git a/problems/jump-game.py b/problems/python/jump-game.py old mode 100644 new mode 100755 similarity index 100% rename from problems/jump-game.py rename to problems/python/jump-game.py diff --git a/problems/python/k-closest-points-to-origin.py b/problems/python/k-closest-points-to-origin.py new file mode 100755 index 0000000..1f536c1 --- /dev/null +++ b/problems/python/k-closest-points-to-origin.py @@ -0,0 +1,131 @@ +class Solution(object): + def kClosest(self, points, k): + h = [] + + for x, y in points: + d = (x**2+y**2)**0.5 + if len(h)>=k and -h[0][0]>d: + heapq.heappop(h) + heapq.heappush(h, (-d, x, y)) + elif len(h)>=k and -h[0][0]<=d: + pass + else: + heapq.heappush(h, (-d, x, y)) + + return [(x, y) for _, x, y in h] + + +""" +1. Process `points` into `distances`. + +2. Binary search the "distance". For every distance: +Split the elements in `distances` by distance +Put the ones smaller to smaller. +Put the ones larger to larger. +If len(smaller)<=k, then all the elements in the smaller must belong to the `ans`, do the same thing to the larger. +Else we ignore the larger and do the same thing to the smaller again. + +Time: O(N) +The binary search range in average is N, N/2, N/4... = 2N +Space: O(N) +""" +class Solution(object): + def kClosest(self, points, K): + #[1] + maxD = float('-inf') + minD = float('inf') + distances = [] + for x, y in points: + distance = ((x**2+y**2)**0.5) + distances.append((distance, x, y)) + maxD = max(maxD, distance) + minD = min(minD, distance) + + #[2] + ans = [] + smaller = [] + larger = [] + while K>0: + #split distances into smaller and larger + distance = (maxD+minD)/2 + for d, x, y in distances: + if d<=distance: + smaller.append((d, x, y)) + else: + larger.append((d, x, y)) + + if len(smaller)<=K: + ans += smaller + K -= len(smaller) + distances = larger + minD = distance + larger = [] + smaller = [] + else: + distances = smaller + maxD = distance + larger = [] + smaller = [] + + return [(x, y) for _, x, y in ans] + + + +""" +Quick Select +Time: O(N) +Space: O(N), can be optimize to O(1). +""" +class Solution(object): + def kClosest(self, points, K): + """ + Start State: + i = s #next element after "SSS"s + t = s #unprocessed elements + j = e #next element after "LLLL"s + + SSSPP?????LLL + i t j + + End State: + SSSPPPPPLLLLL + i jt + """ + def quickSelect(distances, s, e, k): + + pivot = distances[(s+e)/2][0] + i = s + j = e + t = s + + while t<=j: + if pivotdistances[t][0]: + distances[t], distances[i] = distances[i], distances[t] + i += 1 + t += 1 + else: + t += 1 + + if i-s>=k: + return quickSelect(distances, s, i, k) + elif j-s+1>=k: + return pivot + else: + return quickSelect(distances, t, e, k-(t-s)) + + distances = [] + for x, y in points: + distance = ((x**2+y**2)**0.5) + distances.append((distance, x, y)) + + kthSmallestDistance = quickSelect(distances, 0, len(distances)-1, K) + + ans = [] + for d, x, y in distances: + if len(ans)==K: break + if d<=kthSmallestDistance: ans.append((x, y)) + + return ans \ No newline at end of file diff --git a/problems/k-empty-slots.py b/problems/python/k-empty-slots.py old mode 100644 new mode 100755 similarity index 100% rename from problems/k-empty-slots.py rename to problems/python/k-empty-slots.py diff --git a/problems/keys-and-rooms.py b/problems/python/keys-and-rooms.py old mode 100644 new mode 100755 similarity index 100% rename from problems/keys-and-rooms.py rename to problems/python/keys-and-rooms.py diff --git a/problems/knight-dialer.py b/problems/python/knight-dialer.py old mode 100644 new mode 100755 similarity index 100% rename from problems/knight-dialer.py rename to problems/python/knight-dialer.py diff --git a/problems/knight-probability-in-chessboard.py b/problems/python/knight-probability-in-chessboard.py old mode 100644 new mode 100755 similarity index 100% rename from problems/knight-probability-in-chessboard.py rename to problems/python/knight-probability-in-chessboard.py diff --git a/problems/koko-eating-bananas.py b/problems/python/koko-eating-bananas.py old mode 100644 new mode 100755 similarity index 100% rename from problems/koko-eating-bananas.py rename to problems/python/koko-eating-bananas.py diff --git a/problems/kth-largest-element-in-an-array.py b/problems/python/kth-largest-element-in-an-array.py old mode 100644 new mode 100755 similarity index 64% rename from problems/kth-largest-element-in-an-array.py rename to problems/python/kth-largest-element-in-an-array.py index d12f59c..27c4b72 --- a/problems/kth-largest-element-in-an-array.py +++ b/problems/python/kth-largest-element-in-an-array.py @@ -63,4 +63,34 @@ def partition(A, l, r, p): k = len(nums)-k #redefine the problem to find the kth nums when sorted sortRange(nums, 0, len(nums)-1) - return nums[k] \ No newline at end of file + return nums[k] + + +#Quick Select +class Solution(object): + def findKthLargest(self, nums, K): + def quickSelect(A, s, e, K): + pivot = A[(s+e)/2] + i = s + t = s + j = e + + while t<=j: + if A[t]=K: + return quickSelect(A, j+1, e, K) + elif e-i+1>=K: + return pivot + else: + return quickSelect(A, s, i-1, K-(e-(i-1))) + + return quickSelect(nums, 0, len(nums)-1, K) \ No newline at end of file diff --git a/problems/python/kth-smallest-element-in-a-bst.py b/problems/python/kth-smallest-element-in-a-bst.py new file mode 100755 index 0000000..56c355b --- /dev/null +++ b/problems/python/kth-smallest-element-in-a-bst.py @@ -0,0 +1,89 @@ +class Solution(object): + def kthSmallest(self, root, k): + count = 0 + stack = [] + node = root + + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + count += 1 + + if count==k: return node.val + + node = node.right + + return 0 + +""" +Time complexity: O(N). Because we use inorder traversal. +Space complexity: O(N). For stack may contains all the nodes. +""" + +""" +For follow up question. +We might need to keep a sorted list of node. +Every time we inset and delete. We need to update the list. Taking up O(N) of time. +(Originally it was O(LogN) for insert and delete) +But to find the kth smallest element will only takes O(1). +""" + + + + + + + + + + + + + + + +""" +Template for in-order traverse iteratively. +""" +def inOrderTraverse(root): + stack = [] + node = root + + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + + #do something + print node.val + + node = node.right + +""" +Time complexity: O(N). Because we use inorder traversal. +Space complexity: O(N). For stack may contains all the nodes. + +Use the in-order traverse to find the kth smallest element. +""" +class Solution(object): + def kthSmallest(self, root, k): + stack = [] + node = root + + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + + k -= 1 + if k==0: return node.val + + node = node.right + return node.val \ No newline at end of file diff --git a/problems/kth-smallest-element-in-a-sorted-matrix.py b/problems/python/kth-smallest-element-in-a-sorted-matrix.py old mode 100644 new mode 100755 similarity index 100% rename from problems/kth-smallest-element-in-a-sorted-matrix.py rename to problems/python/kth-smallest-element-in-a-sorted-matrix.py diff --git a/problems/largest-1-bordered-square.py b/problems/python/largest-1-bordered-square.py old mode 100644 new mode 100755 similarity index 100% rename from problems/largest-1-bordered-square.py rename to problems/python/largest-1-bordered-square.py diff --git a/problems/python/largest-bst-subtree.py b/problems/python/largest-bst-subtree.py new file mode 100755 index 0000000..82afb93 --- /dev/null +++ b/problems/python/largest-bst-subtree.py @@ -0,0 +1,22 @@ +""" +Time: O(N) for recursively traverse each node once. +Space: O(LogN) for recursive stack. +""" +class Solution(object): + def largestBSTSubtree(self, root): + def helper(node, minVal, maxVal): + if not node: return True, 0, float('-inf'), float('inf') + if not node.left and not node.right: return True, 1, node.val, node.val + + isLeftBST, leftSize, leftMin, leftMax = helper(node.left, minVal, node.val) + isRightBST, rightSize, rightMin, rightMax = helper(node.right, node.val, maxVal) + + currMin = min(leftMin, rightMin, node.val) + currMax = max(leftMax, rightMax, node.val) + + if isLeftBST and isRightBST and leftMaxk, O(N+NLogN+KLogN) ~= O(NLogN) + +Space: O(N) +""" +class Solution(object): + def findLeastNumOfUniqueInts(self, nums, k): + counter = collections.Counter(nums) + h = [] + + for num in counter: + heapq.heappush(h, (counter[num], num)) + + for _ in xrange(k): + count, num = heapq.heappop(h) + count -= 1 + if count>0: heapq.heappush(h, (count, num)) + + return len(h) \ No newline at end of file diff --git a/problems/letter-case-permutation.py b/problems/python/letter-case-permutation.py old mode 100644 new mode 100755 similarity index 100% rename from problems/letter-case-permutation.py rename to problems/python/letter-case-permutation.py diff --git a/problems/letter-combinations-of-a-phone-number.py b/problems/python/letter-combinations-of-a-phone-number.py old mode 100644 new mode 100755 similarity index 100% rename from problems/letter-combinations-of-a-phone-number.py rename to problems/python/letter-combinations-of-a-phone-number.py diff --git a/problems/license-key-formatting.py b/problems/python/license-key-formatting.py old mode 100644 new mode 100755 similarity index 100% rename from problems/license-key-formatting.py rename to problems/python/license-key-formatting.py diff --git a/problems/linked-list-cycle-ii.py b/problems/python/linked-list-cycle-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/linked-list-cycle-ii.py rename to problems/python/linked-list-cycle-ii.py diff --git a/problems/linked-list-cycle.py b/problems/python/linked-list-cycle.py old mode 100644 new mode 100755 similarity index 100% rename from problems/linked-list-cycle.py rename to problems/python/linked-list-cycle.py diff --git a/problems/linked-list-random-node.py b/problems/python/linked-list-random-node.py old mode 100644 new mode 100755 similarity index 100% rename from problems/linked-list-random-node.py rename to problems/python/linked-list-random-node.py diff --git a/problems/python/logger-rate-limiter.py b/problems/python/logger-rate-limiter.py new file mode 100755 index 0000000..0f5f7ad --- /dev/null +++ b/problems/python/logger-rate-limiter.py @@ -0,0 +1,32 @@ +class Logger(object): + def __init__(self): + self.log = collections.Counter() #store the latest timestamp + + def shouldPrintMessage(self, timestamp, message): + if message not in self.log or self.log[message]+10<=timestamp: + self.log[message] = timestamp + return True + else: + return False + + + +class Logger(object): + + def __init__(self): + #stores the messages within 10 seconds + self.q = collections.deque() + self.set = set() + + + def shouldPrintMessage(self, timestamp, message): + while self.q and timestamp-self.q[0][0]>=10: + time, msg = self.q.popleft() + self.set.remove(msg) + + if message not in self.set: + self.q.append((timestamp, message)) + self.set.add(message) + return True + else: + return False \ No newline at end of file diff --git a/problems/longest-common-prefix.py b/problems/python/longest-common-prefix.py old mode 100644 new mode 100755 similarity index 100% rename from problems/longest-common-prefix.py rename to problems/python/longest-common-prefix.py diff --git a/problems/longest-common-subsequence.py b/problems/python/longest-common-subsequence.py old mode 100644 new mode 100755 similarity index 100% rename from problems/longest-common-subsequence.py rename to problems/python/longest-common-subsequence.py diff --git a/problems/longest-consecutive-sequence.py b/problems/python/longest-consecutive-sequence.py old mode 100644 new mode 100755 similarity index 100% rename from problems/longest-consecutive-sequence.py rename to problems/python/longest-consecutive-sequence.py diff --git a/problems/python/longest-increasing-path-in-a-matrix.py b/problems/python/longest-increasing-path-in-a-matrix.py new file mode 100755 index 0000000..cd446d9 --- /dev/null +++ b/problems/python/longest-increasing-path-in-a-matrix.py @@ -0,0 +1,31 @@ +class Solution(object): + def longestIncreasingPath(self, matrix): + def getLongest(i, j): + if (i, j) in longest: return longest[(i, j)] + l = 1 + + #call getLongest to the neighbors that are larger than itself. + if i+1=0 and matrix[i][j]=0 and matrix[i][j]k: + counter[s[l]] -= 1 + l += 1 + + ans = max(ans, r-l+1) + return ans \ No newline at end of file diff --git a/problems/longest-string-chain.py b/problems/python/longest-string-chain.py old mode 100644 new mode 100755 similarity index 62% rename from problems/longest-string-chain.py rename to problems/python/longest-string-chain.py index 79cb2f8..a21ef17 --- a/problems/longest-string-chain.py +++ b/problems/python/longest-string-chain.py @@ -66,3 +66,42 @@ def shortestPredecessor(word): return ans +class Solution(object): + def longestStrChain(self, words): + def getLongestStringChain(word): + if word in history: return history[word] + if not orderedWords[len(word)+1]: return 1 + + temp = 0 + for word2 in orderedWords[len(word)+1]: + if isPredecessor(word2, word): + temp = max(temp, getLongestStringChain(word2)) + + history[word] = temp+1 + return temp+1 + + def isPredecessor(w2, w1): + if len(w2)!=len(w1)+1: return False + + j = 0 + for c in w1: + found = False + while jq.val and node.val>p.val: + node = node.left + elif node.val=2 and not self.ans: self.ans = node + return count + + dfs(root) + return self.ans diff --git a/problems/python/lowest-common-ancestor-of-a-binary-tree-iii.py b/problems/python/lowest-common-ancestor-of-a-binary-tree-iii.py new file mode 100755 index 0000000..cfd14f7 --- /dev/null +++ b/problems/python/lowest-common-ancestor-of-a-binary-tree-iii.py @@ -0,0 +1,56 @@ +""" +Time: O(H), H is the height of the tree. +Space: O(1) +""" +class Solution(object): + def lowestCommonAncestor(self, p, q): + ancestorP = set() + ancestorQ = set() + + temp = p + while temp: + ancestorP.add(temp) + temp = temp.parent + + temp = q + while temp: + ancestorQ.add(temp) + temp = temp.parent + + commonAncestor = ancestorQ.intersection(ancestorP) + temp = q + while temp: + if temp in commonAncestor: return temp + temp = temp.parent + return None + + +""" +Time: O(LogN) +Space: O(LogN) + +Looking from backward, parents1 and parents2 will be the same at first, since they must have a common ancestor. +Find the last the same parents. +""" +class Solution(object): + def lowestCommonAncestor(self, p, q): + parents1 = [] + parents2 = [] + + curr = p + while curr: + parents1.append(curr) + curr = curr.parent + + curr = q + while curr: + parents2.append(curr) + curr = curr.parent + + i = len(parents1)-1 + j = len(parents2)-1 + while i>=0 and j>=0 and parents1[i]==parents2[j]: + i -= 1 + j -= 1 + return parents1[i+1] + diff --git a/problems/python/lowest-common-ancestor-of-a-binary-tree-iv.py b/problems/python/lowest-common-ancestor-of-a-binary-tree-iv.py new file mode 100755 index 0000000..f7458a3 --- /dev/null +++ b/problems/python/lowest-common-ancestor-of-a-binary-tree-iv.py @@ -0,0 +1,18 @@ +class Solution(object): + def __init__(self): + self.ans = None + + def lowestCommonAncestor(self, root, nodes): + def dfs(node): + if not node: return 0 + + count = 0 + if node in nodes: count += 1 + count += dfs(node.left) + count += dfs(node.right) + if count>=len(nodes) and not self.ans: self.ans = node + return count + + nodes = set(nodes) + dfs(root) + return self.ans \ No newline at end of file diff --git a/problems/lowest-common-ancestor-of-a-binary-tree.py b/problems/python/lowest-common-ancestor-of-a-binary-tree.py old mode 100644 new mode 100755 similarity index 88% rename from problems/lowest-common-ancestor-of-a-binary-tree.py rename to problems/python/lowest-common-ancestor-of-a-binary-tree.py index fdbabd3..54546a3 --- a/problems/lowest-common-ancestor-of-a-binary-tree.py +++ b/problems/python/lowest-common-ancestor-of-a-binary-tree.py @@ -129,4 +129,25 @@ def find_lowest_common(a1, a2): find_ancestors() is O(LogN). find_lowest_common is O(LogN). Space complexity is O(N), since `genealogy` may carry all the nodes. -""" \ No newline at end of file +""" + + +# 2021/9/25 +class Solution(object): + def lowestCommonAncestor(self, root, p, q): + def __init__(self): + self.ans = None + + def helper(node): + + if not node: return False + mid = node==q or node==p + left = helper(node.left) + right = helper(node.right) + + if (mid and left) or (mid and right) or (left and right): self.ans = node + + return mid or left or right + + helper(root) + return self.ans \ No newline at end of file diff --git a/problems/python/lowest-common-ancestor-of-deepest-leaves.py b/problems/python/lowest-common-ancestor-of-deepest-leaves.py new file mode 100755 index 0000000..cffade3 --- /dev/null +++ b/problems/python/lowest-common-ancestor-of-deepest-leaves.py @@ -0,0 +1,41 @@ +class Solution(object): + def __init__(self): + self.ans = None + + def lcaDeepestLeaves(self, root): + def checkCount(node, deepestNode): + if not node: + count = 0 + if count==len(deepestNode) and not self.ans: self.ans = node + return 0 + + if node in deepestNode: + count = 1 + if count==len(deepestNode) and not self.ans: self.ans = node + return count + + leftCount = checkCount(node.left, deepestNode) + rightCount = checkCount(node.right, deepestNode) + + if leftCount+rightCount==len(deepestNode) and not self.ans: self.ans = node + return leftCount+rightCount + + q = collections.deque([(root, 0)]) + q2 = collections.deque() + deepestNode = set([node for node, h in q]) + + while q: + node, d = q.popleft() + if node.left: q2.append((node.left, d+1)) + if node.right: q2.append((node.right, d+1)) + if not q: + q = q2 + if q: deepestNode = set([node for node, h in q]) + q2 = collections.deque() + + checkCount(root, deepestNode) + return self.ans + + + + \ No newline at end of file diff --git a/problems/lru-cache.py b/problems/python/lru-cache.py old mode 100644 new mode 100755 similarity index 100% rename from problems/lru-cache.py rename to problems/python/lru-cache.py diff --git a/problems/majority-element-ii.py b/problems/python/majority-element-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/majority-element-ii.py rename to problems/python/majority-element-ii.py diff --git a/problems/majority-element.py b/problems/python/majority-element.py old mode 100644 new mode 100755 similarity index 100% rename from problems/majority-element.py rename to problems/python/majority-element.py diff --git a/problems/python/making-a-large-island.py b/problems/python/making-a-large-island.py new file mode 100755 index 0000000..215aa2a --- /dev/null +++ b/problems/python/making-a-large-island.py @@ -0,0 +1,51 @@ +""" +For each "island" asign them a group id. Also calculate the group's size. +Iterate all the zeros, update the ans. + +Time:O(MN) +Space: O(MN) in the worst case. +""" +class Solution(object): + def largestIsland(self, grid): + def isValid(i, j, M, N): + return 0<=i1: + neiGroupId.add(grid[iNext][jNext]) + + for groupId in list(neiGroupId): + neiSize += groupIdToSize[groupId] + + ans = max(ans, 1+neiSize) + + return ans \ No newline at end of file diff --git a/problems/max-area-of-island.py b/problems/python/max-area-of-island.py old mode 100644 new mode 100755 similarity index 100% rename from problems/max-area-of-island.py rename to problems/python/max-area-of-island.py diff --git a/problems/python/max-consecutive-ones-iii.py b/problems/python/max-consecutive-ones-iii.py new file mode 100755 index 0000000..2ebd05c --- /dev/null +++ b/problems/python/max-consecutive-ones-iii.py @@ -0,0 +1,15 @@ +class Solution(object): + def longestOnes(self, nums, k): + ans = 0 + zeroCount = 0 + i = 0 + + for j, num in enumerate(nums): + if num==0: zeroCount += 1 + + while zeroCount>k: + if nums[i]==0: zeroCount -= 1 + i += 1 + ans = max(ans, j-i+1) + + return ans \ No newline at end of file diff --git a/problems/max-stack.py b/problems/python/max-stack.py old mode 100644 new mode 100755 similarity index 100% rename from problems/max-stack.py rename to problems/python/max-stack.py diff --git a/problems/max-sum-of-rectangle-no-larger-than-k.py b/problems/python/max-sum-of-rectangle-no-larger-than-k.py old mode 100644 new mode 100755 similarity index 100% rename from problems/max-sum-of-rectangle-no-larger-than-k.py rename to problems/python/max-sum-of-rectangle-no-larger-than-k.py diff --git a/problems/python/maximal-square.py b/problems/python/maximal-square.py new file mode 100755 index 0000000..9e6711b --- /dev/null +++ b/problems/python/maximal-square.py @@ -0,0 +1,33 @@ +class Solution(object): + def maximalSquare(self, grid): + if not grid or not grid[0]: return + M, N = len(grid), len(grid[0]) + + dp = [[0 for _ in xrange(N+1)] for _ in xrange(M+1)] + ans = 0 + for i in xrange(M): + for j in xrange(N): + if grid[i][j]=='1': + dp[i+1][j+1] = min(dp[i][j], dp[i][j+1], dp[i+1][j])+1 + ans = max(ans, dp[i+1][j+1]) + return ans**2 + + + +class Solution(object): + def maximalSquare(self, matrix): + """ + dp[i][j] := maximal square length with matrix[i-1][j-1] at the bottom right corner in the square + """ + ans = 0 + N = len(matrix) + M = len(matrix[0]) + + dp = [[0 for _ in xrange(M+1)] for _ in xrange(N+1)] + + for i in xrange(1, N+1): + for j in xrange(1, M+1): + if matrix[i-1][j-1]=='1': + dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1])+1 + ans = max(ans, dp[i][j]**2) + return ans \ No newline at end of file diff --git a/problems/python/maximum-average-subtree.py b/problems/python/maximum-average-subtree.py new file mode 100755 index 0000000..4cef9b5 --- /dev/null +++ b/problems/python/maximum-average-subtree.py @@ -0,0 +1,25 @@ +""" +Recursivly get the average and count of each node. +Update the self.ans before returning. + +Time: O(N), N is the number of nodes. +Space: O(LogN) for the recursive stack, since the tree hight is LogN if the tree is balanced. +""" +class Solution(object): + def __init__(self): + self.ans = float('-inf') + + def maximumAverageSubtree(self, root): + def getAVGAndCount(node): + if not node: return 0, 0 + leftAvg, leftCount = getAVGAndCount(node.left) + rightAvg, rightCount = getAVGAndCount(node.right) + + count = leftCount+rightCount+1 + avg = (leftAvg*leftCount + rightAvg*rightCount + node.val)/float(count) + + self.ans = max(self.ans, avg) + return avg, count + + getAVGAndCount(root) + return self.ans \ No newline at end of file diff --git a/problems/python/maximum-compatibility-score-sum.py b/problems/python/maximum-compatibility-score-sum.py new file mode 100755 index 0000000..a0f744c --- /dev/null +++ b/problems/python/maximum-compatibility-score-sum.py @@ -0,0 +1,44 @@ +""" +Time: O(ELogE), E is the edge of the graph. +Note that state[i] means if the ith student is matched or not. (for example M=4, 0000, 0010, 0111, 1111...) +2^M is the number of states. So in this case E will be 2^M x M. + +Space: O(2^M) +""" +class Solution(object): + def maxCompatibilitySum(self, students, mentors): + M = len(students) + N = len(students[0]) + + #initialize reverseScores + reverseScores = [[0]*M for _ in xrange(M)] + for i in xrange(M): + for j in xrange(M): + reverseScore = 0 + for k in xrange(N): + if students[i][k]!=mentors[j][k]: + reverseScore += 1 + reverseScores[i][j] = reverseScore + + #Dijkstra + startState = '0'*M + endState = '1'*M + visited = set() + pq = [(0, startState)] + + while pq: + cost, state = heapq.heappop(pq) + if state in visited: continue + visited.add(state) + + if state==endState: return M*N-cost + + j = state.count('1') + for i in xrange(M): + if state[i]=='1': continue + + nextState = state[:i]+'1'+state[i+1:] + if nextState in visited: continue + heapq.heappush(pq, (cost+reverseScores[i][j], nextState)) + + return -1 \ No newline at end of file diff --git a/problems/maximum-depth-of-binary-tree.py b/problems/python/maximum-depth-of-binary-tree.py old mode 100644 new mode 100755 similarity index 58% rename from problems/maximum-depth-of-binary-tree.py rename to problems/python/maximum-depth-of-binary-tree.py index 020d153..f114251 --- a/problems/maximum-depth-of-binary-tree.py +++ b/problems/python/maximum-depth-of-binary-tree.py @@ -18,4 +18,28 @@ def maxDepth(self, root): if node.right: stack.append((node.right, depth+1)) return max_depth - \ No newline at end of file + + +""" +Time: O(N) +Space: O(N) + +Standard BFS. +""" +class Solution(object): + def maxDepth(self, root): + if not root: return 0 + + ans = float('-inf') + q = collections.deque([(root, 1)]) + node = root + + while q: + node, d = q.popleft() + + ans = max(ans, d) + + if node.left: q.append((node.left, d+1)) + if node.right: q.append((node.right, d+1)) + + return ans \ No newline at end of file diff --git a/problems/maximum-gap.py b/problems/python/maximum-gap.py old mode 100644 new mode 100755 similarity index 100% rename from problems/maximum-gap.py rename to problems/python/maximum-gap.py diff --git a/problems/maximum-length-of-repeated-subarray.py b/problems/python/maximum-length-of-repeated-subarray.py old mode 100644 new mode 100755 similarity index 100% rename from problems/maximum-length-of-repeated-subarray.py rename to problems/python/maximum-length-of-repeated-subarray.py diff --git a/problems/python/maximum-number-of-events-that-can-be-attended.py b/problems/python/maximum-number-of-events-that-can-be-attended.py new file mode 100755 index 0000000..087489d --- /dev/null +++ b/problems/python/maximum-number-of-events-that-can-be-attended.py @@ -0,0 +1,25 @@ +""" +Time: O(NLogN). Sort the event takes O(NLogN). Each event will get push in and pop out the heap: O(NLogN) +Space: O(N) +""" +class Solution(object): + def maxEvents(self, A): + d = 0 + count = 0 + h = [] #a heap. store the started event + A.sort(reverse=True) + + #for each day, attend the event with smallest endtime, so we can have the most free time in the future. + while A or h: + if not h: d = A[-1][0] + + while A and A[-1][0]<=d: + heapq.heappush(h, A.pop()[1]) + + heapq.heappop(h) #attend the event with smallest endtime + count += 1 + d += 1 + + while h and h[0]=k (Part 1) +points[i][j]+max([dp[i-1][k] - (j-k) for k in xrange(M)]) +points[i][j] - j + max([dp[i-1][k] + k) for k in xrange(M)]) + +if k>=j (Part 2) +points[i][j] + max([dp[i-1][k] - (k-j) for k in xrange(M)]) +points[i][j] + j + max([dp[i-1][k] - k) for k in xrange(M)]) + +Since we cannot do a full scan +why not we update the value from left to right for Part 1 and +right to left for part 2 +With a variable call rollingMax to store the max. + +That way dp[i][j] will be updated as if we do a full scan. + +The time complexity will become O(NM) +""" +class Solution(object): + def maxPoints(self, points): + N = len(points) + M = len(points[0]) + dp = [[0]*M for _ in xrange(N)] + + for j in xrange(M): + dp[0][j] = points[0][j] + + for i in xrange(1, N): + rollingMax = float('-inf') + for j in xrange(M): + rollingMax = max(rollingMax, dp[i-1][j] - j) + dp[i][j] = max(dp[i][j], points[i][j] + j + rollingMax)) + + rollingMax = float('-inf') + for j in xrange(M, -1, -1): + rollingMax = max(rollingMax, dp[i-1][j] + j) + dp[i][j] = max(dp[i][j], points[i][j] - j + rollingMax)) + + return max(dp[N-1]) \ No newline at end of file diff --git a/problems/python/maximum-number-of-visible-points.py b/problems/python/maximum-number-of-visible-points.py new file mode 100755 index 0000000..f247156 --- /dev/null +++ b/problems/python/maximum-number-of-visible-points.py @@ -0,0 +1,54 @@ +""" +Time: O(NLogN), N is the number of points. +Space: O(N) + +1. Get the angle of each point relative to the "location" +2. Sort the angles +3. Do a sliding window to angles to see what the maximum number of angles within the interval "angle" + +[1] For example, angles = [10, 20, 360], angle = 20 we will count 2, but actually it will be 3. +""" +class Solution(object): + def visiblePoints(self, points, angle, location): + def getAngle(x, y): + #4 axis + if x>0 and y==0: + return 0 + elif x==0 and y>0: + return 90 + elif x<0 and y==0: + return 180 + elif x==0 and y<0: + return 270 + + #4 quadrant + if x>0 and y>0: + return math.degrees(math.atan2(abs(y), abs(x))) + elif x<0 and y>0: + return 180-math.degrees(math.atan2(abs(y), abs(x))) + elif x<0 and y<0: + return 180+math.degrees(math.atan2(abs(y), abs(x))) + else: + return 360-math.degrees(math.atan2(abs(y), abs(x))) + + ans = 0 + onLocation = 0 + angles = [] + + for x, y in points: + if x==location[0] and y==location[1]: + onLocation += 1 + else: + a = getAngle(x-location[0], y-location[1]) + angles.append(a) + if a<=angle: angles.append(360+a) #[1] + + angles.sort() + + i = 0 + for j in xrange(len(angles)): + while angles[j]-angles[i]>angle: + i += 1 + ans = max(ans, j-i+1) + + return ans+onLocation \ No newline at end of file diff --git a/problems/maximum-product-of-three-numbers.py b/problems/python/maximum-product-of-three-numbers.py old mode 100644 new mode 100755 similarity index 100% rename from problems/maximum-product-of-three-numbers.py rename to problems/python/maximum-product-of-three-numbers.py diff --git a/problems/python/maximum-product-subarray.py b/problems/python/maximum-product-subarray.py new file mode 100755 index 0000000..b48bf12 --- /dev/null +++ b/problems/python/maximum-product-subarray.py @@ -0,0 +1,57 @@ +""" +Time: O(N) +Space: O(N) + +dp[i][0] := the maximum subarray product that includes nums[i] +dp[i][1] := the minimum subarray product that includes nums[i] +""" +class Solution(object): + def maxProduct(self, nums): + if not nums: return 0 + dp = [[float('-inf'), float('inf')] for _ in xrange(len(nums))] + + dp[0] = [nums[0], nums[0]] + ans = nums[0] + + for i in xrange(1, len(nums)): + if nums[i]==0: + dp[i][0] = 0 + dp[i][1] = 0 + elif nums[i]>0: + dp[i][0] = dp[i-1][0]*nums[i] if dp[i-1][0]>0 else nums[i] + dp[i][1] = dp[i-1][1]*nums[i] if dp[i-1][1]<=0 else nums[i] + else: + dp[i][0] = dp[i-1][1]*nums[i] if dp[i-1][1]<=0 else nums[i] + dp[i][1] = dp[i-1][0]*nums[i] if dp[i-1][0]>0 else nums[i] + + ans = max(ans, dp[i][0]) + + return ans + +""" +The above solution can further reduce the space complexity to O(1). +""" +class Solution(object): + def maxProduct(self, nums): + if not nums: return 0 + + maxLast = nums[0] + minLast = nums[0] + ans = nums[0] + + for i in xrange(1, len(nums)): + if nums[i]==0: + newMax = 0 + newMin = 0 + elif nums[i]>0: + newMax = maxLast*nums[i] if maxLast>0 else nums[i] + newMin = minLast*nums[i] if minLast<=0 else nums[i] + else: + newMax = minLast*nums[i] if minLast<=0 else nums[i] + newMin = maxLast*nums[i] if maxLast>0 else nums[i] + + maxLast = newMax + minLast = newMin + ans = max(ans, maxLast) + + return ans \ No newline at end of file diff --git a/problems/maximum-subarray-sum-with-one-deletion.py b/problems/python/maximum-subarray-sum-with-one-deletion.py old mode 100644 new mode 100755 similarity index 92% rename from problems/maximum-subarray-sum-with-one-deletion.py rename to problems/python/maximum-subarray-sum-with-one-deletion.py index d713751..a7b34f2 --- a/problems/maximum-subarray-sum-with-one-deletion.py +++ b/problems/python/maximum-subarray-sum-with-one-deletion.py @@ -23,7 +23,7 @@ def maximumSum(self, arr): dp[i][1] = arr[i] else: dp[i][0] = max(dp[i-1][0]+arr[i], arr[i]) - dp[i][1] = max(dp[i-1][0], dp[i-1][1]+arr[i], arr[i]) + dp[i][1] = max(dp[i-1][0], dp[i-1][1]+arr[i]) subarrayMaxSum = max(subarrayMaxSum, dp[i][0], dp[i][1]) return subarrayMaxSum \ No newline at end of file diff --git a/problems/maximum-subarray.py b/problems/python/maximum-subarray.py old mode 100644 new mode 100755 similarity index 100% rename from problems/maximum-subarray.py rename to problems/python/maximum-subarray.py diff --git a/problems/python/maximum-swap.py b/problems/python/maximum-swap.py new file mode 100755 index 0000000..ffca66a --- /dev/null +++ b/problems/python/maximum-swap.py @@ -0,0 +1,23 @@ +""" +1. Generate positions. Storing the mapping between number to indices. +2. Iterate from left, for each n1, find the largest number larger than n1 (searching from 9, 8, 7 to n1+1). +3. Since we need to find the max output. There might be multiple the same number, we need to find the index of the rightest number. +4. Remove n1 when it is done. Because right of the n1 should not consider it anymore. +""" +class Solution(object): + def maximumSwap(self, num): + numList = [int(n) for n in str(num)] + positions = collections.defaultdict(list) + for i, n in enumerate(numList): positions[n].append(i) #[1] + + i = 0 + for i, n1 in enumerate(numList): #[2] + n1 = numList[i] + for n2 in xrange(9, n1, -1): + if n2 in positions and len(positions[n2])>0: + j = positions[n2][-1] #[3] + numList[i], numList[j] = numList[j], numList[i] + return int(''.join([str(n) for n in numList])) + positions[n1].pop(0) #[4] + + return num \ No newline at end of file diff --git a/problems/python/maximum-units-on-a-truck.py b/problems/python/maximum-units-on-a-truck.py new file mode 100755 index 0000000..4e0ce10 --- /dev/null +++ b/problems/python/maximum-units-on-a-truck.py @@ -0,0 +1,17 @@ +class Solution(object): + def maximumUnits(self, boxTypes, truckSize): + sortedBox = [] + ans = 0 + + for count, units in boxTypes: + sortedBox.append((units, count)) + + sortedBox.sort() + + while sortedBox and truckSize>0: + units, count = sortedBox.pop() + d = min(count, truckSize) + truckSize -= d + ans += d*units + + return ans \ No newline at end of file diff --git a/problems/median-of-two-sorted-arrays.py b/problems/python/median-of-two-sorted-arrays.py old mode 100644 new mode 100755 similarity index 100% rename from problems/median-of-two-sorted-arrays.py rename to problems/python/median-of-two-sorted-arrays.py diff --git a/problems/meeting-rooms-ii.py b/problems/python/meeting-rooms-ii.py old mode 100644 new mode 100755 similarity index 77% rename from problems/meeting-rooms-ii.py rename to problems/python/meeting-rooms-ii.py index 99d5a28..50cebde --- a/problems/meeting-rooms-ii.py +++ b/problems/python/meeting-rooms-ii.py @@ -83,4 +83,32 @@ def minMeetingRooms(self, intervals): heapq.heappop(h) heapq.heappush(h, end) - return len(h) \ No newline at end of file + return len(h) + + +""" +Time: O(NLogN) +Space: O(N) + +e0, s0 is one of the previous meetings where the end time is the earliest (smallest). +""" +class Solution(object): + def minMeetingRooms(self, intervals): + if not intervals: return 0 + ans = 1 + h = [] + + intervals.sort() + heapq.heappush(h, (intervals[0][1], intervals[0][0])) + + for i in xrange(1, len(intervals)): + s = intervals[i][0] + e = intervals[i][1] + + e0, s0 = h[0] + + if s>=e0: heapq.heappop(h) + heapq.heappush(h, (e, s)) + ans = max(ans, len(h)) + + return ans \ No newline at end of file diff --git a/problems/meeting-rooms.py b/problems/python/meeting-rooms.py old mode 100644 new mode 100755 similarity index 100% rename from problems/meeting-rooms.py rename to problems/python/meeting-rooms.py diff --git a/problems/merge-intervals.py b/problems/python/merge-intervals.py old mode 100644 new mode 100755 similarity index 100% rename from problems/merge-intervals.py rename to problems/python/merge-intervals.py diff --git a/problems/merge-k-sorted-lists.py b/problems/python/merge-k-sorted-lists.py old mode 100644 new mode 100755 similarity index 100% rename from problems/merge-k-sorted-lists.py rename to problems/python/merge-k-sorted-lists.py diff --git a/problems/merge-sorted-array.py b/problems/python/merge-sorted-array.py old mode 100644 new mode 100755 similarity index 100% rename from problems/merge-sorted-array.py rename to problems/python/merge-sorted-array.py diff --git a/problems/merge-two-sorted-lists.py b/problems/python/merge-two-sorted-lists.py old mode 100644 new mode 100755 similarity index 100% rename from problems/merge-two-sorted-lists.py rename to problems/python/merge-two-sorted-lists.py diff --git a/problems/min-cost-climbing-stairs.py b/problems/python/min-cost-climbing-stairs.py old mode 100644 new mode 100755 similarity index 100% rename from problems/min-cost-climbing-stairs.py rename to problems/python/min-cost-climbing-stairs.py diff --git a/problems/min-stack.py b/problems/python/min-stack.py old mode 100644 new mode 100755 similarity index 100% rename from problems/min-stack.py rename to problems/python/min-stack.py diff --git a/problems/minimize-malware-spread.py b/problems/python/minimize-malware-spread.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimize-malware-spread.py rename to problems/python/minimize-malware-spread.py diff --git a/problems/minimum-absolute-difference-in-bst.py b/problems/python/minimum-absolute-difference-in-bst.py old mode 100644 new mode 100755 similarity index 52% rename from problems/minimum-absolute-difference-in-bst.py rename to problems/python/minimum-absolute-difference-in-bst.py index 4ab828c..2fe5f5b --- a/problems/minimum-absolute-difference-in-bst.py +++ b/problems/python/minimum-absolute-difference-in-bst.py @@ -43,4 +43,47 @@ def traverse(node): """ Time complexity: O(N) Space complexity: O(N) -""" \ No newline at end of file +""" + + +""" +Inorder traverse the BST and update the diff. +Time complexity: O(N) +Space complexity: O(N) +""" +class Solution(object): + def getMinimumDifference(self, root): + ans = float('inf') + + stack = [] + curr = root + prevVal = float('-inf') + + while stack or curr: + while curr: + stack.append(curr) + curr = curr.left + + curr = stack.pop() + ans = min(ans, curr.val-prevVal) + prevVal = curr.val + + curr = curr.right + return ans + + +def inorderTraversal(self, root): + stack = [] + curr = root + + while stack or curr: + while curr: + stack.append(curr) + curr = curr.left + + curr = stack.pop() + + # do something + print curr.val + + curr = curr.right \ No newline at end of file diff --git a/problems/minimum-ascii-delete-sum-for-two-strings.py b/problems/python/minimum-ascii-delete-sum-for-two-strings.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimum-ascii-delete-sum-for-two-strings.py rename to problems/python/minimum-ascii-delete-sum-for-two-strings.py diff --git a/problems/python/minimum-cost-to-connect-sticks.py b/problems/python/minimum-cost-to-connect-sticks.py new file mode 100755 index 0000000..81cd617 --- /dev/null +++ b/problems/python/minimum-cost-to-connect-sticks.py @@ -0,0 +1,15 @@ +class Solution(object): + def connectSticks(self, sticks): + if not sticks or len(sticks)==1: return 0 + h = [] + + for n in sticks: + heapq.heappush(h, n) + + cost = 0 + while len(h)>1: + currSum = heapq.heappop(h)+heapq.heappop(h) + cost += currSum + heapq.heappush(h, currSum) + + return cost \ No newline at end of file diff --git a/problems/minimum-cost-to-hire-k-workers.py b/problems/python/minimum-cost-to-hire-k-workers.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimum-cost-to-hire-k-workers.py rename to problems/python/minimum-cost-to-hire-k-workers.py diff --git a/problems/python/minimum-cost-to-make-at-least-one-valid-path-in-a-grid.py b/problems/python/minimum-cost-to-make-at-least-one-valid-path-in-a-grid.py new file mode 100755 index 0000000..2188f4a --- /dev/null +++ b/problems/python/minimum-cost-to-make-at-least-one-valid-path-in-a-grid.py @@ -0,0 +1,32 @@ +class Solution(object): + def minCost(self, grid): + pq = [(0, False, 0, 0, 0)] + visited = set() + M = len(grid) + N = len(grid[0]) + + while pq: + cost, modified, direction, x, y = heapq.heappop(pq) + if x<0 or x>=M or y<0 or y>=N: continue + if (direction, x, y) in visited: continue + visited.add((direction, x, y)) + + if x==M-1 and y==N-1: return cost + + if direction==0: direction = grid[x][y] + + if direction==1: + heapq.heappush(pq, (cost, False, 0, x, y+1)) + elif direction==2: + heapq.heappush(pq, (cost, False, 0, x, y-1)) + elif direction==3: + heapq.heappush(pq, (cost, False, 0, x+1, y)) + elif direction==4: + heapq.heappush(pq, (cost, False, 0, x-1, y)) + + if not modified: + for d in [1,2,3,4]: + if d==grid[x][y]: continue + heapq.heappush(pq, (cost+1, True, d, x, y)) + + return float('inf') \ No newline at end of file diff --git a/problems/python/minimum-cost-to-reach-city-with-discounts.py b/problems/python/minimum-cost-to-reach-city-with-discounts.py new file mode 100755 index 0000000..93205f3 --- /dev/null +++ b/problems/python/minimum-cost-to-reach-city-with-discounts.py @@ -0,0 +1,25 @@ +class Solution(object): + def minimumCost(self, n, highways, discounts): + pq = [(0, discounts, 0)] + visited = set() + + adj = collections.defaultdict(list) + for city1, city2, toll in highways: + adj[city1].append((city2, toll)) + adj[city2].append((city1, toll)) + + + while pq: + toll, d, city = heapq.heappop(pq) + if (d, city) in visited: continue + visited.add((d, city)) + + if city==n-1: return toll + + for nei, toll2 in adj[city]: + if (d, nei) not in visited: + heapq.heappush(pq, (toll+toll2, d, nei)) + if d>0 and (d-1, nei) not in visited: + heapq.heappush(pq, (toll+toll2/2, d-1, nei)) + + return -1 \ No newline at end of file diff --git a/problems/minimum-depth-of-binary-tree.py b/problems/python/minimum-depth-of-binary-tree.py old mode 100644 new mode 100755 similarity index 53% rename from problems/minimum-depth-of-binary-tree.py rename to problems/python/minimum-depth-of-binary-tree.py index b919ab1..1ea7e27 --- a/problems/minimum-depth-of-binary-tree.py +++ b/problems/python/minimum-depth-of-binary-tree.py @@ -18,4 +18,27 @@ def minDepth(self, root): if node.left: q.append((node.left, depth+1)) if node.right: - q.append((node.right, depth+1)) \ No newline at end of file + q.append((node.right, depth+1)) + + + +""" +Time: O(N) +Space: O(N) + +Standard BFS on a binary tree. +""" +class Solution(object): + def minDepth(self, root): + if not root: return 0 + q = collections.deque([(root, 1)]) + + while q: + node, depth = q.popleft() + + if not node.left and not node.right: return depth + + if node.left: q.append((node.left, depth+1)) + if node.right: q.append((node.right, depth+1)) + + return 'ERROR' \ No newline at end of file diff --git a/problems/minimum-difficulty-of-a-job-schedule.py b/problems/python/minimum-difficulty-of-a-job-schedule.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimum-difficulty-of-a-job-schedule.py rename to problems/python/minimum-difficulty-of-a-job-schedule.py diff --git a/problems/minimum-falling-path-sum-ii.py b/problems/python/minimum-falling-path-sum-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimum-falling-path-sum-ii.py rename to problems/python/minimum-falling-path-sum-ii.py diff --git a/problems/python/minimum-knight-moves.py b/problems/python/minimum-knight-moves.py new file mode 100755 index 0000000..84cb400 --- /dev/null +++ b/problems/python/minimum-knight-moves.py @@ -0,0 +1,30 @@ +# Bidirectional BFS +class Solution(object): + def minKnightMoves(self, x, y): + offsets = [(1, 2), (2, 1), (2, -1), (1, -2), (-1, -2), (-2, -1), (-2, 1), (-1, 2)] + q1 = collections.deque([(0, 0)]) + q2 = collections.deque([(x, y)]) + steps1 = {(0, 0): 0} #steps needed starting from (0, 0) + steps2 = {(x, y): 0} #steps needed starting from (x,y) + + while q1 and q2: + i1, j1 = q1.popleft() + if (i1, j1) in steps2: return steps1[(i1, j1)]+steps2[(i1, j1)] + + i2, j2 = q2.popleft() + if (i2, j2) in steps1: return steps1[(i2, j2)]+steps2[(i2, j2)] + + for ox, oy in offsets: + nextI1 = i1+ox + nextJ1 = j1+oy + if (nextI1, nextJ1) not in steps1: + q1.append((nextI1, nextJ1)) + steps1[(nextI1, nextJ1)] = steps1[(i1, j1)]+1 + + nextI2 = i2+ox + nextJ2 = j2+oy + if (nextI2, nextJ2) not in steps2: + q2.append((nextI2, nextJ2)) + steps2[(nextI2, nextJ2)] = steps2[(i2, j2)]+1 + + return float('inf') \ No newline at end of file diff --git a/problems/python/minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix.py b/problems/python/minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix.py new file mode 100755 index 0000000..5559c80 --- /dev/null +++ b/problems/python/minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix.py @@ -0,0 +1,42 @@ +class Solution(object): + def minFlips(self, mat): + def flip(mat, m, n): + mat[m][n] = 0 if mat[m][n]==1 else 1 + if m+1=0: mat[m-1][n] = 0 if mat[m-1][n]==1 else 1 + if n-1>=0: mat[m][n-1] = 0 if mat[m][n-1]==1 else 1 + + def check(mat, state): + for i, b in enumerate(state): + if b=='1': + m = i/len(mat[0]) + n = i%len(mat[0]) + flip(mat, m, n) + + for i in xrange(len(mat)): + for j in xrange(len(mat[0])): + if mat[i][j]==1: return False + + return True + + + M = len(mat) + N = len(mat[0]) + q = collections.deque(['0'*(M*N)]) + visited = set() + + while q: + state = q.popleft() + if state in visited: continue + visited.add(state) + + if check([row[:] for row in mat], state): return state.count('1') + + for i in xrange(len(state)): + if state[i]=='1': continue + nextState = state[:i] + '1' +state[i+1:] + q.append(nextState) + + return -1 + \ No newline at end of file diff --git a/problems/minimum-path-sum.py b/problems/python/minimum-path-sum.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimum-path-sum.py rename to problems/python/minimum-path-sum.py diff --git a/problems/minimum-score-triangulation-of-polygon.py b/problems/python/minimum-score-triangulation-of-polygon.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimum-score-triangulation-of-polygon.py rename to problems/python/minimum-score-triangulation-of-polygon.py diff --git a/problems/minimum-size-subarray-sum.py b/problems/python/minimum-size-subarray-sum.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimum-size-subarray-sum.py rename to problems/python/minimum-size-subarray-sum.py diff --git a/problems/python/minimum-swaps-to-group-all-1s-together.py b/problems/python/minimum-swaps-to-group-all-1s-together.py new file mode 100755 index 0000000..0f44bc6 --- /dev/null +++ b/problems/python/minimum-swaps-to-group-all-1s-together.py @@ -0,0 +1,33 @@ +""" +We know that, after swapping, the length of the continuous 1 will be equal to number of 1 in the data. Let's call it `L`. +So what we need to do is sliding window with length L and see what is the max 1 count within the window. +The window with max 1 count will need to do minimal swap. + +Time: O(N) +Space: O(1) +""" +class Solution(object): + def minSwaps(self, data): + L = 0 #sliding window length + #initialize L + for num in data: + if num==1: L += 1 + + count = 0 #count of 1 within the sliding window + maxCount = 0 #max count of 1 within the sliding window + + #initialize the count from data[0] to data[L-1] + for i in xrange(L): + if data[i]==1: count += 1 + maxCount = count + + #slide the window + for i in xrange(1, len(data)): + #i: start index of the sliding window + j = i+L-1 #end index of the sliding window + if j>=len(data): break + if data[j]==1: count += 1 + if data[i-1]==1: count -= 1 + maxCount = max(maxCount, count) + + return L-maxCount \ No newline at end of file diff --git a/problems/minimum-swaps-to-make-sequences-increasing.py b/problems/python/minimum-swaps-to-make-sequences-increasing.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimum-swaps-to-make-sequences-increasing.py rename to problems/python/minimum-swaps-to-make-sequences-increasing.py diff --git a/problems/python/minimum-time-difference.py b/problems/python/minimum-time-difference.py new file mode 100755 index 0000000..1645692 --- /dev/null +++ b/problems/python/minimum-time-difference.py @@ -0,0 +1,32 @@ +""" +Time: O(1440) +Space: O(1440) +""" +class Solution(object): + def findMinDifference(self, timeStrings): + def timeStringToMinutes(timeString): + time = timeString.split(':') + h = int(time[0]) + m = int(time[1]) + return h*60+m + + ans = float('inf') + minTime = float('inf') + maxTime = float('-inf') + timeSet = set() + for timeString in timeStrings: + t = timeStringToMinutes(timeString) + if t in timeSet: return 0 + minTime = min(minTime, t) + maxTime = max(maxTime, t) + timeSet.add(t) + + + prev = None + for t in xrange(minTime, maxTime+1): + if t not in timeSet: continue + if prev!=None: ans = min(ans, t-prev) + prev = t + + ans = min(ans, 1440+minTime-maxTime) #compare minTime and maxTime + return ans \ No newline at end of file diff --git a/problems/minimum-window-substring.py b/problems/python/minimum-window-substring.py old mode 100644 new mode 100755 similarity index 100% rename from problems/minimum-window-substring.py rename to problems/python/minimum-window-substring.py diff --git a/problems/python/minimum-xor-sum-of-two-arrays.py b/problems/python/minimum-xor-sum-of-two-arrays.py new file mode 100755 index 0000000..7c3a195 --- /dev/null +++ b/problems/python/minimum-xor-sum-of-two-arrays.py @@ -0,0 +1,24 @@ +""" +state[i] := nums1[i] has been matched. +""" +class Solution(object): + def minimumXORSum(self, nums1, nums2): + N = len(nums1) + + pq = [(0, '0'*N)] + visited = set() + while pq: + s, state = heapq.heappop(pq) + if state in visited: continue + visited.add(state) + + j = state.count('1') + if j==N: return s + + for i in xrange(N): + if state[i]=='1': continue + nextState = state[:i]+'1'+state[i+1:] + if nextState in visited: continue + heapq.heappush(pq, ((s+(nums1[i]^nums2[j-1]), nextState))) + + return float('inf') \ No newline at end of file diff --git a/problems/missing-number.py b/problems/python/missing-number.py old mode 100644 new mode 100755 similarity index 100% rename from problems/missing-number.py rename to problems/python/missing-number.py diff --git a/problems/python/most-frequent-subtree-sum.py b/problems/python/most-frequent-subtree-sum.py new file mode 100755 index 0000000..5c1457f --- /dev/null +++ b/problems/python/most-frequent-subtree-sum.py @@ -0,0 +1,40 @@ +from collections import Counter +class Solution(object): + def findFrequentTreeSum(self, root): + def dfs(node): + if not node: return 0 + s = dfs(node.left)+node.val+dfs(node.right) + counter[s] += 1 + return s + + if not root: return [] + counter = Counter() + dfs(root) #start counting + + max_freq = max(counter.values()) + return [v for v, freq in counter.items() if freq==max_freq] + +""" +Time: O(N) +Space: O(N) +""" + +""" +Time: O(N) +Space: O(N) + +Throught the getSubtreeSum(root), we will also count subtreeSum of each node. +""" +class Solution(object): + def findFrequentTreeSum(self, root): + def getSubtreeSum(node): + if not node: return 0 + subtreeSum = node.val+getSubtreeSum(node.left)+getSubtreeSum(node.right) + memo[subtreeSum] += 1 + return subtreeSum + + + memo = collections.Counter() + getSubtreeSum(root) + mostFrequenctCount = max([memo[subtreeSum] for subtreeSum in memo]) + return [subtreeSum for subtreeSum in memo if memo[subtreeSum]==mostFrequenctCount] \ No newline at end of file diff --git a/problems/python/most-stones-removed-with-same-row-or-column.py b/problems/python/most-stones-removed-with-same-row-or-column.py new file mode 100755 index 0000000..b10b373 --- /dev/null +++ b/problems/python/most-stones-removed-with-same-row-or-column.py @@ -0,0 +1,49 @@ +""" +Max stones can be remove = number of stones - number of groups +The stone in the same group has the same row or col with any stone in the group. +Since finding stones in the same group will take O(N^2) +Instead, we union the row or col itself. + +So initially each col (c1, c2...) and row (r1, r2...)'s parent is itself. +We iterate through the stones and union the col and row. +For example, the stone (1, 2) will union r2 and c1. +the stone (0, 3) will union r3 and c0. +... + +This way, we can also get the number of groups. + +See better explanation in https://www.youtube.com/watch?v=beOCN7G4h-M + +Time: O(N). +Space: O(N) +""" +class Solution(object): + def removeStones(self, stones): + def union(x, y): + row = 'r'+str(x) + col = 'c'+str(y) + p1 = find(row) + p2 = find(col) + if p1==p2: return False + parents[p2] = p1 + return True + + def find(c): + p = parents[c] + while p!=parents[p]: + p = find(p) + parents[c] = p + return p + + parents = {} + for x, y in stones: + row = 'r'+str(x) + col = 'c'+str(y) + parents[row] = row + parents[col] = col + + groupCount = len(parents) + for x, y in stones: + if union(x, y): groupCount -= 1 + + return len(stones)-groupCount \ No newline at end of file diff --git a/problems/move-zeroes.py b/problems/python/move-zeroes.py old mode 100644 new mode 100755 similarity index 100% rename from problems/move-zeroes.py rename to problems/python/move-zeroes.py diff --git a/problems/moving-average-from-data-stream.py b/problems/python/moving-average-from-data-stream.py old mode 100644 new mode 100755 similarity index 100% rename from problems/moving-average-from-data-stream.py rename to problems/python/moving-average-from-data-stream.py diff --git a/problems/my-calendar-ii.py b/problems/python/my-calendar-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/my-calendar-ii.py rename to problems/python/my-calendar-ii.py diff --git a/problems/n-ary-tree-level-order-traversal.py b/problems/python/n-ary-tree-level-order-traversal.py old mode 100644 new mode 100755 similarity index 100% rename from problems/n-ary-tree-level-order-traversal.py rename to problems/python/n-ary-tree-level-order-traversal.py diff --git a/problems/n-ary-tree-postorder-traversal.py b/problems/python/n-ary-tree-postorder-traversal.py old mode 100644 new mode 100755 similarity index 100% rename from problems/n-ary-tree-postorder-traversal.py rename to problems/python/n-ary-tree-postorder-traversal.py diff --git a/problems/n-ary-tree-preorder-traversal.py b/problems/python/n-ary-tree-preorder-traversal.py old mode 100644 new mode 100755 similarity index 100% rename from problems/n-ary-tree-preorder-traversal.py rename to problems/python/n-ary-tree-preorder-traversal.py diff --git a/problems/nested-list-weight-sum.py b/problems/python/nested-list-weight-sum.py old mode 100644 new mode 100755 similarity index 100% rename from problems/nested-list-weight-sum.py rename to problems/python/nested-list-weight-sum.py diff --git a/problems/python/network-delay-time.py b/problems/python/network-delay-time.py new file mode 100755 index 0000000..aa67bb0 --- /dev/null +++ b/problems/python/network-delay-time.py @@ -0,0 +1,213 @@ +""" +Dijkstra's algorithm heap implementation +first, we build an adjacent list. [0] +if you don't know what is adjacent list, see https://www.khanacademy.org/computing/computer-science/algorithms/graph-representation/a/representing-graphs + +we use a hash-map 'dis' to keep track of every node's distance to K. [1] +and we use a priority queue 'pq' to store all the node we encounter and its distance to K. [2] +which we use a tuple (distance to K, node) + +for every node we visit, if its distance to K is determined, we don't need to look at it anymore. [3] +because we always pop the nearest one to the K in the priority queue, we can be sure that the distance in 'dis' is the shortest. +then from the node, which we know the shortest path from K to the node, we keep on explore its neighbors. [4] + +then if we didn't visit every node we return -1, else we return the node which it takes the longest time to reach. [5] + +for time complexity +we use O(E) to make an adjacency list from edge list. +we will go through the while loop V-1 times because we need to determined the distance of V-1 other nodes. +for every loop + we pop from priority queue (here need O(logE) to get the nearest node). + we push the node's neighbor to the priority queue d times (which is the degree of the node) + push the node takes O(logE), we assume all the node is in the queue. + so every loop we use O(d*logE+logE)~=O(d*logE) +so total for total, it is O((V-1)*(d*logE))+O(E), we know that V-1~=V and V*d=E +so it is O(ElogE)+O(E)~=O(ElogE) +E is the number of edges, which is the length of 'times' of the input +V is the number of node, which is the 'N' of the inout + +for space complexity +we used O(V) and O(E) in 'dis' and 'pq', and O(E) on the adjacency list. +so, it is O(V+E). +""" +class Solution(object): + def networkDelayTime(self, times, N, K): + aj_list = collections.defaultdict(list) #[0] + for u, v, w in times: + aj_list[u].append((w, v)) + + dis = {} #[1] + pq = [(0, K)] #[2] + + while pq: + if len(dis)==N: break + + d, node = heapq.heappop(pq) + if node in dis: continue #[3] + + dis[node] = d + + for d2, nb in aj_list[node]: #[4] + if nb not in dis: #[3] + heapq.heappush(pq, (d+d2, nb)) #[2] + + return max(dis.values()) if len(dis)==N else -1 #[5] + + +#2020/10/17 +class Solution(object): + def networkDelayTime(self, times, N, K): + ans = float('-inf') + h = [(0, K)] + visited = set() + G = collections.defaultdict(list) + + for u, v, w in times: + G[u].append((v, w)) + + while h: + time, node = heapq.heappop(h) + + if node in visited: continue + visited.add(node) + ans = max(ans, time) + + if len(visited)==N: return time + for nei, time_to_nei in G[node]: + heapq.heappush(h, (time+time_to_nei, nei)) + + return -1 + + + + + + + +class Solution(object): + def networkDelayTime(self, times, n, k): + G = collections.defaultdict(list) #adjacency list + D = [float('inf')]*(n+1) #D[n] store the distance between n and k + D[k] = 0 + h = [(0, k)] + visited = set() + + #form the djacency list + for u, v, w in times: + G[u].append((v, w)) + + #dijkstra + while h: + d1, node = heapq.heappop(h) + if node in visited: continue + visited.add(node) + + for nei, d2 in G[node]: + if d1+d29 else '0'+str(h) + if all([c in digits for c in hs]): return hs + return '' + + def getSmallestHour(): + for h in xrange(0, 24): + hs = str(h) if h>9 else '0'+str(h) + if all([c in digits for c in hs]): return hs + return '' + + def getNextMinute(): + currentMinute = int(time[3]+time[4]) if time[3]!='0' else int(time[4]) + + for m in xrange(currentMinute+1, 60): + ms = str(m) if m>9 else '0'+str(m) + if all([c in digits for c in ms]): return ms + return '' + + def getSmallestMinute(): + for m in xrange(0, 60): + ms = str(m) if m>9 else '0'+str(m) + if all([c in digits for c in ms]): return ms + return '' + + digits = set([time[0], time[1], time[3], time[4]]) + nextMinute = getNextMinute() + if nextMinute: + return time[:3]+nextMinute + else: + return (getNextHour() or getSmallestHour())+':'+getSmallestMinute() \ No newline at end of file diff --git a/problems/python/next-permutation.py b/problems/python/next-permutation.py new file mode 100755 index 0000000..0a816ee --- /dev/null +++ b/problems/python/next-permutation.py @@ -0,0 +1,61 @@ +""" +This answer is the python version of the offical answer. + +Time: O(N) +Space: O(1) +""" +class Solution(object): + def nextPermutation(self, nums): + def reverse(start): + end = len(nums)-1 + + while start=0 and nums[i+1]<=nums[i]: + i -= 1 + + if i>=0: + j = len(nums)-1 + while nums[j]<=nums[i]: j -= 1 + swap(i, j) + + reverse(i+1) + return nums + +""" +Next Permutation means find the next (slightly) larger number using nums. + +1. Iterate from right, find the first num that is smaller. That's the one we are going to swap. => nums[i] +2. From the nums right to nums[i], find the smallest num that is larger than nums[i] => nums[j] +Since the right of the i must be an increasing sequence (looking from right), the first one that larger than nums[i] is the smallest one that is larger than nums[i] +3. Swap nums[i] and nums[j] +4. sort nums[i+1:] it will be the smallest permutaion. +5. Note that when an list is in increasing order looking from right, we can use `reverse` to sort it. +""" +class Solution(object): + def nextPermutation(self, nums): + def reverse(nums, l, r): + while l<=r: + nums[l], nums[r] = nums[r], nums[l] + l += 1 + r -= 1 + + i = len(nums)-2 + while i>=0 and nums[i]>=nums[i+1]: i -= 1 #[1] + + if i==-1: + return reverse(nums, 0, len(nums)-1) #nums is the largest permutation, sort nums + else: + j = len(nums)-1 + while j>i and nums[j]<=nums[i]: j -= 1 #[2] + nums[i], nums[j] = nums[j], nums[i] #[3] + reverse(nums, i+1, len(nums)-1) #[4] \ No newline at end of file diff --git a/problems/number-complement.py b/problems/python/number-complement.py old mode 100644 new mode 100755 similarity index 100% rename from problems/number-complement.py rename to problems/python/number-complement.py diff --git a/problems/python/number-of-connected-components-in-an-undirected-graph.py b/problems/python/number-of-connected-components-in-an-undirected-graph.py new file mode 100755 index 0000000..dbd37c2 --- /dev/null +++ b/problems/python/number-of-connected-components-in-an-undirected-graph.py @@ -0,0 +1,50 @@ +class Solution(object): + def countComponents(self, N, edges): + def dfs(start): + stack = [start] + + while stack: + node = stack.pop() + if node in visited: continue + visited.add(node) + + for nei in g[node]: + stack.append(nei) + + g = collections.defaultdict(list) + visited = set() + count = 0 + + for n1, n2 in edges: + g[n1].append(n2) + g[n2].append(n1) + + for n in xrange(N): + if n in visited: continue + dfs(n) + count += 1 + + return count + + +class Solution(object): + def countComponents(self, n, edges): + def find(n): + p = parents[n] + while p!=parents[p]: + p = find(p) + parents[n] = p + return p + + def union(n1, n2): + p1, p2 = find(n1), find(n2) + if p1==p2: return False + parents[p2] = p1 + return True + + count = n + parents = [n for n in xrange(n)] + + for n1, n2 in edges: + if union(n1, n2): count -= 1 + return count \ No newline at end of file diff --git a/problems/python/number-of-islands-ii.py b/problems/python/number-of-islands-ii.py new file mode 100755 index 0000000..a3893ae --- /dev/null +++ b/problems/python/number-of-islands-ii.py @@ -0,0 +1,50 @@ +""" +For each new land, the count will -= ((number of neighbors) - 1). +Because some for this new land introduced, some of the separate islands will be connected. +Note that, "number of neighbors", islands connected does not count. So for each neighbor, we use "find" to find its root. +""" +class Solution(object): + def numIslands2(self, M, N, positions): + def coorToNum(r, c): + return N*r+c + + def find(n): + p = parents[n] + while p!=parents[p]: + p = find(p) + parents[n] = p + return parents[n] + + def union(n1, n2): + p1 = find(n1) + p2 = find(n2) + + if p1==p2: return + parents[p1] = p2 + + count = 0 + lands = set() + ans = [] + parents = [n for n in xrange(M*N)] + + for r, c in positions: + n = coorToNum(r, c) + + if n in lands: + ans.append(count) + continue + + neis = set() + if r+10 and n==remains[i-1]: continue + if isSquare(last+n): + helper(per+[n], remains[:i]+remains[i+1:]) + + nums.sort() + for i, n in enumerate(nums): + if i>0 and n==nums[i-1]: continue + helper([nums[i]], nums[:i]+nums[i+1:]) + + return self.ans \ No newline at end of file diff --git a/problems/python/number-of-substrings-containing-all-thre.py b/problems/python/number-of-substrings-containing-all-thre.py new file mode 100755 index 0000000..66d43cc --- /dev/null +++ b/problems/python/number-of-substrings-containing-all-thre.py @@ -0,0 +1,22 @@ +class Solution(object): + def numberOfSubstrings(self, s): + #number of subarrays that at most have k unique char + def atMost(k): + counter = collections.Counter() + uniqueCount = 0 + ans = 0 + i = 0 + + for j, c in enumerate(s): + counter[c] += 1 + if counter[c]==1: uniqueCount += 1 + + while uniqueCount>k: + counter[s[i]] -= 1 + if counter[s[i]]==0: uniqueCount-= 1 + i += 1 + ans += j-i+1 + return ans + + n = len(s) + return atMost(3) - atMost(2) \ No newline at end of file diff --git a/problems/python/number-of-ways-to-arrive-at-destination.py b/problems/python/number-of-ways-to-arrive-at-destination.py new file mode 100755 index 0000000..3231254 --- /dev/null +++ b/problems/python/number-of-ways-to-arrive-at-destination.py @@ -0,0 +1,33 @@ +class Solution(object): + def countPaths(self, n, roads): + def countWaysToReach(node): + if node==0: return 1 + if node in history: return history[node] + c = 0 + for nei, t in adj[node]: + if nei in times and times[nei]+t==times[node]: + c += countWaysToReach(nei) + history[node] = c + return c + + history = {} #cache for countWaysToReach() + times = {} #min times to reach node n-1 + pq = [(0, 0)] + + adj = collections.defaultdict(list) + for u, v, t in roads: + adj[u].append((v, t)) + adj[v].append((u, t)) + + while pq: + t, node = heapq.heappop(pq) + if node in times: continue + times[node] = t + + if node==n-1: break + + for nei, t2 in adj[node]: + if nei in times: continue + heapq.heappush(pq, (t+t2, nei)) + + return countWaysToReach(n-1)%(10**9 + 7) \ No newline at end of file diff --git a/problems/odd-even-jump.py b/problems/python/odd-even-jump.py old mode 100644 new mode 100755 similarity index 100% rename from problems/odd-even-jump.py rename to problems/python/odd-even-jump.py diff --git a/problems/ones-and-zeroes.py b/problems/python/ones-and-zeroes.py old mode 100644 new mode 100755 similarity index 100% rename from problems/ones-and-zeroes.py rename to problems/python/ones-and-zeroes.py diff --git a/problems/open-the-lock.py b/problems/python/open-the-lock.py old mode 100644 new mode 100755 similarity index 100% rename from problems/open-the-lock.py rename to problems/python/open-the-lock.py diff --git a/problems/out-of-boundary-paths.py b/problems/python/out-of-boundary-paths.py old mode 100644 new mode 100755 similarity index 100% rename from problems/out-of-boundary-paths.py rename to problems/python/out-of-boundary-paths.py diff --git a/problems/python/pacific-atlantic-water-flow.py b/problems/python/pacific-atlantic-water-flow.py new file mode 100755 index 0000000..15186ff --- /dev/null +++ b/problems/python/pacific-atlantic-water-flow.py @@ -0,0 +1,30 @@ +class Solution(object): + def pacificAtlantic(self, heights): + def bfs(q, ocian): + while q: + i0, j0 = q.popleft() + if (i0, j0) in ocian: continue + ocian.add((i0, j0)) + + for i, j in [(i0+1, j0), (i0-1, j0), (i0, j0+1), (i0, j0-1)]: + if i>=len(heights) or i<0 or j>=len(heights[0]) or j<0: continue + if heights[i][j]>=heights[i0][j0]: q.append((i, j)) + + pacific = set() + altalantic = set() + q1 = collections.deque() + q2 = collections.deque() + + #add top, left to pacific + for j in xrange(len(heights[0])): q1.append((0, j)) + for i in xrange(len(heights)): q1.append((i, 0)) + + #add right, bottom to atalantic + for j in xrange(len(heights[0])): q2.append((len(heights)-1, j)) + for i in xrange(len(heights)): q2.append((i, len(heights[0])-1)) + + bfs(q1, pacific) + bfs(q2, altalantic) + + return pacific.intersection(altalantic) + \ No newline at end of file diff --git a/problems/python/pairs-of-songs-with-total-durations-divisible-by-60.py b/problems/python/pairs-of-songs-with-total-durations-divisible-by-60.py new file mode 100755 index 0000000..d4d7fa6 --- /dev/null +++ b/problems/python/pairs-of-songs-with-total-durations-divisible-by-60.py @@ -0,0 +1,10 @@ +class Solution(object): + def numPairsDivisibleBy60(self, times): + counter = collections.Counter() + ans = 0 + + for time in times: + time = time%60 + ans += counter[60-time if time!=0 else 0] + counter[time] += 1 + return ans \ No newline at end of file diff --git a/problems/palindrome-number.py b/problems/python/palindrome-number.py old mode 100644 new mode 100755 similarity index 100% rename from problems/palindrome-number.py rename to problems/python/palindrome-number.py diff --git a/problems/palindrome-pairs.py b/problems/python/palindrome-pairs.py old mode 100644 new mode 100755 similarity index 100% rename from problems/palindrome-pairs.py rename to problems/python/palindrome-pairs.py diff --git a/problems/palindrome-partitioning-iii.py b/problems/python/palindrome-partitioning-iii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/palindrome-partitioning-iii.py rename to problems/python/palindrome-partitioning-iii.py diff --git a/problems/palindrome-partitioning.py b/problems/python/palindrome-partitioning.py old mode 100644 new mode 100755 similarity index 100% rename from problems/palindrome-partitioning.py rename to problems/python/palindrome-partitioning.py diff --git a/problems/python/palindromic-substrings.py b/problems/python/palindromic-substrings.py new file mode 100755 index 0000000..a375af6 --- /dev/null +++ b/problems/python/palindromic-substrings.py @@ -0,0 +1,61 @@ +""" +dp[i][j] := if s[i:j+1] is palindrome. +1. all length==1 string is palindrome. +2. all length==2 string is palindrome if two are the same. +3. all length>=3 string is palindrome if the outer most two char is the same and have an palindrome string inside. + +Time: O(N^2) +Space: 0(N^2) +""" +class Solution(object): + def countSubstrings(self, s): + N = len(s) + dp = [[False]*N for _ in xrange(N)] + ans = 0 + + #1 + for i in xrange(N): + dp[i][i] = True + ans += 1 + + #2 + for i in xrange(N-1): + if s[i]==s[i+1]: + dp[i][i+1] = True + ans += 1 + + #3 + for l in xrange(3, N+1): + for i in xrange(N): + j = i+l-1 + if j>=N: continue + if dp[i+1][j-1] and s[i]==s[j]: + dp[i][j] = True + ans += 1 + return ans + +""" +Iterate through the string, expand around each character. +Like this solution more since it is more intuitive. + +Time: O(N^2) +Space: O(1) +""" +class Solution(object): + def countSubstrings(self, s): + def count(l, r, N): + count = 0 + + while l>=0 and r0: return False + + N = len(nums) + visited = [False]*N + return dfs(target, k) \ No newline at end of file diff --git a/problems/python/path-sum-ii.py b/problems/python/path-sum-ii.py new file mode 100755 index 0000000..2bfcda5 --- /dev/null +++ b/problems/python/path-sum-ii.py @@ -0,0 +1,42 @@ +class Solution(object): + def pathSum(self, root, S): + if not root: return False + + opt = [] + stack = [] + stack.append((root, 0, [])) + + while stack: + node, s, path = stack.pop() + s += node.val + path = path + [node.val] + if not node.left and not node.right and s==S: opt.append(path) + if node.left: stack.append((node.left, s, path)) + if node.right: stack.append((node.right, s, path)) + return opt +""" +Time complexity is O(N), because we traverse all the nodes. +Space complexity is O(N^2), because in the worst case, all node could carry all the other nodes in the `path`. +""" + + + +""" +Time complexity is O(N), because we traverse all the nodes. +Space complexity is O(N^2), because in the worst case, all node could carry all the other nodes in the `path`. +""" +class Solution(object): + def pathSum(self, root, targetSum): + if not root: return [] + + ans = [] + stack = [(root, root.val, [root.val])] + + while stack: + node, total, path = stack.pop() + + if not node.left and not node.right and total==targetSum: ans.append(path) + if node.left: stack.append((node.left, total+node.left.val, path+[node.left.val])) + if node.right: stack.append((node.right, total+node.right.val, path+[node.right.val])) + + return ans \ No newline at end of file diff --git a/problems/path-sum-iii.py b/problems/python/path-sum-iii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/path-sum-iii.py rename to problems/python/path-sum-iii.py diff --git a/problems/python/path-sum.py b/problems/python/path-sum.py new file mode 100755 index 0000000..ce50d04 --- /dev/null +++ b/problems/python/path-sum.py @@ -0,0 +1,37 @@ +class Solution(object): + def hasPathSum(self, root, S): + if not root: return False + + stack = [] + stack.append((root, 0)) + + while stack: + node, s = stack.pop() + s += node.val + if not node.left and not node.right and s==S: return True + if node.left: stack.append((node.left, s)) + if node.right: stack.append((node.right, s)) + return False + + +""" +Time: O(N) +Space: O(N) + +DFS +""" +class Solution(object): + def hasPathSum(self, root, targetSum): + if not root: return False + + stack = [(root, root.val)] + + while stack: + node, total = stack.pop() + + if not node.left and not node.right and total==targetSum: return True + if node.left: stack.append((node.left, total+node.left.val)) + if node.right: stack.append((node.right, total+node.right.val)) + + return False + \ No newline at end of file diff --git a/problems/python/path-with-maximum-probability.py b/problems/python/path-with-maximum-probability.py new file mode 100755 index 0000000..db1b1d4 --- /dev/null +++ b/problems/python/path-with-maximum-probability.py @@ -0,0 +1,24 @@ +class Solution(object): + def maxProbability(self, n, edges, succProb, start, end): + pq = [(-1, start)] + visited = set() + adj = collections.defaultdict(list) + for i in xrange(len(edges)): + a, b = edges[i] + p = succProb[i] + adj[a].append((b, p)) + adj[b].append((a, p)) + + while pq: + p, node = heapq.heappop(pq) + p = p*-1 + if node in visited: continue + visited.add(node) + + if node==end: return p + + for nei, p2 in adj[node]: + if nei in visited: continue + heapq.heappush(pq, (-1*p*p2, nei)) + + return 0 \ No newline at end of file diff --git a/problems/peak-index-in-a-mountain-array.PY b/problems/python/peak-index-in-a-mountain-array.py old mode 100644 new mode 100755 similarity index 100% rename from problems/peak-index-in-a-mountain-array.PY rename to problems/python/peak-index-in-a-mountain-array.py diff --git a/problems/perfect-squares.py b/problems/python/perfect-squares.py old mode 100644 new mode 100755 similarity index 100% rename from problems/perfect-squares.py rename to problems/python/perfect-squares.py diff --git a/problems/permutation-in-string.py b/problems/python/permutation-in-string.py old mode 100644 new mode 100755 similarity index 100% rename from problems/permutation-in-string.py rename to problems/python/permutation-in-string.py diff --git a/problems/python/permutation-sequence.py b/problems/python/permutation-sequence.py new file mode 100755 index 0000000..f7e5be2 --- /dev/null +++ b/problems/python/permutation-sequence.py @@ -0,0 +1,14 @@ +class Solution(object): + def getPermutation(self, N, K): + K = K-1 #make it 0-index + nums = range(1, N+1) + ans = '' + + while N>0: + a = K/math.factorial(N-1) + ans += str(nums[a]) + nums.pop(a) + + K -= math.factorial(N-1)*(a+1) + N -= 1 + return ans \ No newline at end of file diff --git a/problems/python/permutations-ii.py b/problems/python/permutations-ii.py new file mode 100755 index 0000000..a6c9ff7 --- /dev/null +++ b/problems/python/permutations-ii.py @@ -0,0 +1,80 @@ +""" +This is the extended question of the [first one](https://leetcode.com/problems/permutations/submissions/). +The differece is there will be duplicates in the `nums`. + +```python +if i>0 and options[i]==options[i-1]: continue +``` +Above lets us skip exploring the same path. +We can directly skip the `i` if the value is the same with `i-1`, because `dfs()` on `i-1` has already cover up all the possiblities. + +The time complexity is `O(N!)`. +The space complexity is `O(N!)`, too. +""" +class Solution(object): + def permuteUnique(self, nums): + def dfs(path, options): + if len(path)==len(nums): + opt.append(path) + return + for i, n in enumerate(options): + if i>0 and options[i]==options[i-1]: continue + dfs(path+[n], options[:i]+options[i+1:]) + opt = [] + nums.sort() + dfs([], nums) + return opt + + +""" +Time: O(N!) +Space: O(N!) + +Since we are iterating the key of the counter, we will only place "each kind of number" at the first place once. +For example, [1,1,2], it will not happend that +We place the first "1" at index 0, and keep exploring... +And place the second "1" at index 0, and keep exploring... +""" +class Solution(object): + def permuteUnique(self, nums): + def helper(path): + if len(path)==len(nums): ans.append(path[:]) + + for num in counter: + if counter[num]>0: + path.append(num) + counter[num] -= 1 + + helper(path) + + path.pop() + counter[num] += 1 + ans = [] + counter = collections.Counter(nums) + helper([]) + return ans + + +""" +差板法 +""" +class Solution(object): + def permuteUnique(self, nums): + if not nums: return [] + + permutations = collections.deque([[nums[0]]]) + + for i in xrange(1, len(nums)): + num = nums[i] + l = len(permutations) + + while l: + permutation = permutations.popleft() + for j in xrange(len(permutation)+1): + if 0=N: ans.append(nums[:]) + + for j in xrange(i, N): + nums[i], nums[j] = nums[j], nums[i] + helper(i+1) + nums[i], nums[j] = nums[j], nums[i] + + N = len(nums) + ans = [] + helper(0) + return ans + + +""" +差板法 +""" +class Solution(object): + def permute(self, nums): + if not nums: return [] + + permutations = collections.deque([[nums[0]]]) + + for i in xrange(1, len(nums)): + num = nums[i] + l = len(permutations) + + while l: + permutation = permutations.popleft() + for j in xrange(len(permutation)+1): + newPermutaion = permutation[:] + newPermutaion.insert(j, num) + permutations.append(newPermutaion) + l -= 1 + + return permutations \ No newline at end of file diff --git a/problems/python/populating-next-right-pointers-in-each-node-ii.py b/problems/python/populating-next-right-pointers-in-each-node-ii.py new file mode 100755 index 0000000..718d84d --- /dev/null +++ b/problems/python/populating-next-right-pointers-in-each-node-ii.py @@ -0,0 +1,50 @@ +""" +Time: O(N) +Space: O(1) + +View each level as a link list. +Traverse the tree level by level. For each level: +Starting from the leftmost (the head of the link list), we establish the next pointer of the children. + +curr.left.next = curr.right or (curr.next.left or curr.next.right) or (curr.next.next.left or curr.next.next.right) or (...) +curr.right.next = (curr.next.left or curr.next.right) or (curr.next.next.left or curr.next.next.right) or (...) +""" +class Solution(object): + def connect(self, root): + if not root: return root + + leftmost = root + + while leftmost: + curr = leftmost + + while curr: + if curr.left: + if curr.right: + curr.left.next = curr.right + else: + n = curr.next + while n: + curr.left.next = n.left or n.right + if curr.left.next: break + n = n.next + + if curr.right: + n = curr.next + while n: + curr.right.next = n.left or n.right + if curr.right.next: break + n = n.next + + curr = curr.next + + nextLevelLeftmost = None + n = leftmost + while n: + if n.left or n.right: + nextLevelLeftmost = n.left or n.right + break + n = n.next + leftmost = nextLevelLeftmost + + return root \ No newline at end of file diff --git a/problems/python/populating-next-right-pointers-in-each-node.py b/problems/python/populating-next-right-pointers-in-each-node.py new file mode 100755 index 0000000..0f01cf2 --- /dev/null +++ b/problems/python/populating-next-right-pointers-in-each-node.py @@ -0,0 +1,56 @@ +""" +Time: O(N) +Space: O(N) + +Traverse the tree level by level. +For each level connect the node to the "next node". +""" +class Solution(object): + def connect(self, root): + if not root: return root + level = collections.deque([root]) + nextLevel = collections.deque() + + while level: + node = level.popleft() + + if node.left: nextLevel.append(node.left) + if node.right: nextLevel.append(node.right) + + if not level: + if not nextLevel: break + for i, c in enumerate(nextLevel): + if i==len(nextLevel)-1: continue #skip last + c.next = nextLevel[i+1] + level = nextLevel + nextLevel = collections.deque() + + return root + + +""" +Time: O(N) +Space: O(1) + +View each level as a link list. +Traverse the tree level by level. For each level: +Starting from the leftmost (the head of the link list), we establish the next pointer of the children. +curr.left.next = curr.right +curr.right.next = curr.next.left +""" +class Solution(object): + def connect(self, root): + if not root: return root + + leftmost = root + while leftmost: + + curr = leftmost + while curr: + if curr.left: curr.left.next = curr.right + if curr.right and curr.next: curr.right.next = curr.next.left + curr = curr.next + + leftmost = leftmost.left + + return root \ No newline at end of file diff --git a/problems/python/powx-n.py b/problems/python/powx-n.py new file mode 100755 index 0000000..d0fac2b --- /dev/null +++ b/problems/python/powx-n.py @@ -0,0 +1,12 @@ +class Solution(object): + def myPow(self, x, n): + if n==0: + return 1 + elif n<0: + return 1/self.myPow(x, -n) + elif n%2>0: + half = self.myPow(x, n-1) + return x*half + elif n%2==0: + half = self.myPow(x, n/2) + return half*half \ No newline at end of file diff --git a/problems/product-of-array-except-self.py b/problems/python/product-of-array-except-self.py old mode 100644 new mode 100755 similarity index 100% rename from problems/product-of-array-except-self.py rename to problems/python/product-of-array-except-self.py diff --git a/problems/python/profitable-schemes.py b/problems/python/profitable-schemes.py new file mode 100755 index 0000000..2c16d85 --- /dev/null +++ b/problems/python/profitable-schemes.py @@ -0,0 +1,45 @@ +""" +TLE +dp[i][n][p] := considering profit[:i], what is the number of ways produce profit p with n people. +""" +class Solution(object): + def profitableSchemes(self, maxMember, minProfit, group, profit): + P = sum(profit) + N = sum(group) + dp = [[[0 for _ in xrange(P+1)] for _ in xrange(N+1)] for _ in xrange(len(profit)+1)] + dp[0][0][0] = 1 + + count = 0 + for i in xrange(1, len(profit)+1): + for n in xrange(N+1): + for p in xrange(P+1): + dp[i][n][p] = (dp[i-1][n-group[i-1]][p-profit[i-1]] if p-profit[i-1]>=0 and n-group[i-1]>=0 else 0) + dp[i-1][n][p] + if i==len(profit) and p>=minProfit and n<=maxMember: count += dp[i][n][p] + return count + + + +""" +dp[i][g][p] := consider only crime[:i] the scheme that can generate profit p using man power g. +""" +class Solution(object): + def profitableSchemes(self, n, minProfit, group, profit): + N = len(profit) + + dp = [[[0]*(n+2) for _ in xrange(minProfit+1)] for _ in xrange(N+1)] + dp[0][0][0] = 1 + + for i in xrange(1, N+1): + for p in xrange(minProfit+1): + for g in xrange(n+1): + pi = profit[i-1] + gi = group[i-1] + + #considerting last round using p and g + dp[i][p][g] += dp[i-1][p][g] + dp[i][min(pi+p, minProfit)][min(gi+g, n+1)] += dp[i-1][p][g] + + ans = 0 + for g in xrange(n+1): + ans += dp[N][minProfit][g] + return ans % (10**9 + 7) \ No newline at end of file diff --git a/problems/queue-reconstruction-by-height.py b/problems/python/queue-reconstruction-by-height.py old mode 100644 new mode 100755 similarity index 100% rename from problems/queue-reconstruction-by-height.py rename to problems/python/queue-reconstruction-by-height.py diff --git a/problems/python/random-pick-with-weight.py b/problems/python/random-pick-with-weight.py new file mode 100755 index 0000000..9c88bd0 --- /dev/null +++ b/problems/python/random-pick-with-weight.py @@ -0,0 +1,29 @@ +""" +For W = [1,3,2,4] (total 10, 1+2+3+4) +Think of a line _,___,__,____ ([1,4,6,10]) +Now we randomly throw a ball on the line the probability to land on the first section will be 1/10. +the second section, 3/10. +the third section, 2/10. +the forth section, 4/10. + +Above is equivilant to we randomly pick a number [0~10) and see which section it is in. +And since the cumulative probability in the "line" is strictly increasing, we can use binary search. + +Time: O(LogN), N is the number of W. +Space: O(N). +""" +from random import randrange + +class Solution(object): + + def __init__(self, W): + self.line = [] #cumulative probability distribution + self.total = 0 + + for w in W: + self.total += w + self.line.append(self.total) + + def pickIndex(self): + rand = random.randrange(self.total) + return bisect.bisect(self.line, rand) \ No newline at end of file diff --git a/problems/range-addition.py b/problems/python/range-addition.py old mode 100644 new mode 100755 similarity index 100% rename from problems/range-addition.py rename to problems/python/range-addition.py diff --git a/problems/python/range-sum-of-bst.py b/problems/python/range-sum-of-bst.py new file mode 100755 index 0000000..0ca07ed --- /dev/null +++ b/problems/python/range-sum-of-bst.py @@ -0,0 +1,11 @@ +class Solution(object): + def rangeSumBST(self, root, low, high): + def helper(node, low, high): + total = 0 + if not node: return total + if low<=node.val<=high: total += node.val + if node.val<=high: total += helper(node.right, low, high) + if node.val>=low: total += helper(node.left, low, high) + return total + + return helper(root, low, high) \ No newline at end of file diff --git a/problems/range-sum-query-immutable.py b/problems/python/range-sum-query-immutable.py old mode 100644 new mode 100755 similarity index 100% rename from problems/range-sum-query-immutable.py rename to problems/python/range-sum-query-immutable.py diff --git a/problems/python/range-sum-query-mutable.py b/problems/python/range-sum-query-mutable.py new file mode 100755 index 0000000..2502a15 --- /dev/null +++ b/problems/python/range-sum-query-mutable.py @@ -0,0 +1,67 @@ +""" +Time: O(LogN) for all operations. +Space: O(N) for the segment tree. + +Build a segment tree from the array `nums`. Each node store the info of the segment in nums (from `node.start` to `node.end`) +And `node.val` is equal to sum from nums[start] to nums[end] +""" +class NumArray(object): + + def __init__(self, nums): + def buildSegmentTree(start, end): + if start>end: return None + node = Node(start, end) + + if start==end: + node.val = nums[end] + else: + node.left = buildSegmentTree(start, node.mid) + node.right = buildSegmentTree(node.mid+1, end) + node.val = (node.left.val if node.left else 0) + (node.right.val if node.right else 0) + + return node + + self.root = buildSegmentTree(0, len(nums)-1) + + + def update(self, i, val): + def helper(node, i, val): + if node.start==node.end==i: + node.val = val + return + + if node.mid=2: break + + #swap the values + memo[diff[0]].val = diff[1] + memo[diff[1]].val = diff[0] + + return root + + +""" +Time: O(N) +Space: O(N) +""" +class Solution(object): + def recoverTree(self, root): + stack = [] + node = root + prev = TreeNode(float('-inf')) + swap1 = swap2 = None + + #inorder traversal and find the swapped values + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + + if swap1==None and prev.val>node.val: swap1 = prev + if swap1!=None and prev.val>node.val: swap2 = node + + prev = node + + node = node.right + + #swap the values + swap1.val, swap2.val = swap2.val, swap1.val + + return root \ No newline at end of file diff --git a/problems/redundant-connection.py b/problems/python/redundant-connection.py old mode 100644 new mode 100755 similarity index 100% rename from problems/redundant-connection.py rename to problems/python/redundant-connection.py diff --git a/problems/python/remove-all-adjacent-duplicates-in-string.py b/problems/python/remove-all-adjacent-duplicates-in-string.py new file mode 100755 index 0000000..f4bb906 --- /dev/null +++ b/problems/python/remove-all-adjacent-duplicates-in-string.py @@ -0,0 +1,12 @@ +class Solution(object): + def removeDuplicates(self, s): + stack = [] + i = 0 + while i2: return False + if len(rowStrings)==1: return True + + s1 = rowStrings.pop() + s2 = rowStrings.pop() + + for i in xrange(len(s1)): + if (s1[i]=='0' and s2[i]=='1') or (s1[i]=='1' and s2[i]=='0'): continue + return False + return True \ No newline at end of file diff --git a/problems/remove-duplicates-from-sorted-array-ii.py b/problems/python/remove-duplicates-from-sorted-array-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/remove-duplicates-from-sorted-array-ii.py rename to problems/python/remove-duplicates-from-sorted-array-ii.py diff --git a/problems/remove-duplicates-from-sorted-array.py b/problems/python/remove-duplicates-from-sorted-array.py old mode 100644 new mode 100755 similarity index 100% rename from problems/remove-duplicates-from-sorted-array.py rename to problems/python/remove-duplicates-from-sorted-array.py diff --git a/problems/remove-duplicates-from-sorted-list.py b/problems/python/remove-duplicates-from-sorted-list.py old mode 100644 new mode 100755 similarity index 100% rename from problems/remove-duplicates-from-sorted-list.py rename to problems/python/remove-duplicates-from-sorted-list.py diff --git a/problems/remove-element.py b/problems/python/remove-element.py old mode 100644 new mode 100755 similarity index 100% rename from problems/remove-element.py rename to problems/python/remove-element.py diff --git a/problems/python/remove-invalid-parentheses.py b/problems/python/remove-invalid-parentheses.py new file mode 100755 index 0000000..0cb7695 --- /dev/null +++ b/problems/python/remove-invalid-parentheses.py @@ -0,0 +1,38 @@ +class Solution(object): + def removeInvalidParentheses(self, s): + def dfs(s, curr, i, count): + if count<0: return + if len(curr)>self.maxLen: return + + if i>=len(s): + if len(curr)==self.maxLen and count==0: + self.ans.append(curr) + return + + if s[i]!='(' and s[i]!=')': + dfs(s, curr+s[i], i+1, count) + elif not curr or s[i]!=curr[-1]: + dfs(s, curr+s[i], i+1, count + (1 if s[i]=='(' else -1)) + dfs(s, curr, i+1, count) + elif s[i]==curr[-1]: + dfs(s, curr+s[i], i+1, count + (1 if s[i]=='(' else -1)) + + def getMaxLen(s): + openCount = removeCount = 0 + for c in s: + if c=='(': + openCount += 1 + elif c==')': + openCount -= 1 + + if openCount<0: + removeCount += abs(openCount) + openCount = 0 + removeCount += openCount + return len(s)-removeCount + + self.ans = [] + self.maxLen = getMaxLen(s) + + dfs(s, "", 0, 0) + return self.ans \ No newline at end of file diff --git a/problems/remove-linked-list-elements.py b/problems/python/remove-linked-list-elements.py old mode 100644 new mode 100755 similarity index 100% rename from problems/remove-linked-list-elements.py rename to problems/python/remove-linked-list-elements.py diff --git a/problems/python/remove-nth-node-from-end-of-list.py b/problems/python/remove-nth-node-from-end-of-list.py new file mode 100755 index 0000000..d92ddd7 --- /dev/null +++ b/problems/python/remove-nth-node-from-end-of-list.py @@ -0,0 +1,25 @@ +class Solution(object): + def removeNthFromEnd(self, head, n): + def getCount(node0): + curr = node0 + count = 0 + while curr: + curr = curr.next + count += 1 + return count + + def removeNext(node0): + nextNode = node0.next + if not nextNode: return + node0.next = nextNode.next + + k = getCount(head)-n-1 #need to "curr = curr.next" k times to reach the node that we can call removeNext(curr) + if k==-1: return head.next #remove head + + curr = head + while k>0: + k -= 1 + curr = curr.next + + removeNext(curr) + return head \ No newline at end of file diff --git a/problems/reorder-list.py b/problems/python/reorder-list.py old mode 100644 new mode 100755 similarity index 100% rename from problems/reorder-list.py rename to problems/python/reorder-list.py diff --git a/problems/repeated-string-match.py b/problems/python/repeated-string-match.py old mode 100644 new mode 100755 similarity index 100% rename from problems/repeated-string-match.py rename to problems/python/repeated-string-match.py diff --git a/problems/python/replace-the-substring-for-balanced-string.py b/problems/python/replace-the-substring-for-balanced-string.py new file mode 100755 index 0000000..0ce1250 --- /dev/null +++ b/problems/python/replace-the-substring-for-balanced-string.py @@ -0,0 +1,15 @@ +class Solution(object): + def balancedString(self, s): + n = len(s) + counter = collections.Counter(s) + i = 0 + ans = n + + for j, c in enumerate(s): + counter[c] -= 1 + + while ival: + node = node.left + else: + node = node.right + + return None + +""" +Time complexity: O(LogN) +Space complexity: O(LogN) +""" +class Solution(object): + def searchBST(self, node, val): + if not node: return None + if node.val==val: + return node + elif node.valval: + return self.searchBST(node.left, val) \ No newline at end of file diff --git a/problems/search-in-rotated-sorted-array-ii.py b/problems/python/search-in-rotated-sorted-array-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/search-in-rotated-sorted-array-ii.py rename to problems/python/search-in-rotated-sorted-array-ii.py diff --git a/problems/search-in-rotated-sorted-array.py b/problems/python/search-in-rotated-sorted-array.py old mode 100644 new mode 100755 similarity index 100% rename from problems/search-in-rotated-sorted-array.py rename to problems/python/search-in-rotated-sorted-array.py diff --git a/problems/search-insert-position.py b/problems/python/search-insert-position.py old mode 100644 new mode 100755 similarity index 100% rename from problems/search-insert-position.py rename to problems/python/search-insert-position.py diff --git a/problems/python/search-suggestions-system.py b/problems/python/search-suggestions-system.py new file mode 100755 index 0000000..46e88b2 --- /dev/null +++ b/problems/python/search-suggestions-system.py @@ -0,0 +1,99 @@ +""" +Time: O(NLogN) for sorting the products, N is the number of product. O(SLogN) for search, S is the number of char of the searchWord. +Space: O(1) + +Sort the products then binary search it. +""" +class Solution(object): + def suggestedProducts(self, products, searchWord): + def search(products, word): + l = 0 + r = len(products)-1 + n = len(word) + + while l<=r: + m = (l+r)/2 + if products[l].startswith(word): return getTop3(products, word, l) + if products[r].startswith(word): return getTop3(products, word, r) + if products[m].startswith(word): return getTop3(products, word, m) + + if products[m][:n]>word: + r = m-1 + else: + l = m+1 + + return [] + + def getTop3(products, word, i): + suggestion = [] + + while i-1>=0 and products[i-1].startswith(word): + i -= 1 + + for j in xrange(i, min(len(products), i+3)): + if not products[j].startswith(word): break + suggestion.append(products[j]) + return suggestion + + + ans = [] + products.sort() + + currSearchWord = '' + for c in searchWord: + currSearchWord += c + ans.append(search(products, currSearchWord)) + + return ans + + + +""" +Time: O(M) + O(S^2) ~= O(M). M is the number of char in products. S is number of char of a searchWord. +O(M) to build trie. O(len(prefix)) (~= O(S/2)) to get to the prefix node, and it will perform S time. O(1) to get top 3. +Space: O(M) + +Use Trie to store the relation and search it. +""" +class Node(object): + def __init__(self, c): + self.c = c + self.nexts = {} + + +class Solution(object): + def suggestedProducts(self, products, searchWord): + ans = [] + root = Node('') + + #build trie + for product in products: + curr = root + for c in product+'.': + if c not in curr.nexts: + curr.nexts[c] = Node(c) + curr = curr.nexts[c] + + # search top3 for each prefix + curr = root + for i, c in enumerate(searchWord): + if c not in curr.nexts: break + curr = curr.nexts[c] + ans.append(self.getTop3(searchWord[:i+1], curr)) + + ans += [[] for _ in xrange(len(searchWord)-len(ans))] + return ans + + def getTop3(self, prefix, node): + def helper(prefix, node): + if len(top3)>=3: return + if '.' in node.nexts: top3.append(prefix) + + for c in 'abcdefghijklmnopqrstuvwxyz': + if c in node.nexts: + helper(prefix+c, node.nexts[c]) + + top3 = [] + helper(prefix, node) + return top3 + diff --git a/problems/second-highest-salary.sql b/problems/python/second-highest-salary.sql old mode 100644 new mode 100755 similarity index 100% rename from problems/second-highest-salary.sql rename to problems/python/second-highest-salary.sql diff --git a/problems/python/sell-diminishing-valued-colored-balls.py b/problems/python/sell-diminishing-valued-colored-balls.py new file mode 100755 index 0000000..e871597 --- /dev/null +++ b/problems/python/sell-diminishing-valued-colored-balls.py @@ -0,0 +1,25 @@ +class Solution(object): + def maxProfit(self, I, orders): + I.sort(reverse=True) + I.append(0) + + profit = 0 + w = 1 + + for i in xrange(len(I)-1): + if orders==0: break + if I[i]>I[i+1]: + if w*(I[i]-I[i+1])=len(data): continue + if data[i]!='#': + node.right = TreeNode(data[i]) + q.append(node.right) + i += 1 + + return root + + + +#Preorder traversal +class Codec: + def serialize(self, root): + if not root: return '#,' + return str(root.val) + ',' + self.serialize(root.left) + self.serialize(root.right) + + def deserialize(self, data): + data = data.split(',')[::-1] + return self.getNode(data) + + def getNode(self, data): + if not data: return None + val = data.pop() + if val=='#': return None + node = TreeNode(val) + node.left = self.getNode(data) + node.right = self.getNode(data) + return node \ No newline at end of file diff --git a/problems/serialize-and-deserialize-bst.py b/problems/python/serialize-and-deserialize-bst.py old mode 100644 new mode 100755 similarity index 100% rename from problems/serialize-and-deserialize-bst.py rename to problems/python/serialize-and-deserialize-bst.py diff --git a/problems/python/set-matrix-zeroes.py b/problems/python/set-matrix-zeroes.py new file mode 100755 index 0000000..94e512d --- /dev/null +++ b/problems/python/set-matrix-zeroes.py @@ -0,0 +1,35 @@ +""" +Space: O(1) +Time: O(N^2) + +Mark matrix[i][0] as string if all matrix[i][?] needs to be set to 0. +Mark matrix[0][j] as string if all matrix[?][j] needs to be set to 0. +There is an edje case where matrix[0][0] cannot represent for row and col at the same time. +So we need an additional variable, firstRowToZero. + +Explanation: https://www.youtube.com/watch?v=T41rL0L3Pnw +""" +class Solution(object): + def setZeroes(self, matrix): + firstRowToZero = False + + for i in xrange(len(matrix)): + for j in xrange(len(matrix[0])): + if matrix[i][j]==0: + if i!=0: + matrix[i][0] = str(matrix[i][0]) + else: + firstRowToZero = True + matrix[0][j] = str(matrix[0][j]) + + + for i in xrange(1, len(matrix)): + if type(matrix[i][0])!=type(''): continue + for j in xrange(len(matrix[0])): matrix[i][j] = 0 + + for j in xrange(len(matrix[0])): + if type(matrix[0][j])!=type(''): continue + for i in xrange(len(matrix)): matrix[i][j] = 0 + + if firstRowToZero: + for j in xrange(len(matrix[0])): matrix[0][j] = 0 \ No newline at end of file diff --git a/problems/shortest-bridge.py b/problems/python/shortest-bridge.py old mode 100644 new mode 100755 similarity index 100% rename from problems/shortest-bridge.py rename to problems/python/shortest-bridge.py diff --git a/problems/shortest-common-supersequence.py b/problems/python/shortest-common-supersequence.py old mode 100644 new mode 100755 similarity index 100% rename from problems/shortest-common-supersequence.py rename to problems/python/shortest-common-supersequence.py diff --git a/problems/python/shortest-distance-from-all-buildings.py b/problems/python/shortest-distance-from-all-buildings.py new file mode 100755 index 0000000..490e6a2 --- /dev/null +++ b/problems/python/shortest-distance-from-all-buildings.py @@ -0,0 +1,39 @@ +class Solution(object): + def shortestDistance(self, grid): + def bfs(i0, j0): + q = collections.deque([(i0, j0, 0)]) + visited = set() + while q: + i, j, dis = q.popleft() + if (i, j) in visited: continue + visited.add((i, j)) + + reach[i][j] += 1 + distances[i][j] += dis + + for iNext, jNext in ((i+1, j), (i-1, j), (i, j+1), (i, j-1)): + if not (0<=iNext=N or i<0 or j>=M or j<0: continue + if grid[i][j]==1 and k0>0: + qNext.append((step+1, i, j, k0-1)) + elif grid[i][j]==0: + qNext.append((step+1, i, j, k0)) + if not q: q = qNext + + return -1 \ No newline at end of file diff --git a/problems/python/shortest-path-in-binary-matrix.py b/problems/python/shortest-path-in-binary-matrix.py new file mode 100755 index 0000000..6811948 --- /dev/null +++ b/problems/python/shortest-path-in-binary-matrix.py @@ -0,0 +1,17 @@ +class Solution(object): + def shortestPathBinaryMatrix(self, grid): + q = collections.deque([(0, 0, 1)]) + seen = 2 + + while q: + i, j, step = q.popleft() + if not (0<=i0: + skip -= 1 + continue + else: + ans = '/'+directory+ans + + return ans if ans!='' else '/' \ No newline at end of file diff --git a/problems/python/single-threaded-cpu.py b/problems/python/single-threaded-cpu.py new file mode 100755 index 0000000..fe30e21 --- /dev/null +++ b/problems/python/single-threaded-cpu.py @@ -0,0 +1,22 @@ +class Solution(object): + def getOrder(self, tasks): + ans = [] + tasks = sorted([(task[0], task[1], i) for i, task in enumerate(tasks)], reverse=True) + pq = [] #tasks available + now = 0 + + + while tasks or pq: + #check if the task is availiable, if yes, add to pq + while tasks and tasks[-1][0]<=now: + startTime, processTime, i = tasks.pop() + heapq.heappush(pq, (processTime, i)) + + if pq: + processTime, i = heapq.heappop(pq) + ans.append(i) + now += processTime + else: + now = tasks[-1][0] + + return ans \ No newline at end of file diff --git a/problems/sliding-window-maximum.py b/problems/python/sliding-window-maximum.py old mode 100644 new mode 100755 similarity index 100% rename from problems/sliding-window-maximum.py rename to problems/python/sliding-window-maximum.py diff --git a/problems/python/snapshot-array.py b/problems/python/snapshot-array.py new file mode 100755 index 0000000..baa9f15 --- /dev/null +++ b/problems/python/snapshot-array.py @@ -0,0 +1,19 @@ +class SnapshotArray(object): + + def __init__(self, length): + self.data = [[(-1, 0)] for _ in xrange(length)] + self.snapId = 0 + + + def set(self, index, val): + self.data[index].append((self.snapId, val)) + + + def snap(self): + self.snapId += 1 + return self.snapId - 1 + + + def get(self, index, snapId): + j = bisect.bisect_right(self.data[index], (snapId, float('inf'))) - 1 + return self.data[index][j][1] \ No newline at end of file diff --git a/problems/sort-colors.py b/problems/python/sort-colors.py old mode 100644 new mode 100755 similarity index 100% rename from problems/sort-colors.py rename to problems/python/sort-colors.py diff --git a/problems/sort-list.py b/problems/python/sort-list.py old mode 100644 new mode 100755 similarity index 100% rename from problems/sort-list.py rename to problems/python/sort-list.py diff --git a/problems/python/spiral-matrix.py b/problems/python/spiral-matrix.py new file mode 100755 index 0000000..96965d2 --- /dev/null +++ b/problems/python/spiral-matrix.py @@ -0,0 +1,42 @@ +class Solution(object): + def spiralOrder(self, matrix): + ans = [] + i, j = 0, 0 + direction = 'right' + count = 0 + + while count<=len(matrix)*len(matrix[0]): + iNext, jNext = i, j + + if direction=='right': + if j+1=0 and matrix[i][j-1]!='v': + jNext = j-1 + else: + direction = 'up' + elif direction=='up': + if i-1>=0 and matrix[i-1][j]!='v': + iNext = i-1 + else: + direction = 'right' + + if (iNext, jNext)!=(i, j) or count>=len(matrix)*len(matrix[0])-1: + ans.append(matrix[i][j]) + matrix[i][j] = 'v' #visited + count += 1 + i = iNext + j = jNext + + return ans[:-1] + + + \ No newline at end of file diff --git a/problems/split-array-into-fibonacci-sequence.py b/problems/python/split-array-into-fibonacci-sequence.py old mode 100644 new mode 100755 similarity index 100% rename from problems/split-array-into-fibonacci-sequence.py rename to problems/python/split-array-into-fibonacci-sequence.py diff --git a/problems/split-array-largest-sum.py b/problems/python/split-array-largest-sum.py old mode 100644 new mode 100755 similarity index 100% rename from problems/split-array-largest-sum.py rename to problems/python/split-array-largest-sum.py diff --git a/problems/sqrtx.py b/problems/python/sqrtx.py old mode 100644 new mode 100755 similarity index 100% rename from problems/sqrtx.py rename to problems/python/sqrtx.py diff --git a/problems/squares-of-a-sorted-array.py b/problems/python/squares-of-a-sorted-array.py old mode 100644 new mode 100755 similarity index 100% rename from problems/squares-of-a-sorted-array.py rename to problems/python/squares-of-a-sorted-array.py diff --git a/problems/python/step-by-step-directions-from-a-binary-tree-node-to-another.py b/problems/python/step-by-step-directions-from-a-binary-tree-node-to-another.py new file mode 100755 index 0000000..a9177d4 --- /dev/null +++ b/problems/python/step-by-step-directions-from-a-binary-tree-node-to-another.py @@ -0,0 +1,36 @@ +class Solution(object): + def __init__(self): + self.lca = None + + def getDirections(self, root, startValue, destValue): + def findPath(node, path, val): + if not node: return False + if node.val==val: return True + + path.append('L') + if findPath(node.left, path, val): return True + path.pop() + + path.append('R') + if findPath(node.right, path, val): return True + path.pop() + + return False + + def findCount(node, startValue, destValue): + if not node: return 0 + count = 0 + if node.val==startValue or node.val==destValue: count += 1 + count += findCount(node.left, startValue, destValue) + count += findCount(node.right, startValue, destValue) + if count>=2 and not self.lca: self.lca = node + return count + + findCount(root, startValue, destValue) + + path1 = [] + findPath(self.lca, path1, startValue) + path2 = [] + findPath(self.lca, path2, destValue) + + return 'U'*len(path1) + ''.join(path2) \ No newline at end of file diff --git a/problems/python/stock-price-fluctuation.py b/problems/python/stock-price-fluctuation.py new file mode 100755 index 0000000..e75beb2 --- /dev/null +++ b/problems/python/stock-price-fluctuation.py @@ -0,0 +1,37 @@ +""" +[0] priceToTime tracks price to timestamps, but since for each timestamp the price might be overridden, the price in priceToTime might not be exsit. So we need to check if the price still have any valid timestamps. +[1] SortedDict is a BST like structure. + +Time: O(LogN) +Space: O(N) +""" +from sortedcontainers import SortedDict #[1] + +class StockPrice(object): + + def __init__(self): + self.timeToPrice = SortedDict() #time will be sorted + self.priceToTime = SortedDict() #the price will be sorted + + + def update(self, timestamp, price): + if timestamp in self.timeToPrice: + prevPrice = self.timeToPrice[timestamp] + self.priceToTime[prevPrice].remove(timestamp) + if len(self.priceToTime[prevPrice])==0: self.priceToTime.pop(prevPrice) #[0] + + if price not in self.priceToTime: self.priceToTime[price] = set() #initialized + self.priceToTime[price].add(timestamp) + self.timeToPrice[timestamp] = price + + + def current(self): + return self.timeToPrice.peekitem(-1)[1] + + + def maximum(self): + return self.priceToTime.peekitem(-1)[0] + + + def minimum(self): + return self.priceToTime.peekitem(0)[0] \ No newline at end of file diff --git a/problems/stone-game-ii.py b/problems/python/stone-game-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/stone-game-ii.py rename to problems/python/stone-game-ii.py diff --git a/problems/python/student-attendance-record-ii.py b/problems/python/student-attendance-record-ii.py new file mode 100755 index 0000000..efc2923 --- /dev/null +++ b/problems/python/student-attendance-record-ii.py @@ -0,0 +1,27 @@ +class Solution(object): + def checkRecord(self, n): + """ + dp[l] := number of eligible combination of a length l record without A + """ + + ans = 0 + M = 1000000007 + + dp = [0]*max(n+1, 4) + dp[0] = 1 + dp[1] = 2 + dp[2] = 4 + dp[3] = 7 + + for i in xrange(4, n+1): + dp[i] += dp[i-1]%M #ends at P + dp[i] += (dp[i-1]%M - dp[i-4]%M) #ends at L. All posiblity but the end cannot be PLL + + ans += dp[n] + + for i in xrange(n): + ans += dp[i] * dp[n-i-1] + ans %= M + + return ans + \ No newline at end of file diff --git a/problems/subarray-sum-equals-k.py b/problems/python/subarray-sum-equals-k.py old mode 100644 new mode 100755 similarity index 97% rename from problems/subarray-sum-equals-k.py rename to problems/python/subarray-sum-equals-k.py index 3220cff..0158b04 --- a/problems/subarray-sum-equals-k.py +++ b/problems/python/subarray-sum-equals-k.py @@ -39,7 +39,6 @@ def subarraySum(self, nums, k): """ class Solution(object): def subarraySum(self, nums, k): - N = len(nums) ans = 0 prefixSumCount = collections.Counter() @@ -52,4 +51,4 @@ def subarraySum(self, nums, k): ans += prefixSumCount[I] #there are "prefixSumCount[I]" combinations of J-I that equals to k. prefixSumCount[J] += 1 - return ans \ No newline at end of file + return ans diff --git a/problems/python/subarrays-with-k-different-integers.py b/problems/python/subarrays-with-k-different-integers.py new file mode 100755 index 0000000..5753427 --- /dev/null +++ b/problems/python/subarrays-with-k-different-integers.py @@ -0,0 +1,25 @@ +class Solution(object): + def subarraysWithKDistinct(self, nums, K): + #number of subarray of nums which have at most k different numbers. + def atMost(k): + i = 0 + ans = 0 + counter = collections.Counter() + uniqueCount = 0 + + for j in xrange(len(nums)): + counter[nums[j]] += 1 + if counter[nums[j]]==1: uniqueCount += 1 + + while uniqueCount>k: + counter[nums[i]] -= 1 + if counter[nums[i]]==0: uniqueCount -= 1 + i += 1 + + # the logest subarray that ends at j is nums[i:j+1] + # nums[i:j+1] can produce j-i+1 subarrays that at most has k different number. + ans += j-i+1 + + return ans + + return atMost(K)-atMost(K-1) \ No newline at end of file diff --git a/problems/subdomain-visit-count.py b/problems/python/subdomain-visit-count.py old mode 100644 new mode 100755 similarity index 100% rename from problems/subdomain-visit-count.py rename to problems/python/subdomain-visit-count.py diff --git a/problems/subsets-ii.py b/problems/python/subsets-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/subsets-ii.py rename to problems/python/subsets-ii.py diff --git a/problems/subsets.py b/problems/python/subsets.py old mode 100644 new mode 100755 similarity index 100% rename from problems/subsets.py rename to problems/python/subsets.py diff --git a/problems/substring-with-concatenation-of-all-words.py b/problems/python/substring-with-concatenation-of-all-words.py old mode 100644 new mode 100755 similarity index 99% rename from problems/substring-with-concatenation-of-all-words.py rename to problems/python/substring-with-concatenation-of-all-words.py index b11bb81..e3e6420 --- a/problems/substring-with-concatenation-of-all-words.py +++ b/problems/python/substring-with-concatenation-of-all-words.py @@ -47,4 +47,6 @@ def test(self, s, wl, countExpected): return True - \ No newline at end of file + + + diff --git a/problems/subtree-of-another-tree.py b/problems/python/subtree-of-another-tree.py old mode 100644 new mode 100755 similarity index 100% rename from problems/subtree-of-another-tree.py rename to problems/python/subtree-of-another-tree.py diff --git a/problems/python/sum-of-subarray-minimums.py b/problems/python/sum-of-subarray-minimums.py new file mode 100755 index 0000000..87eabf4 --- /dev/null +++ b/problems/python/sum-of-subarray-minimums.py @@ -0,0 +1,51 @@ +""" +[0] +Instead of listing iterate through all the subarrays (Which is O(N^2) in Time). +We think reversely. +For each num, how many subarrays such that the num is the minimum. +For each num, the count is (i-l) * (r-i) +Where l is the index of the first left number smaller to num. +Where r is the index of the first right number smaller to num. + +[1] +To construct nextSmaller, we iterate through the number, if it is incremental, we put it in the stack. +Once we encounter a number that is "not" incremental, then it is the next smaller number of stack[-1]. + +[2] +Since there are some the same number in the arr. If we use nextSmaller and prevSmaller to calculate, there will be some subarrays repeatly counted. +So we need to set up boundaries. So that even with the same number, they will not count the same subarray. + +Time: O(N) +Space: O(N) +""" +class Solution(object): + def sumSubarrayMins(self, arr): + N = len(arr) + ans = 0 + nextSmaller = [N]*N + prevSmallerOrEqual = [-1]*N #[2] + + #construct nextSmaller [1] + stack = [] + for i in xrange(N): + n = arr[i] + while stack and nendRIndex[i]: + return False + + return True \ No newline at end of file diff --git a/problems/swap-nodes-in-pairs.py b/problems/python/swap-nodes-in-pairs.py old mode 100644 new mode 100755 similarity index 100% rename from problems/swap-nodes-in-pairs.py rename to problems/python/swap-nodes-in-pairs.py diff --git a/problems/swim-in-rising-water.py b/problems/python/swim-in-rising-water.py old mode 100644 new mode 100755 similarity index 100% rename from problems/swim-in-rising-water.py rename to problems/python/swim-in-rising-water.py diff --git a/problems/symmetric-tree.py b/problems/python/symmetric-tree.py old mode 100644 new mode 100755 similarity index 58% rename from problems/symmetric-tree.py rename to problems/python/symmetric-tree.py index 86330bb..bb5c717 --- a/problems/symmetric-tree.py +++ b/problems/python/symmetric-tree.py @@ -33,4 +33,35 @@ def isSymmetric(self, root): # if left_node.val!=right_node.val: return False # stack.append((left_node.right, right_node.left)) # stack.append((left_node.left, right_node.right)) -# return True \ No newline at end of file +# return True + +""" +Time: O(N) +Space: O(N) + +Use 2 DFS to traverse the tree at the same time. +s1 will go left first. +s2 will fo right first. +Make sure every element popping out is the same. +""" +class Solution(object): + def isSymmetric(self, root): + s1 = [root] + s2 = [root] + + while s1 and s2: + node1 = s1.pop() + node2 = s2.pop() + + if not node1 and not node2: continue + if not node1 and node2: return False + if node1 and not node2: return False + if node1.val!=node2.val: return False + + s1.append(node1.left) + s1.append(node1.right) + s2.append(node2.right) + s2.append(node2.left) + + if s1 or s2: return False #if there are something left in one of the stack + return True \ No newline at end of file diff --git a/problems/tallest-billboard.py b/problems/python/tallest-billboard.py old mode 100644 new mode 100755 similarity index 100% rename from problems/tallest-billboard.py rename to problems/python/tallest-billboard.py diff --git a/problems/target-sum.py b/problems/python/target-sum.py old mode 100644 new mode 100755 similarity index 72% rename from problems/target-sum.py rename to problems/python/target-sum.py index 75d9abe..c216522 --- a/problems/target-sum.py +++ b/problems/python/target-sum.py @@ -55,4 +55,23 @@ def findTargetSumWays(self, nums, target): for t in xrange(minTarget, maxTarget+1): dp[i][t] = (dp[i-1][t-nums[i-1]] if t-nums[i-1]>=minTarget else 0) + (dp[i-1][t+nums[i-1]] if t+nums[i-1]<=maxTarget else 0) - return dp[N][target] \ No newline at end of file + return dp[N][target] + + +""" +dp[i][s] := considerting nums[:i] how many expressions can sum up to s. +dp[0][0] = 1 +""" +class Solution(object): + def findTargetSumWays(self, nums, target): + S = sum(nums) + if not -S<=target<=S: return 0 + + dp = [{s:0 for s in xrange(-S, S+1)} for _ in xrange(len(nums)+1)] + dp[0][0] = 1 + + for i in xrange(1, len(nums)+1): + for s in xrange(-S, S+1): + dp[i][s] = (dp[i-1][s+nums[i-1]] if s+nums[i-1]<=S else 0) + (dp[i-1][s-nums[i-1]] if s-nums[i-1]>=-S else 0) + + return dp[-1][target] \ No newline at end of file diff --git a/problems/task-scheduler.py b/problems/python/task-scheduler.py old mode 100644 new mode 100755 similarity index 100% rename from problems/task-scheduler.py rename to problems/python/task-scheduler.py diff --git a/problems/python/text-justification.py b/problems/python/text-justification.py new file mode 100755 index 0000000..98eb0be --- /dev/null +++ b/problems/python/text-justification.py @@ -0,0 +1,68 @@ +""" +The description of this problem is unclear. +"Extra spaces between words should be distributed as evenly as possible." Should be "Extra spaces between words IN EACH LINE should be distributed as evenly as possible." +So the problem in short is: +1. Every line should have as many as word as possible. But they should be separate by space or spaces. +2. Mid align all the line. Except the last line and the lines with one word. +3. For mid align, distribute space evenly, if there are extra spaces, it should be located left. For example 5 spaces distribute to 2 places, it should be 3 2, not 2 3. 7 spaces distribute to 3 places, it should be 3 2 2, not 2 3 2, not 2 2 3. + +Time: O(N), N is the number of words. +Space: O(W), W is maxWidth for keeping strings in currLine. Can further reduce to O(1) using only index. +""" +class Solution(object): + def fullJustify(self, words, maxWidth): + def canAddToCurrLine(word, currLineStringLength): + currWidth = len(currLine)-1 + currLineStringLength # space+currLineStringLength + return currWidth+len(word)+1<=maxWidth + + def leftAlign(currLine): + line = '' + + for word in currLine: + line += word + ' ' + line = line[:-1] #remove last space + + line += ' '*(maxWidth-len(line)) + return line + + def midAlign(currLine, currLineStringLength): + line = '' + + totalSpaceCount = maxWidth-currLineStringLength + extraSpaceCount = totalSpaceCount%(len(currLine)-1) + spaceCount = totalSpaceCount-extraSpaceCount + spaces = ' '*(totalSpaceCount/(len(currLine)-1)) + + for word in currLine: + line += word + if spaceCount>0: + line += spaces + spaceCount -= len(spaces) + + if extraSpaceCount>0: + line += ' ' + extraSpaceCount -= 1 + + return line + + currLineStringLength = 0 + currLine = [] + ans = [] + + i = 0 + while i=N or j<0 or j>=M: continue + if maze[i][j]==1: continue + + if i==destination[0] and j==destination[1] and direction=='stop': return dis + + if direction=='stop': + heapq.heappush(pq, (dis+1, i-1, j, 'left')) + heapq.heappush(pq, (dis+1, i+1, j, 'right')) + heapq.heappush(pq, (dis+1, i, j-1, 'down')) + heapq.heappush(pq, (dis+1, i, j+1, 'up')) + elif direction=='left': + if i-1<0 or i-1>=N or j<0 or j>=M or maze[i-1][j]==1: + heapq.heappush(pq, (dis, i, j, 'stop')) + else: + heapq.heappush(pq, (dis+1, i-1, j, direction)) + elif direction=='right': + if i+1<0 or i+1>=N or j<0 or j>=M or maze[i+1][j]==1: + heapq.heappush(pq, (dis, i, j, 'stop')) + else: + heapq.heappush(pq, (dis+1, i+1, j, direction)) + elif direction=='down': + if i<0 or i>=N or j-1<0 or j-1>=M or maze[i][j-1]==1: + heapq.heappush(pq, (dis, i, j, 'stop')) + else: + heapq.heappush(pq, (dis+1, i, j-1, direction)) + elif direction=='up': + if i<0 or i>=N or j+1<0 or j+1>=M or maze[i][j+1]==1: + heapq.heappush(pq, (dis, i, j, 'stop')) + else: + heapq.heappush(pq, (dis+1, i, j+1, direction)) + + return -1 \ No newline at end of file diff --git a/problems/time-based-key-value-store.py b/problems/python/time-based-key-value-store.py old mode 100644 new mode 100755 similarity index 100% rename from problems/time-based-key-value-store.py rename to problems/python/time-based-key-value-store.py diff --git a/problems/to-lower-case.py b/problems/python/to-lower-case.py old mode 100644 new mode 100755 similarity index 100% rename from problems/to-lower-case.py rename to problems/python/to-lower-case.py diff --git a/problems/python/toeplitz-matrix.py b/problems/python/toeplitz-matrix.py new file mode 100755 index 0000000..431ce27 --- /dev/null +++ b/problems/python/toeplitz-matrix.py @@ -0,0 +1,22 @@ +class Solution(object): + def isToeplitzMatrix(self, matrix): + def check(i0, j0, matrix): + i = i0 + j = j0 + + while 0<=i=k: break + + return ans + + +# Quick Select +class Solution(object): + def topKFrequent(self, nums, K): + def quickselect(A, s, e, K): + i = s + t = s + j = e + + pivot = A[(s+e)/2][0] + while t<=j: + if A[t][0]=K: + return quickselect(A, j+1, e, K) + elif e-(i-1)>=K: + return pivot + else: + return quickselect(A, s, i-1, K-(e-i+1)) + + ans = [] + counter = collections.Counter(nums) + freqs = [(counter[num], num) for num in counter] + + KthLargestFreq = quickselect(freqs, 0, len(freqs)-1, K) + + for freq, num in freqs: + if freq>=KthLargestFreq: ans.append(num) + return ans \ No newline at end of file diff --git a/problems/python/trapping-rain-water-ii.py b/problems/python/trapping-rain-water-ii.py new file mode 100755 index 0000000..dd14bc7 --- /dev/null +++ b/problems/python/trapping-rain-water-ii.py @@ -0,0 +1,28 @@ +class Solution(object): + def trapRainWater(self, heightMap): + pq = [] + N = len(heightMap) + M = len(heightMap[0]) + visited = set() + ans = 0 + curr = float('-inf') + + for i in xrange(N): + for j in xrange(M): + if i==0 or i==N-1 or j==0 or j==M-1: + heapq.heappush(pq, (heightMap[i][j], i, j)) + + while pq: + h, i, j = heapq.heappop(pq) + if (i, j) in visited: continue + visited.add((i, j)) + + if h>curr: curr = h + ans += (curr-h) + + for iNext, jNext in [(i+1, j), (i-1, j), (i, j+1), (i, j-1)]: + if iNext<0 or iNext>=N or jNext<0 or jNext>=M: continue + if (iNext, jNext) in visited: continue + heapq.heappush(pq, (heightMap[iNext][jNext], iNext, jNext)) + + return ans \ No newline at end of file diff --git a/problems/trapping-rain-water.py b/problems/python/trapping-rain-water.py old mode 100644 new mode 100755 similarity index 100% rename from problems/trapping-rain-water.py rename to problems/python/trapping-rain-water.py diff --git a/problems/trim-a-binary-search-tree.py b/problems/python/trim-a-binary-search-tree.py old mode 100644 new mode 100755 similarity index 100% rename from problems/trim-a-binary-search-tree.py rename to problems/python/trim-a-binary-search-tree.py diff --git a/problems/python/two-out-of-three.py b/problems/python/two-out-of-three.py new file mode 100755 index 0000000..8518903 --- /dev/null +++ b/problems/python/two-out-of-three.py @@ -0,0 +1,17 @@ +class Solution(object): + def twoOutOfThree(self, nums1, nums2, nums3): + ans = [] + + counter = collections.Counter() + + for num in list(set(nums1)): + counter[num] += 1 + for num in list(set(nums2)): + counter[num] += 1 + for num in list(set(nums3)): + counter[num] += 1 + + for num in counter: + if counter[num]>=2: ans.append(num) + + return ans \ No newline at end of file diff --git a/problems/two-sum-ii-input-array-is-sorted.py b/problems/python/two-sum-ii-input-array-is-sorted.py old mode 100644 new mode 100755 similarity index 100% rename from problems/two-sum-ii-input-array-is-sorted.py rename to problems/python/two-sum-ii-input-array-is-sorted.py diff --git a/problems/two-sum.py b/problems/python/two-sum.py old mode 100644 new mode 100755 similarity index 100% rename from problems/two-sum.py rename to problems/python/two-sum.py diff --git a/problems/ugly-number-ii.py b/problems/python/ugly-number-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/ugly-number-ii.py rename to problems/python/ugly-number-ii.py diff --git a/problems/ugly-number.py b/problems/python/ugly-number.py old mode 100644 new mode 100755 similarity index 100% rename from problems/ugly-number.py rename to problems/python/ugly-number.py diff --git a/problems/python/umber-of-islands-ii.py b/problems/python/umber-of-islands-ii.py new file mode 100755 index 0000000..6879201 --- /dev/null +++ b/problems/python/umber-of-islands-ii.py @@ -0,0 +1,56 @@ +class Solution(object): + def numIslands2(self, m, n, positions): + ans = [] + uf = UnionFind(m*n) + + for x, y in positions: + i = x*n+y + if uf.isLand(i): + ans.append(uf.count) + continue + + connected = [] + if (x-1>=0 and uf.isLand((x-1)*n+y)): connected.append((x-1)*n+y) + if (x+1=0 and uf.isLand(x*n+y-1)): connected.append(x*n+y-1) + if (y+1=0 + + def setLand(self, i): + self.parents[i] = i + self.count += 1 + + def find(self, i): + if self.parents[i]!=i: self.parents[i] = self.find(self.parents[i]) + return self.parents[i] + + def union(self, i1, i2): + parent1 = self.find(i1) + parent2 = self.find(i2) + + if parent1!=parent2: + if self.ranks[parent1]>self.ranks[parent2]: + self.parents[parent2] = parent1 + elif self.ranks[parent1]maxDots: return False + if digitCount==0: return False + + return True + + #get e's position. Also check if all char is in validChar + validChar = set(['1','2', '3', '4', '5', '6', '7', '8', '9', '0', 'e', 'E', '.', '+', '-']) + eCount = 0 + ePos = 0 + for i, c in enumerate(s): + if c not in validChar: + return False + if c=='e' or c=='E': + eCount += 1 + ePos = i + + if eCount>1: + return False + elif eCount==1: + return isOK(s[:ePos], 1) and isOK(s[ePos+1:], 0) + else: + return isOK(s, 1) diff --git a/problems/python/valid-palindrome-ii.py b/problems/python/valid-palindrome-ii.py new file mode 100755 index 0000000..84bce52 --- /dev/null +++ b/problems/python/valid-palindrome-ii.py @@ -0,0 +1,29 @@ +""" +When seeing the first different chars in the string. +Check if anyone of the string is palindrome if we remove one of them. + +Time: O(N) +Space: O(1) +""" +class Solution(object): + def validPalindrome(self, s): + def isPalindrome(s): + i = 0 + j = len(s)-1 + while i=max_val:return False if not helper(node.left, min_val, node.val): return False if not helper(node.right, node.val, max_val): return False return True - return helper(root, float('-inf'), float('inf')) + return helper(node, float('-inf'), float('inf')) #iterative - def isValidBST(self, root): - if root==None: return True - stack = [(root, float('-inf'), float('inf'))] + def isValidBST(self, node): + if node==None: return True + stack = [(node, float('-inf'), float('inf'))] while stack: node, min_val, max_val = stack.pop() if node.val<=min_val or node.val>=max_val: @@ -51,18 +51,47 @@ def isValidBST(self, root): return True #in-order traversal - def isValidBST(self, root): + def isValidBST(self, node): stack = [] last_val = float('-inf') - while root or stack: - while root: - stack.append(root) - root = root.left - root = stack.pop() + while node or stack: + while node: + stack.append(node) + node = node.left + node = stack.pop() - if root.val<=last_val: + if node.val<=last_val: return False - last_val = root.val - root = root.right #if root.right: root = root.right + last_val = node.val + node = node.right #if node.right: node = node.right return True + + + +#2021/9/14 +""" +Time: O(N) +Space: O(N) + +Use in-order traversal to check. If the bst is valid, the node.val should always be larger than the prev. +""" +class Solution(object): + def isValidBST(self, root): + node = root + stack = [] + prev = float('-inf') + + while node or stack: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + + if prev>=node.val: return False + prev = node.val + + node = node.right + + return True \ No newline at end of file diff --git a/problems/python/verify-preorder-serialization-of-a-binary-tree.py b/problems/python/verify-preorder-serialization-of-a-binary-tree.py new file mode 100755 index 0000000..39930bf --- /dev/null +++ b/problems/python/verify-preorder-serialization-of-a-binary-tree.py @@ -0,0 +1,14 @@ +""" +Time: O(N) +Space: O(N) for creating the list preorder.split(',') +""" +class Solution(object): + def isValidSerialization(self, preorder): + slot = 1 + + for c in preorder.split(','): + slot -= 1 #each elemet consumes a slot + if slot<0: return False + if c!='#': slot += 2 #each non-null node also create 2 slot + + return slot==0 #all slots should be fill \ No newline at end of file diff --git a/problems/python/vertical-order-traversal-of-a-binary-tree.py b/problems/python/vertical-order-traversal-of-a-binary-tree.py new file mode 100755 index 0000000..bda15ad --- /dev/null +++ b/problems/python/vertical-order-traversal-of-a-binary-tree.py @@ -0,0 +1,67 @@ +""" +The description wants us to report from left to right, top to bottom. +Thus, we keep add value into `temp` with `x_position` as the key and `(y_position, node.val)` as the value. +And maintain `min_x`, `max_x` at the same time. So we know how to iterate the `temp`. +""" +from collections import defaultdict + +class Solution(object): + def verticalTraversal(self, root): + temp = defaultdict(list) + min_x = 0 + max_x = 0 + + stack = [] + stack.append((root, 0, 0)) + while stack: + node, x, y = stack.pop() + temp[x].append((y, node.val)) #append the value of height, so we can sort by height later on + min_x = min(min_x, x) + max_x = max(max_x, x) + if node.left: stack.append((node.left, x-1, y+1)) + if node.right: stack.append((node.right, x+1, y+1)) + + opt = [] + for i in range(min_x, max_x+1): + opt.append([v for y, v in sorted(temp[i])]) #the temp[i] will be sorted by y_position then sorted by node.val + + return opt + + +""" +Time: O(Nk) in average case. +dfs takes O(N). +Double for-loop take nearly O(N), in the double for-loop we need to spend some time on sorting. +But the number of elements in the same column and row is very very small. Lets say, O(k) + +Space: O(N) +""" +class Solution(object): + def verticalTraversal(self, root): + def dfs(node, x, y): + if not node: return + self.maxX = max(self.maxX, x) + self.minX = min(self.minX, x) + self.minY = min(self.minY, y) + data[(x, y)].append(node.val) + dfs(node.left, x-1, y-1) + dfs(node.right, x+1, y-1) + + data = collections.defaultdict(list) + self.maxX = float('-inf') + self.minX = float('inf') + self.minY = float('inf') + + dfs(root, 0, 0) + + ans = [] + for x in xrange(self.minX, self.maxX+1): + temp = [] + for y in xrange(0, self.minY-1, -1): + if (x, y) not in data: continue + if len(data[(x, y)])>1: + temp += sorted(data[(x, y)]) + else: + temp.append(data[(x, y)][0]) + ans.append(temp) + return ans \ No newline at end of file diff --git a/problems/wiggle-subsequence.py b/problems/python/wiggle-subsequence.py old mode 100644 new mode 100755 similarity index 100% rename from problems/wiggle-subsequence.py rename to problems/python/wiggle-subsequence.py diff --git a/problems/word-break.py b/problems/python/word-break.py old mode 100644 new mode 100755 similarity index 100% rename from problems/word-break.py rename to problems/python/word-break.py diff --git a/problems/word-ladder-ii.py b/problems/python/word-ladder-ii.py old mode 100644 new mode 100755 similarity index 100% rename from problems/word-ladder-ii.py rename to problems/python/word-ladder-ii.py diff --git a/problems/word-ladder.py b/problems/python/word-ladder.py old mode 100644 new mode 100755 similarity index 100% rename from problems/word-ladder.py rename to problems/python/word-ladder.py diff --git a/problems/python/word-search-ii.py b/problems/python/word-search-ii.py new file mode 100755 index 0000000..e62eba3 --- /dev/null +++ b/problems/python/word-search-ii.py @@ -0,0 +1,96 @@ +# TLE +class Solution(object): + def findWords(self, board, words): + def getNeighbor(i, j): + opt = [] + if i+1=0: opt.append((i-1, j)) + if j-1>=0: opt.append((i, j-1)) + return opt + + def dfs(i, j, l, word): + if word in found: return + if board[i][j]!=word[l]: return + + if l==len(word)-1 and board[i][j]==word[l]: + opt.append(word) + found.add(word) + return + + char = board[i][j] + board[i][j] = '#' + + for ni, nj in getNeighbor(i, j): + dfs(ni, nj, l+1, word) + + board[i][j] = char + + opt = [] + M = len(board) + N = len(board[0]) + found = set() + + for i in xrange(M): + for j in xrange(N): + for word in words: + if word not in found: + dfs(i, j, 0, word) + return opt + + +board = [ + ['o','a','a','n'], + ['e','t','a','e'], + ['i','h','k','r'], + ['i','f','l','v'] +] +words = ["oath","pea","eat","rain"] + +print Solution().findWords(board, words) + + + + +class Solution(object): + def findWords(self, board, words): + def dfs(i, j, parent): + c = board[i][j] + node = parent[c] + + if '.' in node: ans.append(node.pop('.')) + + board[i][j] = '#' + + for (rowOffset, colOffset) in [(-1, 0), (0, 1), (1, 0), (0, -1)]: + newRow, newCol = i + rowOffset, j + colOffset + if newRow<0 or newRow>=len(board) or newCol<0 or newCol>=len(board[0]): continue + if not board[newRow][newCol] in node: continue + dfs(newRow, newCol, node) + + board[i][j] = c + if not node: parent.pop(c) + + trie = {} + ans = [] + + #build trie + for word in words: + node = trie + for c in word: + if c not in node: + node[c] = {} + node = node[c] + node['.'] = word + print trie + + for i in xrange(len(board)): + for j in xrange(len(board[0])): + if board[i][j] in trie: dfs(i, j, trie) + + + return ans + + + + \ No newline at end of file diff --git a/problems/word-search.py b/problems/python/word-search.py old mode 100644 new mode 100755 similarity index 100% rename from problems/word-search.py rename to problems/python/word-search.py diff --git a/problems/python3/3sum-closest.py b/problems/python3/3sum-closest.py new file mode 100755 index 0000000..472c80f --- /dev/null +++ b/problems/python3/3sum-closest.py @@ -0,0 +1,24 @@ +""" +Time: O(N^2) +Space: O(1) +""" +class Solution: + def threeSumClosest(self, nums: List[int], target: int) -> int: + nums.sort() + + N = len(nums) + ans = float('inf') + + for i in range(N): + j = i+1 + k = N-1 + + while jabs(target-total): ans = total + if total>target: + k -= 1 + elif total int: + N = len(nums) + ans = 0 + + nums.sort() + + + for i in range(N): + j = i+1 + k = N-1 + while j List[List[int]]: + ans = [] + nums.sort() + + for i in range(len(nums)): + if 00: break #[1] + + j, k = i+1, len(nums)-1 + while j0: + k -= 1 + elif nums[i]+nums[j]+nums[k]<0: + j += 1 + else: + ans.append((nums[i], nums[j], nums[k])) + while 0 List[List[int]]: + ans = set() + seen = set() + N = len(nums) + + for i, v1 in enumerate(nums): + if v1 in seen: continue + seen.add(v1) + needed = set() + for j, v2 in enumerate(nums[i+1:]): + if v2 in needed: + ans.add(tuple(sorted((v1, v2, -v1-v2)))) + needed.add(-v1-v2) + return ans + + + +class Solution: + def threeSum(self, nums: List[int]) -> List[List[int]]: + ans = [] + + nums.sort() + + for i in range(len(nums)): + if nums[i]>0: break + if i>0 and nums[i]==nums[i-1]: continue + + j = i+1 + k = len(nums)-1 + while j0: + k -= 1 + elif nums[j]+nums[k]+nums[i]<0: + j += 1 + else: + ans.append((nums[i], nums[j], nums[k])) + + while j O(N^3) +Space: O(K) -> O(1) +""" +class Solution: + def fourSum(self, nums: List[int], target: int) -> List[List[int]]: + def kSum(k, start, target): + if k>2: + for i in range(start, len(nums)-k+1): + if i!=start and nums[i]==nums[i-1]: continue + temp.append(nums[i]) + kSum(k-1, i+1, target-nums[i]) + temp.pop() + else: + l, r = start, len(nums)-1 + + while ltarget: + r -= 1 + elif nums[l]+nums[r] Optional[ListNode]: + carry = 0 + dummy = ListNode() + curr = dummy + + while l1 or l2 or carry: + n = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry + if n>=10: + curr.next = ListNode(n-10) + carry = 1 + else: + curr.next = ListNode(n) + carry = 0 + curr = curr.next + if l1: l1 = l1.next + if l2: l2 = l2.next + + return dummy.next \ No newline at end of file diff --git a/problems/python3/alien-dictionary.py b/problems/python3/alien-dictionary.py new file mode 100644 index 0000000..23ca1ee --- /dev/null +++ b/problems/python3/alien-dictionary.py @@ -0,0 +1,36 @@ +class Solution: + def alienOrder(self, words: List[str]) -> str: + adj = collections.defaultdict(list) + inbounds = collections.Counter() + q = collections.deque() + ans = '' + + adj = {c: set() for word in words for c in word} + for i in range(len(words)-1): + w1, w2 = words[i], words[i+1] + minLen = min(len(w1), len(w2)) + if w1[:minLen]==w2[:minLen] and len(w1)>len(w2): return "" + + for j in range(minLen): + if w1[j]!=w2[j]: + adj[w1[j]].add(w2[j]) + break + + for c in adj: + for nc in list(adj[c]): + inbounds[nc] += 1 + + for c in adj: + if inbounds[c]==0: q.append(c) + + while q: + c = q.popleft() + + ans += c + + for nc in adj[c]: + inbounds[nc] -= 1 + if inbounds[nc]==0: q.append(nc) + + return ans if len(ans)==len(adj) else '' + diff --git a/problems/python3/balanced-binary-tree.py b/problems/python3/balanced-binary-tree.py new file mode 100755 index 0000000..65c749e --- /dev/null +++ b/problems/python3/balanced-binary-tree.py @@ -0,0 +1,18 @@ +""" +Time: O(N) +Space: O(LogN) Recursion stack. If the tree is balanced. +""" +class Solution: + def isBalanced(self, root: Optional[TreeNode]) -> bool: + def getHeight(node) -> (bool, int): + if not node: return True, 0 + leftIsBalanced, leftHeight = getHeight(node.left) + if not leftIsBalanced: return False, 0 + + rightIsBalanced, rightHeight = getHeight(node.right) + if not rightIsBalanced: return False, 0 + + return abs(leftHeight-rightHeight)<2, 1+max(leftHeight, rightHeight) + + isBalanced, h = getHeight(root) + return isBalanced \ No newline at end of file diff --git a/problems/python3/best-time-to-buy-and-sell-stock-with-cooldown.py b/problems/python3/best-time-to-buy-and-sell-stock-with-cooldown.py new file mode 100644 index 0000000..06644e3 --- /dev/null +++ b/problems/python3/best-time-to-buy-and-sell-stock-with-cooldown.py @@ -0,0 +1,16 @@ +#dp[i][0] := max profit when last action is buy +#dp[i][1] := max profit when last action is sell +class Solution: + def maxProfit(self, prices: List[int]) -> int: + if not prices or len(prices)<=1: return 0 + + N = len(prices) + dp = [[0, 0] for _ in range(N)] + dp[0] = [-prices[0], 0] + dp[1][0] = max(-prices[1], -prices[0]) + dp[1][1] = max(prices[1]+dp[0][0], dp[0][1]) + + for i in range(2, N): + dp[i][0] = max(dp[i-2][1]-prices[i], dp[i-1][0]) + dp[i][1] = max(prices[i]+dp[i-1][0], dp[i-1][1]) + return max(dp[-1][0], dp[-1][1], 0) \ No newline at end of file diff --git a/problems/python3/best-time-to-buy-and-sell-stock.py b/problems/python3/best-time-to-buy-and-sell-stock.py new file mode 100644 index 0000000..f04ad91 --- /dev/null +++ b/problems/python3/best-time-to-buy-and-sell-stock.py @@ -0,0 +1,10 @@ +class Solution: + def maxProfit(self, prices: List[int]) -> int: + minPrice = prices[0] + ans = 0 + + for i in range(1, len(prices)): + ans = max(ans, prices[i]-minPrice) + minPrice = min(prices[i], minPrice) + + return ans \ No newline at end of file diff --git a/problems/python3/binary-search.py b/problems/python3/binary-search.py new file mode 100644 index 0000000..2e65a63 --- /dev/null +++ b/problems/python3/binary-search.py @@ -0,0 +1,16 @@ +class Solution: + def search(self, nums: List[int], target: int) -> int: + l = 0 + r = len(nums)-1 + + while l<=r: + m = l + int((r-l)/2) + + if nums[m]>target: + r = m-1 + elif nums[m] List[List[int]]: + def helper(node, level): + if not node: return + if level==len(ans): ans.append([]) + ans[level].append(node.val) + helper(node.left, level+1) + helper(node.right, level+1) + + ans = [] + helper(root, 0) + + return ans + + +""" +Time: O(N), since we need to go through all the nodes. +Space: O(N) for queue. +""" +class Solution: + def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: + if not root: return [] + + q = collections.deque([(root, 0)]) + ans = [] + + while q: + node, level = q.popleft() + + if level==len(ans): ans.append([]) + ans[level].append(node.val) + if node.left: q.append((node.left, level+1)) + if node.right: q.append((node.right, level+1)) + + return ans \ No newline at end of file diff --git a/problems/python3/binary-tree-maximum-path-sum.py b/problems/python3/binary-tree-maximum-path-sum.py new file mode 100644 index 0000000..e024506 --- /dev/null +++ b/problems/python3/binary-tree-maximum-path-sum.py @@ -0,0 +1,27 @@ +""" +helper(node) will return the max path sum from the node to the leaf. [0] + +Along the way we update the ans. [1] +The path of ans must pass through one of the node in the tree. +`node.val+left+right` means the max path sum that pass through the node. +So after all `helper()` is executed, we tested all the nodes. + +Since there are some negative values in the tree, each node can decided not to take the negative path sum into account. [3] + +Time: O(N) +Space: O(LogN) for the recursive stack (if the tree is balanced). +""" +class Solution: + def maxPathSum(self, root: Optional[TreeNode]) -> int: + def helper(node): + nonlocal ans + + if not node: return 0 + left = max(helper(node.left), 0) #[3] + right = max(helper(node.right), 0) #[3] + ans = max(ans, node.val+left+right) #[1] + return node.val+max(left, right) #[0] + + ans = float('-inf') + helper(root) + return ans \ No newline at end of file diff --git a/problems/python3/binary-tree-right-side-view.py b/problems/python3/binary-tree-right-side-view.py new file mode 100755 index 0000000..53b4d13 --- /dev/null +++ b/problems/python3/binary-tree-right-side-view.py @@ -0,0 +1,18 @@ +""" +Time: O(N) +Space: O(LogN) if the tree is balanced. + +BFS is a more intuitive way to do this, but the time complexity will be at least O(N). +Using recursion instead, will take O(LogN) for the recursion stack size. +""" +class Solution: + def rightSideView(self, root: Optional[TreeNode]) -> List[int]: + def helper(node, level): + if not node: return + if len(ans) int: + def dfs(l, r)->int: + if l>r: return 0 + if (l, r) in dp: return dp[(l, r)] + + dp[(l, r)] = 0 + for i in range(l, r+1): + dp[(l, r)] = max(dp[(l, r)], nums[l-1]*nums[i]*nums[r+1] + dfs(l, i-1) + dfs(i+1, r)) + return dp[(l, r)] + + nums = [1]+nums+[1] + dp = {} + return dfs(1, len(nums)-2) \ No newline at end of file diff --git a/problems/python3/car-fleet.py b/problems/python3/car-fleet.py new file mode 100644 index 0000000..24ccba8 --- /dev/null +++ b/problems/python3/car-fleet.py @@ -0,0 +1,16 @@ +class Solution: + def carFleet(self, target: int, position: List[int], speed: List[int]) -> int: + N = len(position) + + timeInfo = [] + for i in range(N): + timeInfo.append((position[i], (target-position[i])/speed[i])) + timeInfo.sort(reverse=True) + + stack = [] + for _, time in timeInfo: + stack.append(time) + if len(stack)>=2 and stack[-2]>=stack[-1]: + stack.pop() + + return len(stack) \ No newline at end of file diff --git a/problems/python3/cheapest-flights-within-k-stops.py b/problems/python3/cheapest-flights-within-k-stops.py new file mode 100644 index 0000000..4595802 --- /dev/null +++ b/problems/python3/cheapest-flights-within-k-stops.py @@ -0,0 +1,17 @@ +""" +Bellman-Ford. +Time: O(KE) +""" +class Solution: + def findCheapestPrice(self, N: int, flights: List[List[int]], src: int, dst: int, K: int) -> int: + prices = {n:float('inf') for n in range(N)} + prices[src] = 0 + + for k in range(K+1): + temp = prices.copy() + for source, destination, price in flights: + if prices[source]==float('inf'): continue + if prices[source]+price int: + dp = [0]*(N+1) + dp[0] = 1 + for i in range(len(dp)): + if i-1>=0: + dp[i] += dp[i-1] + if i-2>=0: + dp[i] += dp[i-2] + return dp[-1] \ No newline at end of file diff --git a/problems/python3/clone-graph.py b/problems/python3/clone-graph.py new file mode 100644 index 0000000..9a6765f --- /dev/null +++ b/problems/python3/clone-graph.py @@ -0,0 +1,16 @@ +class Solution: + def cloneGraph(self, start: 'Node') -> 'Node': + def dfs(node): + if node in clones: return clones[node] + + copy = Node(node.val) + clones[node] = copy + + for neighbor in node.neighbors: + copy.neighbors.append(dfs(neighbor)) + + return clones[node] + if not start: return start + clones = {} + dfs(start) + return clones[start] \ No newline at end of file diff --git a/problems/python3/coin-change-ii.py b/problems/python3/coin-change-ii.py new file mode 100644 index 0000000..8b4ad26 --- /dev/null +++ b/problems/python3/coin-change-ii.py @@ -0,0 +1,12 @@ +class Solution: + def change(self, amount: int, coins: List[int]) -> int: + dp = [0]*(amount+1) + dp[0] = 1 + + coins.sort() + + for coin in coins: + for a in range(1, amount+1): + if a-coin<0: continue + dp[a] += dp[a-coin] + return dp[-1] \ No newline at end of file diff --git a/problems/python3/coin-change.py b/problems/python3/coin-change.py new file mode 100644 index 0000000..19571e4 --- /dev/null +++ b/problems/python3/coin-change.py @@ -0,0 +1,14 @@ +class Solution: + def coinChange(self, coins: List[int], amount: int) -> int: + dp = [float('inf')]*(amount+1) + dp[0] = 0 + for coin in coins: + if coin List[List[int]]: + def helper(i, target): + if target==0: + ans.append(combination.copy()) + return + + if i==len(candidates) or candidates[i]>target: + return + + combination.append(candidates[i]) + helper(i+1, target-candidates[i]) + combination.pop() + + while i+1 List[List[int]]: + def helper(i, currSum, target): + if currSum>target: + return + + if currSum==target: + ans.append(combination.copy()) + return + + for j in range(i, len(candidates)): + combination.append(candidates[j]) + helper(j, currSum+candidates[j], target) + combination.pop() + + ans = [] + combination = [] + helper(0, 0, target) + return ans \ No newline at end of file diff --git a/problems/python3/construct-binary-tree-from-preorder-and-inorder-traversal.py b/problems/python3/construct-binary-tree-from-preorder-and-inorder-traversal.py new file mode 100644 index 0000000..eb7f19c --- /dev/null +++ b/problems/python3/construct-binary-tree-from-preorder-and-inorder-traversal.py @@ -0,0 +1,34 @@ +""" +preorder: [root][left][right] +inorder: [left][root][right] + +[0] +As you can see, the root is always located at the first index of the preorder list. + +[2] +Using the same logic, we can recursively get the left and right node +Inorder to do that, we need to define left and right nodes' range. +And we can get the length of the left subtree by locating the index of the root in the in order list. [1] +i and j is the startIndex and the (endIndex+1) of the preorder list. +k and l is the startIndex and the (endIndex+1) of the inorder list. + +Time: O(N) +Space: O(N) +""" +class Solution: + def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]: + def helper(i, j, k, l): + if i==j or k==l: return None + + root = TreeNode(preorder[i]) #[0] + + rootInorderIndex = getInorderIndex[root.val] #[1] + leftLen = rootInorderIndex-k + + root.left = helper(i+1, i+1+leftLen, k, k+leftLen) #[2] + root.right = helper(i+1+leftLen, j, rootInorderIndex+1, l) + return root + + getInorderIndex = {} + for i, v in enumerate(inorder): getInorderIndex[v] = i + return helper(0, len(preorder), 0, len(inorder)) \ No newline at end of file diff --git a/problems/python3/container-with-most-water.py b/problems/python3/container-with-most-water.py new file mode 100755 index 0000000..a560441 --- /dev/null +++ b/problems/python3/container-with-most-water.py @@ -0,0 +1,42 @@ +""" +Time: O(N) +Space: O(1) + +i and j starts from the leftest and rightest. Move the one that has less height. + +Why we won't miss any i and j? +For example, currently i is heigher than j. +If between i~j, there are no height that is larger or equal to i, than since the area is `min(height[i], height[j]) * (j-i)`, you cannot find any area that is larger than the current one. +If between i~j, there is a height that is larger or equal to i, j will on it, and it will be tested. +Thus, given any i and j, any other future i and j that have the potential of forming larger area will be tested. +""" +class Solution: + def maxArea(self, height: List[int]) -> int: + i = 0 + j = len(height)-1 + ans = 0 + + while iheight[j]: + j -= 1 + else: + i += 1 + return ans + + + +class Solution: + def maxArea(self, height: List[int]) -> int: + i = 0 + j = len(height)-1 + ans = 0 + + while iheight[j]: + j -= 1 + else: + i += 1 + return ans \ No newline at end of file diff --git a/problems/python3/contains-duplicate.py b/problems/python3/contains-duplicate.py new file mode 100644 index 0000000..7849e84 --- /dev/null +++ b/problems/python3/contains-duplicate.py @@ -0,0 +1,7 @@ +class Solution: + def containsDuplicate(self, nums: List[int]) -> bool: + seen = set() + for num in nums: + if num in seen: return True + seen.add(num) + return False \ No newline at end of file diff --git a/problems/python3/copy-list-with-random-pointer.py b/problems/python3/copy-list-with-random-pointer.py new file mode 100644 index 0000000..4436b78 --- /dev/null +++ b/problems/python3/copy-list-with-random-pointer.py @@ -0,0 +1,34 @@ +""" +Time: O(N) +Space: O(1) + +The easiest way would be maintaining a hash map for original node to the copy and the other way around. Then the rest is easy. +But this will take us O(N) of extra space. + +Two pass solution with constant space. +First pass. +Create a copy of the original and store it inside "random". +The copy points to original's next and original's random. + +Second pass. +Iterate through the nodes again. +This time we adjust the copy to point to the other copies. +""" +class Solution: + def copyRandomList(self, head: 'Optional[Node]') -> 'Optional[Node]': + if not head: return head + + node = head + while node: + copy = Node(node.val, node.next, node.random) + node.random = copy + node = node.next + + node = head + while node: + newNode = node.random + if node.next: newNode.next = node.next.random + if newNode.random: newNode.random = newNode.random.random + node = node.next + + return head.random \ No newline at end of file diff --git a/problems/python3/count-good-nodes-in-binary-tree.py b/problems/python3/count-good-nodes-in-binary-tree.py new file mode 100755 index 0000000..2288d12 --- /dev/null +++ b/problems/python3/count-good-nodes-in-binary-tree.py @@ -0,0 +1,16 @@ +""" +Time: O(N) +Space: O(LogN) for recursion stack if the tree is balanced. +""" +class Solution: + def goodNodes(self, root: TreeNode) -> int: + def helper(node, maxVal): + nonlocal count + if not node: return + if node.val>=maxVal: count += 1 + helper(node.left, max(maxVal, node.val)) + helper(node.right, max(maxVal, node.val)) + + count = 0 + helper(root, float('-inf')) + return count \ No newline at end of file diff --git a/problems/python3/counting-bits.py b/problems/python3/counting-bits.py new file mode 100644 index 0000000..7325596 --- /dev/null +++ b/problems/python3/counting-bits.py @@ -0,0 +1,9 @@ +class Solution: + def countBits(self, n: int) -> List[int]: + ans = [0]*(n+1) + offset = 1 + + for i in range(1, n+1): + if offset*2==i: offset = offset*2 + ans[i] = 1+ans[i-offset] + return ans \ No newline at end of file diff --git a/problems/python3/course-schedule-ii.py b/problems/python3/course-schedule-ii.py new file mode 100644 index 0000000..00725b4 --- /dev/null +++ b/problems/python3/course-schedule-ii.py @@ -0,0 +1,30 @@ +#Topological Sort +class Solution: + def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]: + sortedCourse = [] + G = collections.defaultdict(list) #graph + inbounds = collections.Counter() + q = collections.deque() + + #build graph + for c1, c2 in prerequisites: + G[c2].append(c1) + inbounds[c1] += 1 + + #add the starting point to the q. (the ones that have 0 inbounds) + for course in range(numCourses): + if inbounds[course]==0: q.append(course) + + #add the course that have 0 inbounds to the sortedCourse. + #after that, imagine we remove it from the graph, so nextCourse inbound will -1 + #add to q if nextCourse have 0 inbounds + while q: + course = q.popleft() + + sortedCourse.append(course) + + for nextCourse in G[course]: + inbounds[nextCourse] -= 1 + if inbounds[nextCourse]==0: q.append(nextCourse) + + return sortedCourse if len(sortedCourse)==numCourses else [] \ No newline at end of file diff --git a/problems/python3/course-schedule.py b/problems/python3/course-schedule.py new file mode 100644 index 0000000..f52d07d --- /dev/null +++ b/problems/python3/course-schedule.py @@ -0,0 +1,23 @@ +class Solution: + def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: + sortedCourse = [] + inbounds = collections.Counter() + G = collections.defaultdict(list) + q = collections.deque() + + for c1, c2 in prerequisites: + G[c2].append(c1) + inbounds[c1] += 1 + + for c in range(numCourses): + if inbounds[c]==0: q.append(c) + + while q: + c = q.popleft() + + for c2 in G[c]: + inbounds[c2] -= 1 + if inbounds[c2]==0: q.append(c2) + sortedCourse.append(c) + + return len(sortedCourse)==numCourses \ No newline at end of file diff --git a/problems/python3/daily-temperatures.py b/problems/python3/daily-temperatures.py new file mode 100644 index 0000000..d6e4c98 --- /dev/null +++ b/problems/python3/daily-temperatures.py @@ -0,0 +1,12 @@ +class Solution: + def dailyTemperatures(self, temperatures: List[int]) -> List[int]: + ans = [0]*len(temperatures) + stack = [] + + for i, temp in enumerate(temperatures): + while stack and stack[-1][0] int: + mapping = set([str(n) for n in range(1, 27)]) + N = len(s) + dp = [0]*(N+1) + dp[0] = 1 + + for i in range(1, N+1): + if i-1>=0 and s[i-1] in mapping: dp[i] += dp[i-1] + if i-2>=0 and s[i-2:i] in mapping: dp[i] += dp[i-2] + return dp[-1] \ No newline at end of file diff --git a/problems/python3/design-add-and-search-words-data-structure.py b/problems/python3/design-add-and-search-words-data-structure.py new file mode 100644 index 0000000..52effc8 --- /dev/null +++ b/problems/python3/design-add-and-search-words-data-structure.py @@ -0,0 +1,33 @@ +class WordDictionary: + + def __init__(self): + self.root = {} + + def addWord(self, word: str) -> None: + node = self.root + for c in word: + if c not in node: node[c] = {} + node = node[c] + node['$'] = {} #'.' means the end of the word + + def search(self, word: str) -> bool: + def helper(node, i): + if i==len(word): return '$' in node + + if word[i]=='.': + for c in node: + if helper(node[c], i+1): return True + return False + else: + if word[i] not in node: return False + return helper(node[word[i]], i+1) + return helper(self.root, 0) + + + + + + + + + \ No newline at end of file diff --git a/problems/python3/design-twitter.py b/problems/python3/design-twitter.py new file mode 100644 index 0000000..a2781ae --- /dev/null +++ b/problems/python3/design-twitter.py @@ -0,0 +1,49 @@ +class Twitter: + + def __init__(self): + self.followData = collections.defaultdict(set) + self.tweetData = collections.defaultdict(list) + self.count = 0 + + + def postTweet(self, userId: int, tweetId: int) -> None: + self.tweetData[userId].append((self.count, tweetId)) + self.count -= 1 + + + def getNewsFeed(self, userId: int) -> List[int]: + newsFeed = [] + h = [] + + self.followData[userId].add(userId) + for followeeId in self.followData[userId]: + if followeeId not in self.tweetData: continue + index = len(self.tweetData[followeeId])-1 + count, tweetId = self.tweetData[followeeId][index] + h.append((count, tweetId, followeeId, index-1)) + heapq.heapify(h) + + while h and len(newsFeed)<10: + _, tweetId, userId, index = heapq.heappop(h) + newsFeed.append(tweetId) + if index>=0: + count, tweetId2 = self.tweetData[userId][index] + heapq.heappush(h, (count, tweetId2, userId, index-1)) + return newsFeed + + + + def follow(self, followerId: int, followeeId: int) -> None: + self.followData[followerId].add(followeeId) + + def unfollow(self, followerId: int, followeeId: int) -> None: + if followerId not in self.followData: return + self.followData[followerId].remove(followeeId) + + +# Your Twitter object will be instantiated and called as such: +# obj = Twitter() +# obj.postTweet(userId,tweetId) +# param_2 = obj.getNewsFeed(userId) +# obj.follow(followerId,followeeId) +# obj.unfollow(followerId,followeeId) \ No newline at end of file diff --git a/problems/python3/detect-squares.py b/problems/python3/detect-squares.py new file mode 100644 index 0000000..af86946 --- /dev/null +++ b/problems/python3/detect-squares.py @@ -0,0 +1,22 @@ +class DetectSquares: + + def __init__(self): + self.store = collections.Counter() + + def add(self, point: List[int]) -> None: + self.store[tuple(point)] += 1 + + def count(self, point: List[int]) -> int: + x, y = point + ans = 0 + + for dx, dy in self.store: + if abs(x-dx)!=abs(y-dy) or x==dx or y==dy: continue + ans += self.store[(dx, dy)]*self.store[(dx, y)]*self.store[(x, dy)] + return ans + + +# Your DetectSquares object will be instantiated and called as such: +# obj = DetectSquares() +# obj.add(point) +# param_2 = obj.count(point) \ No newline at end of file diff --git a/problems/python3/diameter-of-binary-tree.py b/problems/python3/diameter-of-binary-tree.py new file mode 100755 index 0000000..1f9704d --- /dev/null +++ b/problems/python3/diameter-of-binary-tree.py @@ -0,0 +1,17 @@ +""" +Time: O(N) +Space: O(LogN) for the recursion stack. If the tree is balanced. +""" +class Solution: + def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int: + def helper(node): + if not node: return 0 + l = helper(node.left) + r = helper(node.right) + self.ans = max(self.ans, 1+l+r) + return 1+max(l, r) + + self.ans = 0 + + helper(root) + return self.ans-1 \ No newline at end of file diff --git a/problems/python3/distinct-subsequences.py b/problems/python3/distinct-subsequences.py new file mode 100644 index 0000000..d84fcf6 --- /dev/null +++ b/problems/python3/distinct-subsequences.py @@ -0,0 +1,17 @@ +class Solution: + def numDistinct(self, s: str, t: str) -> int: + def dfs(i, j): + if (i, j) in visited: return visited[(i, j)] + + if j>=len(t): return 1 + if i>=len(s): return 0 + + if s[i]==t[j]: + visited[(i, j)] = dfs(i+1, j+1)+dfs(i+1, j) + else: + visited[(i, j)] = dfs(i+1, j) + return visited[(i, j)] + + visited = {} + dfs(0, 0) + return dfs(0, 0) \ No newline at end of file diff --git a/problems/python3/edit-distance.py b/problems/python3/edit-distance.py new file mode 100644 index 0000000..c8fa8f4 --- /dev/null +++ b/problems/python3/edit-distance.py @@ -0,0 +1,37 @@ +""" +dfs(i, j) +i being the unprocessed index in word1. +j, word2. + +MAIN LOGIC: +if word1[i]==word2[j], no operation need, return dfs(i+1, j+1) +if not, need 1 operation, so +replace: dfs(i+1, j+1) +insert: dfs(i, j+1) +delete: dfs(i+1, j) + +BASE CASE: +If both string are empty (i==N and j==M), no operation needed. +If one string are empty, then the remain operation is the length of the non-empty one. +""" +class Solution: + def minDistance(self, word1: str, word2: str) -> int: + N = len(word1) + M = len(word2) + def dfs(i, j)->int: + if i==N and j==M: return 0 + if i==N: return M-j + if j==M: return N-i + + if (i, j) in history: + return history[(i, j)] + + if word1[i]==word2[j]: + history[(i, j)] = dfs(i+1, j+1) + else: + history[(i, j)] = 1+min(dfs(i+1, j+1), dfs(i+1, j), dfs(i, j+1)) + + return history[(i, j)] + + history = {} + return dfs(0, 0) \ No newline at end of file diff --git a/problems/python3/encode-and-decode-strings.py b/problems/python3/encode-and-decode-strings.py new file mode 100644 index 0000000..f38c4e4 --- /dev/null +++ b/problems/python3/encode-and-decode-strings.py @@ -0,0 +1,23 @@ +class Codec: + def encode(self, strs: List[str]) -> str: + output = '' + for string in strs: + output += str(len(string))+'#'+string + return output + + + def decode(self, s: str) -> List[str]: + output = [] + i = 0 + + while i int: + operators = set(['+', '-', '*', '/']) + stack = [] + + for c in tokens: + if c in operators: + n2 = stack.pop() + n1 = stack.pop() + + if c=='+': + stack.append(n1+n2) + elif c=='-': + stack.append(n1-n2) + elif c=='*': + stack.append(n1*n2) + elif c=='/': + stack.append(int(n1/n2)) + else: + stack.append(int(c)) + + return stack.pop() \ No newline at end of file diff --git a/problems/python3/find-median-from-data-stream.py b/problems/python3/find-median-from-data-stream.py new file mode 100644 index 0000000..f83ee40 --- /dev/null +++ b/problems/python3/find-median-from-data-stream.py @@ -0,0 +1,32 @@ +class MedianFinder: + + def __init__(self): + self.large = [] #store nums larger or equal to the median + self.small = [] #store nums samaller to the median + + def addNum(self, num: int) -> None: + if not self.large and not self.small: + heapq.heappush(self.large, num) + elif num>=self.findMedian(): + heapq.heappush(self.large, num) + self.balance() + else: + heapq.heappush(self.small, -num) + self.balance() + + def balance(self) -> None: + #make the length of two heaps as even as posible + if len(self.large)>len(self.small)+1: + num = heapq.heappop(self.large) + heapq.heappush(self.small, -num) + + if len(self.small)>len(self.large): + num = -heapq.heappop(self.small) + heapq.heappush(self.large, num) + + + def findMedian(self) -> float: + if (len(self.large)+len(self.small))%2==0: + return (self.large[0]-self.small[0])/2 + else: + return self.large[0] \ No newline at end of file diff --git a/problems/python3/find-minimum-in-rotated-sorted-array.py b/problems/python3/find-minimum-in-rotated-sorted-array.py new file mode 100644 index 0000000..c9af305 --- /dev/null +++ b/problems/python3/find-minimum-in-rotated-sorted-array.py @@ -0,0 +1,22 @@ +""" +Usually the binary search problem, the hard part is to clear all the edge cases. +What will be the edge case? It will be when the recursive function executed at the deepest level. Usually the list with only one or two element. +So I will suggest before submit, try out the case like [0,1] or [1,0] to make sure it will not run unstop. +""" +class Solution: + def findMin(self, nums: List[int]) -> int: + def helper(l, r): + nonlocal ans + if l>r: return + + m = l + int((r-l)/2) + if nums[l]<=nums[m]: + ans = min(ans, nums[l]) + helper(m+1, r) + else: + ans = min(ans, nums[m+1]) + helper(l, m) + + ans = float('inf') + helper(0, len(nums)-1) + return ans \ No newline at end of file diff --git a/problems/python3/find-the-duplicate-number.py b/problems/python3/find-the-duplicate-number.py new file mode 100644 index 0000000..41636df --- /dev/null +++ b/problems/python3/find-the-duplicate-number.py @@ -0,0 +1,16 @@ +class Solution: + def findDuplicate(self, nums: List[int]) -> int: + head = nums[0] + slow = head + fast = head + + while True: + slow = nums[slow] + fast = nums[nums[fast]] + if slow==fast: break + + slow = head + while slow!=fast: + slow = nums[slow] + fast = nums[fast] + return slow \ No newline at end of file diff --git a/problems/python3/gas-station.py b/problems/python3/gas-station.py new file mode 100644 index 0000000..017eaa6 --- /dev/null +++ b/problems/python3/gas-station.py @@ -0,0 +1,23 @@ +""" +First thing, you must understand that if sum(gas)>=sum(cost) there must be an answer. Guaranteed. +If we know there is an answer, we can simply test all the index. +If the currGas ever drops to below 0, it means that we need to switch a "start". + +Time: O(N) +Space: O(1) +""" +class Solution: + def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int: + if sum(gas) List[str]: + def helper(parentheses, left, right): + if len(parentheses)==n*2: + ans.append("".join(parentheses)) + return + + if leftright: + parentheses.append(')') + helper(parentheses, left, right+1) + parentheses.pop() + + ans = [] + helper([], 0, 0) + return ans \ No newline at end of file diff --git a/problems/python3/graph-valid-tree.py b/problems/python3/graph-valid-tree.py new file mode 100644 index 0000000..2168e9a --- /dev/null +++ b/problems/python3/graph-valid-tree.py @@ -0,0 +1,33 @@ +class Solution: + def validTree(self, N: int, edges: List[List[int]]) -> bool: + def union(n1, n2) -> bool: + p1 = find(n1) + p2 = find(n2) + + if p1==p2: + return False + elif p1 int: + p = parents[n] + while p!=parents[p]: p = find(p) + parents[n] = p + return p + + parents = [n for n in range(N)] + + for n1, n2 in edges: + if not union(n1, n2): return False + + #check if all node trace back to the same root + root = find(0) + for n in range(1, N): + if root!=find(n): return False + + return True \ No newline at end of file diff --git a/problems/python3/group-anagrams.py b/problems/python3/group-anagrams.py new file mode 100644 index 0000000..7430f48 --- /dev/null +++ b/problems/python3/group-anagrams.py @@ -0,0 +1,23 @@ +class Solution: + def groupAnagrams(self, strs: List[str]) -> List[List[str]]: + def normalize(string: str) -> str: + counter = collections.Counter() + ans = '' + + for c in string: + counter[c] += 1 + + for c in 'abcdefghijklmnopqrstuvwxyz': + if counter[c]>0: + ans += c+str(counter[c]) + return ans + + group = collections.defaultdict(list) + ans = [] + + for string in strs: + group[normalize(string)].append(string) + + for normalizedString in group: + ans.append(group[normalizedString]) + return ans \ No newline at end of file diff --git a/problems/python3/hand-of-straights.py b/problems/python3/hand-of-straights.py new file mode 100644 index 0000000..f20c32d --- /dev/null +++ b/problems/python3/hand-of-straights.py @@ -0,0 +1,18 @@ +class Solution: + def isNStraightHand(self, hand: List[int], groupSize: int) -> bool: + if len(hand)%groupSize!=0: return False + + counter = collections.Counter(hand) + h = list(counter.keys()) + heapq.heapify(h) + + while h: + minNum = h[0] + for n in range(minNum, minNum+groupSize): + if counter[n]<=0: return False + counter[n] -= 1 + if counter[n]==0: + if h[0]!=n: return False + heapq.heappop(h) + + return True \ No newline at end of file diff --git a/problems/python3/happy-number.py b/problems/python3/happy-number.py new file mode 100644 index 0000000..7756197 --- /dev/null +++ b/problems/python3/happy-number.py @@ -0,0 +1,17 @@ +class Solution: + def isHappy(self, n: int) -> bool: + def digitSquare(n) -> int: + ans = 0 + while n>0: + ans += (n%10)**2 + n = n//10 + return ans + + visited = set() + visited.add(1) + + while n not in visited: + visited.add(n) + n = digitSquare(n) + + return n==1 diff --git a/problems/python3/house-robber-ii.py b/problems/python3/house-robber-ii.py new file mode 100644 index 0000000..1ed9b82 --- /dev/null +++ b/problems/python3/house-robber-ii.py @@ -0,0 +1,18 @@ +class Solution: + def rob(self, nums: List[int]) -> int: + if len(nums)<=1: return max(nums) + + N = len(nums) + dp = [[0, 0] for _ in range(N)] + dp[0][0] = nums[0] + + for i in range(1, N): + dp[i][0] = nums[i]+dp[i-1][1] + dp[i][1] = max(dp[i-1]) + + dp2 = [[0, 0] for _ in range(N)] + for i in range(1, N): + dp2[i][0] = nums[i]+dp2[i-1][1] + dp2[i][1] = max(dp2[i-1]) + + return max(dp[-1][1], dp2[-1][0]) \ No newline at end of file diff --git a/problems/python3/house-robber.py b/problems/python3/house-robber.py new file mode 100644 index 0000000..fbd86a5 --- /dev/null +++ b/problems/python3/house-robber.py @@ -0,0 +1,17 @@ +""" +Time: O(N) +Space: O(N), can reduece to O(1). + +dp[i][0] := max revenue if house i robbed +dp[i][1] := max revenue if house i not robbed +""" +class Solution: + def rob(self, nums: List[int]) -> int: + N = len(nums) + dp = [[0, 0] for _ in range(N)] + dp[0][0] = nums[0] + + for i in range(1, N): + dp[i][0] = nums[i]+dp[i-1][1] + dp[i][1] = max(dp[i-1]) + return max(dp[-1]) \ No newline at end of file diff --git a/problems/python3/implement-trie-prefix-tree.py b/problems/python3/implement-trie-prefix-tree.py new file mode 100644 index 0000000..1118e97 --- /dev/null +++ b/problems/python3/implement-trie-prefix-tree.py @@ -0,0 +1,26 @@ +class Trie: + + def __init__(self): + self.root = {} + + def insert(self, word: str) -> None: + node = self.root + for c in word: + if c not in node: node[c] = {} + node = node[c] + node['.'] = {} #'.' means the end of the word + + + def search(self, word: str) -> bool: + node = self.root + for c in word: + if c not in node: return False + node = node[c] + return '.' in node + + def startsWith(self, prefix: str) -> bool: + node = self.root + for c in prefix: + if c not in node: return False + node = node[c] + return True diff --git a/problems/python3/insert-interval.py b/problems/python3/insert-interval.py new file mode 100644 index 0000000..667ba04 --- /dev/null +++ b/problems/python3/insert-interval.py @@ -0,0 +1,23 @@ +class Solution: + def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]: + ans = [] + i = 0 + + #add intervals before newInterval + while i bool: + def dfs(i, j): + if (i, j) in history: return False + if i+j==len(s3): return True + if i Optional[TreeNode]: + if not root: return root + left = root.left + right = root.right + root.left = self.invertTree(right) + root.right = self.invertTree(left) + return root + + +""" +Iterative +Time: O(N) +Space: O(N) +""" +class Solution: + def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]: + if not root: return root + q = collections.deque([root]) + + while q: + node = q.popleft() + left = node.left + right = node.right + node.right = left + node.left = right + + if node.left: q.append(node.left) + if node.right: q.append(node.right) + + return root \ No newline at end of file diff --git a/problems/python3/jump-game-ii.py b/problems/python3/jump-game-ii.py new file mode 100644 index 0000000..35be09f --- /dev/null +++ b/problems/python3/jump-game-ii.py @@ -0,0 +1,22 @@ +""" +l and r is the index range we can reach within "steps". +So while r can not reach the end (`r int: + l = r = 0 + steps = 0 + + while r bool: + maxIndex = 0 + + for i, num in enumerate(nums): + if maxIndex List[List[int]]: + h = [] + + for x, y in points: + dis = (x**2+y**2)**0.5 + heapq.heappush(h, (-dis, x, y)) + if len(h)>k: heapq.heappop(h) + + return [(x, y) for dis, x, y in h] \ No newline at end of file diff --git a/problems/python3/koko-eating-bananas.py b/problems/python3/koko-eating-bananas.py new file mode 100644 index 0000000..fe03197 --- /dev/null +++ b/problems/python3/koko-eating-bananas.py @@ -0,0 +1,21 @@ +class Solution: + def minEatingSpeed(self, piles: List[int], h: int) -> int: + def canFinish(k): + timeNeeded = 0 + for pile in piles: + timeNeeded += math.ceil(pile/k) + return timeNeeded<=h + + kMin = 1 + kMax = max(piles) + ans = kMax + + while kMin<=kMax: + k = kMin + int((kMax-kMin)/2) + if canFinish(k): + ans = min(ans, k) + kMax = k-1 + else: + kMin = k+1 + + return ans \ No newline at end of file diff --git a/problems/python3/kth-largest-element-in-a-stream.py b/problems/python3/kth-largest-element-in-a-stream.py new file mode 100644 index 0000000..1eb9095 --- /dev/null +++ b/problems/python3/kth-largest-element-in-a-stream.py @@ -0,0 +1,14 @@ +class KthLargest: + + def __init__(self, k: int, nums: List[int]): + self.k = k + self.nums = nums + + heapq.heapify(self.nums) + while len(self.nums)>self.k: heapq.heappop(self.nums) + + + def add(self, val: int) -> int: + heapq.heappush(self.nums, val) + if len(self.nums)>self.k: heapq.heappop(self.nums) + return self.nums[0] \ No newline at end of file diff --git a/problems/python3/kth-smallest-element-in-a-bst.py b/problems/python3/kth-smallest-element-in-a-bst.py new file mode 100755 index 0000000..c8a1555 --- /dev/null +++ b/problems/python3/kth-smallest-element-in-a-bst.py @@ -0,0 +1,21 @@ +""" +Time: O(N) for the inorder traversal +Space: O(LogN) if the tree is balanced. +""" +class Solution: + def kthSmallest(self, root: Optional[TreeNode], k: int) -> int: + count = 0 + stack = [] + node = root + + while stack or node: + while node: + stack.append(node) + node = node.left + node = stack.pop() + + count += 1 + if count==k: return node.val + + node = node.right + return 0 \ No newline at end of file diff --git a/problems/python3/largest-rectangle-in-histogram.py b/problems/python3/largest-rectangle-in-histogram.py new file mode 100644 index 0000000..1cf8508 --- /dev/null +++ b/problems/python3/largest-rectangle-in-histogram.py @@ -0,0 +1,21 @@ +""" +[0] For each height, if the it is lower than the previous one, it means that the previous are not able to extend anymore. So we calculate its area. + +[1] If the previous area, is larger than the current one, it means that the current one are able to extand backward. +""" +class Solution: + def largestRectangleArea(self, heights: List[int]) -> int: + maxArea = 0 + stack = [] + + heights.append(0) #dummy for the ending + + for i, h in enumerate(heights): + start = i + while stack and h int: + stones = [-stone for stone in stones] + heapq.heapify(stones) + + while len(stones)>=2: + w1 = -heapq.heappop(stones) + w2 = -heapq.heappop(stones) + + if w1-w2>0: heapq.heappush(stones, -(w1-w2)) + + return -stones[0] if stones else 0 \ No newline at end of file diff --git a/problems/python3/letter-combinations-of-a-phone-number.py b/problems/python3/letter-combinations-of-a-phone-number.py new file mode 100644 index 0000000..582106c --- /dev/null +++ b/problems/python3/letter-combinations-of-a-phone-number.py @@ -0,0 +1,25 @@ +""" +Time: O(4^N * N), there are around 4^N of combination. Each taking O(N) to form. +Space: O(N). O(N) for recursion stack. O(N). O(N) for `combination`. O(N+N) ~= O(N). +""" +class Solution: + def letterCombinations(self, digits: str) -> List[str]: + def helper(i): + if i==len(digits): + ans.append(''.join(combination)) + return + + for c in mapping[digits[i]]: + combination.append(c) + helper(i+1) + combination.pop() + + mapping = {'2': ('a', 'b', 'c'), '3': ('d', 'e', 'f'), + '4': ('g', 'h', 'i'), '5': ('j', 'k', 'l'), '6': ('m', 'n', 'o'), + '7': ('p', 'q', 'r', 's'), '8': ('t', 'u', 'v'), '9': ('w', 'x', 'y', 'z')} + + if not digits: return [] + ans = [] + combination = [] + helper(0) + return ans \ No newline at end of file diff --git a/problems/python3/linked-list-cycle.py b/problems/python3/linked-list-cycle.py new file mode 100644 index 0000000..d2e5395 --- /dev/null +++ b/problems/python3/linked-list-cycle.py @@ -0,0 +1,16 @@ +class Solution: + def hasCycle(self, head: Optional[ListNode]) -> bool: + if not head: return False + + slow = head + fast = head + + while fast: + slow = slow.next + + if not fast.next: return False + fast = fast.next.next + + if slow==fast: return True + + return False \ No newline at end of file diff --git a/problems/python3/longest-common-subsequence.py b/problems/python3/longest-common-subsequence.py new file mode 100644 index 0000000..5ba3a68 --- /dev/null +++ b/problems/python3/longest-common-subsequence.py @@ -0,0 +1,16 @@ +class Solution: + def longestCommonSubsequence(self, text1: str, text2: str) -> int: + #dp[i][j] := number of longest Common Subsequence with text2[:i] and text2[:j] + + N = len(text1) + M = len(text2) + + dp = [[0]*(M+1) for _ in range(N+1)] + + for i in range(1, N+1): + for j in range(1, M+1): + if text1[i-1]==text2[j-1]: + dp[i][j] = dp[i-1][j-1]+1 + else: + dp[i][j] = max(dp[i][j-1], dp[i-1][j]) + return dp[-1][-1] \ No newline at end of file diff --git a/problems/python3/longest-consecutive-sequence.py b/problems/python3/longest-consecutive-sequence.py new file mode 100644 index 0000000..dec2cfa --- /dev/null +++ b/problems/python3/longest-consecutive-sequence.py @@ -0,0 +1,15 @@ +class Solution: + def longestConsecutive(self, nums: List[int]) -> int: + numSet = set(nums) + ans = 0 + + for num in nums: + isStart = num-1 not in numSet + if isStart: + count = 0 + temp = num + while temp in numSet: + count += 1 + temp += 1 + ans = max(count, ans) + return ans \ No newline at end of file diff --git a/problems/python3/longest-increasing-path-in-a-matrix.py b/problems/python3/longest-increasing-path-in-a-matrix.py new file mode 100644 index 0000000..99fed7d --- /dev/null +++ b/problems/python3/longest-increasing-path-in-a-matrix.py @@ -0,0 +1,26 @@ +""" +Time: O(MN) since the memo at most has MN index. +Space: O(MN) +""" +class Solution: + def longestIncreasingPath(self, matrix: List[List[int]]) -> int: + def dfs(i0, j0): + if (i0, j0) in memo: return memo[(i0, j0)] + ans = 1 + + for i, j in ((i0+1, j0), (i0-1, j0), (i0, j0+1),(i0, j0-1)): + if i<0 or i>=N or j<0 or j>=M: continue + if matrix[i][j]<=matrix[i0][j0]: continue + ans = max(ans, 1+dfs(i, j)) + + memo[(i0, j0)] = ans + return ans + + N = len(matrix) + M = len(matrix[0]) + memo = {} + ans = 0 + for i in range(N): + for j in range(M): + ans = max(ans, dfs(i, j)) + return ans \ No newline at end of file diff --git a/problems/python3/longest-palindromic-substring.py b/problems/python3/longest-palindromic-substring.py new file mode 100644 index 0000000..9edb477 --- /dev/null +++ b/problems/python3/longest-palindromic-substring.py @@ -0,0 +1,45 @@ +""" +Time: O(N^2) +Space: O(N^2) + +DP, TLE +""" +class Solution: + def longestPalindrome(self, s: str) -> str: + ans = s[0] + N = len(s) + dp = [[False]*N for _ in range(N)] + + for i in range(N): dp[i][i] = True + + for l in range(2, N+1): + for i in range(N): + j = i+l-1 + if j>=N: continue + dp[i][j] = s[i]==s[j] and (dp[i+1][j-1] or j-1 str: + N = len(s) + ans = s[0] + + for i in range(N): + l, r = i, i + while l>=0 and rlen(ans): ans = s[l:r+1] + l -= 1 + r += 1 + + l, r = i, i+1 + while l>=0 and rlen(ans): ans = s[l:r+1] + l -= 1 + r += 1 + return ans \ No newline at end of file diff --git a/problems/python3/longest-repeating-character-replacement.py b/problems/python3/longest-repeating-character-replacement.py new file mode 100644 index 0000000..714d8f1 --- /dev/null +++ b/problems/python3/longest-repeating-character-replacement.py @@ -0,0 +1,13 @@ +class Solution: + def characterReplacement(self, s: str, k: int) -> int: + counter = collections.Counter() + l = 0 + ans = 0 + + for r in range(len(s)): + counter[s[r]] += 1 + while (r-l+1)-max(counter.values()) > k: + counter[s[l]] -= 1 + l += 1 + ans = max(ans, r-l+1) + return ans \ No newline at end of file diff --git a/problems/python3/longest-substring-without-repeating-char.py b/problems/python3/longest-substring-without-repeating-char.py new file mode 100644 index 0000000..9902bb5 --- /dev/null +++ b/problems/python3/longest-substring-without-repeating-char.py @@ -0,0 +1,14 @@ +class Solution: + def lengthOfLongestSubstring(self, s: str) -> int: + ans = 0 + l = 0 + seen = set() + + for r in range(len(s)): + while s[r] in seen: + seen.remove(s[l]) + l += 1 + seen.add(s[r]) + ans = max(ans, r-l+1) + + return ans \ No newline at end of file diff --git a/problems/python3/lowest-common-ancestor-of-a-binary-search-tree.py b/problems/python3/lowest-common-ancestor-of-a-binary-search-tree.py new file mode 100755 index 0000000..12e1743 --- /dev/null +++ b/problems/python3/lowest-common-ancestor-of-a-binary-search-tree.py @@ -0,0 +1,28 @@ +""" +Recursive +Time: O(LogN) if the tree is balanced. +Space: O(LogN) for the recursion stack. +""" +class Solution: + def lowestCommonAncestor(self, node: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + if q.val<=node.val<=p.val or p.val<=node.val<=q.val: + return node + elif q.val 'TreeNode': + while not (q.val<=node.val<=p.val or p.val<=node.val<=q.val): + if q.val int: + if key in self.dic: + node = self.dic[key] + self.remove(node) + self.promote(node) + return node.val + return -1 + + def put(self, key: int, value: int) -> None: + if key in self.dic: + self.remove(self.dic[key]) + node = Node(key, value) + self.promote(node) + self.dic[key] = node + + if len(self.dic)>self.capacity: #[2] + del self.dic[self.tail.prev.key] + self.remove(self.tail.prev) diff --git a/problems/python3/max-area-of-island.py b/problems/python3/max-area-of-island.py new file mode 100644 index 0000000..fd55a94 --- /dev/null +++ b/problems/python3/max-area-of-island.py @@ -0,0 +1,23 @@ +class Solution: + def maxAreaOfIsland(self, grid: List[List[int]]) -> int: + def dfs(i, j) -> int: + if i<0 or j<0 or i>=MAX_ROW or j>=MAX_COL: return 0 + if grid[i][j]==0 or grid[i][j]==2: return 0 + + grid[i][j] = 2 #mark as visited + + area = 1 + area += dfs(i+1, j) + area += dfs(i-1, j) + area += dfs(i, j+1) + area += dfs(i, j-1) + return area + + ans = 0 + MAX_ROW = len(grid) + MAX_COL = len(grid[0]) + + for i in range(MAX_ROW): + for j in range(MAX_COL): + ans = max(ans, dfs(i, j)) + return ans \ No newline at end of file diff --git a/problems/python3/maximum-depth-of-binary-tree.py b/problems/python3/maximum-depth-of-binary-tree.py new file mode 100755 index 0000000..3146dcf --- /dev/null +++ b/problems/python3/maximum-depth-of-binary-tree.py @@ -0,0 +1,8 @@ +""" +Time: O(N) +Space:O(LogN), for recursion stack space. +""" +class Solution: + def maxDepth(self, root: Optional[TreeNode]) -> int: + if not root: return 0 + return 1+max(self.maxDepth(root.left), self.maxDepth(root.right)) \ No newline at end of file diff --git a/problems/python3/maximum-product-subarray.py b/problems/python3/maximum-product-subarray.py new file mode 100644 index 0000000..243bf41 --- /dev/null +++ b/problems/python3/maximum-product-subarray.py @@ -0,0 +1,25 @@ +""" +Time: O(N) +Space: O(N), can be reduce to O(1) + +dp[i][0] := The max product from subarray that end with nums[i] +dp[i][1] := The min product from subarray that end with nums[i] +""" +class Solution: + def maxProduct(self, nums: List[int]) -> int: + N = len(nums) + dp = [[1, 1] for _ in range(N+1)] + ans = float('-inf') + + for i in range(1, N+1): + dp[i][0] = dp[i][1] = nums[i-1] + + if nums[i-1]>0: + dp[i][0] = max(dp[i][0], nums[i-1]*dp[i-1][0]) + dp[i][1] = min(dp[i][1], nums[i-1]*dp[i-1][1]) + else: + dp[i][0] = max(dp[i][0], nums[i-1]*dp[i-1][1]) + dp[i][1] = min(dp[i][1], nums[i-1]*dp[i-1][0]) + ans = max(ans, dp[i][0]) + + return ans \ No newline at end of file diff --git a/problems/python3/maximum-subarray.py b/problems/python3/maximum-subarray.py new file mode 100644 index 0000000..6f749fa --- /dev/null +++ b/problems/python3/maximum-subarray.py @@ -0,0 +1,17 @@ +""" +Time: O(N) +Space: O(1) + +Calculate the prefix sum. Whenever the prefix is negative, we ignore all the value before. +Update the ans along the way. +""" +class Solution: + def maxSubArray(self, nums: List[int]) -> int: + ans = float('-inf') + currSum = 0 + + for num in nums: + if currSum<0: currSum = 0 + currSum += num + ans = max(ans, currSum) + return ans \ No newline at end of file diff --git a/problems/python3/median-of-two-sorted-arrays.py b/problems/python3/median-of-two-sorted-arrays.py new file mode 100644 index 0000000..0fa31da --- /dev/null +++ b/problems/python3/median-of-two-sorted-arrays.py @@ -0,0 +1,40 @@ +""" +Try to find a i on array A and corespoding j on array B. + +Aleft = A[i] +Aright = A[i+1] +Bleft = B[j] +Bright = B[j+1] + +Such that Aleft<=Bright and Bleft<=Aright + +This means that A[:i+1] and B[:j+1] holds all the elements that is less than the median. +Aright A[i+1:] and B[j+1:] holds all the elements that is larger or equal to the median. +""" +class Solution: + def findMedianSortedArrays(self, A: List[int], B: List[int]) -> float: + if len(A)>len(B): A, B = B, A + + total = len(A)+len(B) + half = total//2 + l = 0 + r = len(A)-1 + + while True: + i = l + (r-l)//2 + j = half-(i+1)-1 + + Aleft = A[i] if i>=0 else float('-inf') + Aright = A[i+1] if i+1=0 else float('-inf') + Bright = B[j+1] if j+1Bright: + r = i-1 + else: + l = i+1 \ No newline at end of file diff --git a/problems/python3/meeting-rooms-ii.py b/problems/python3/meeting-rooms-ii.py new file mode 100644 index 0000000..1eb4eb4 --- /dev/null +++ b/problems/python3/meeting-rooms-ii.py @@ -0,0 +1,12 @@ +class Solution: + def minMeetingRooms(self, intervals: List[List[int]]) -> int: + intervals.sort() + h = [] + ans = 0 + + for s, e in intervals: + while h and s>=h[0][0]: + heapq.heappop(h) + heapq.heappush(h, (e, s)) + ans = max(ans, len(h)) + return ans \ No newline at end of file diff --git a/problems/python3/meeting-rooms.py b/problems/python3/meeting-rooms.py new file mode 100644 index 0000000..60cf429 --- /dev/null +++ b/problems/python3/meeting-rooms.py @@ -0,0 +1,10 @@ +class Solution: + def canAttendMeetings(self, intervals: List[List[int]]) -> bool: + intervals.sort() + lastEnd = float('-inf') + for s, e in intervals: + if lastEnd>s: + return False + else: + lastEnd = e + return True \ No newline at end of file diff --git a/problems/python3/merge-intervals.py b/problems/python3/merge-intervals.py new file mode 100644 index 0000000..c8a4678 --- /dev/null +++ b/problems/python3/merge-intervals.py @@ -0,0 +1,17 @@ +""" +Time: O(NLogN) for sorting +Space: O(1) excluding the output. +""" +class Solution: + def merge(self, intervals: List[List[int]]) -> List[List[int]]: + ans = [] + intervals.sort() + + for s, e in intervals: + if not ans: + ans.append([s, e]) + elif ans[-1][1]>=s: + ans[-1][1] = max(ans[-1][1], e) + else: + ans.append([s, e]) + return ans \ No newline at end of file diff --git a/problems/python3/merge-triplets-to-form-target-triplet.py b/problems/python3/merge-triplets-to-form-target-triplet.py new file mode 100644 index 0000000..4708e4f --- /dev/null +++ b/problems/python3/merge-triplets-to-form-target-triplet.py @@ -0,0 +1,15 @@ +""" +If any element in the triplets is larger than the element in target, it cannot be used. +Check if we have all 3 index found the same value. +""" +class Solution: + def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -> bool: + okIndex = set() + + for a, b, c in triplets: + if a>target[0] or b>target[1] or c>target[2]: continue + if a==target[0]: okIndex.add(0) + if b==target[1]: okIndex.add(1) + if c==target[2]: okIndex.add(2) + + return len(okIndex)==3 \ No newline at end of file diff --git a/problems/python3/merge-two-sorted-lists.py b/problems/python3/merge-two-sorted-lists.py new file mode 100644 index 0000000..2b31352 --- /dev/null +++ b/problems/python3/merge-two-sorted-lists.py @@ -0,0 +1,15 @@ +class Solution: + def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: + head = ListNode() #dummy + node = head + while list1 and list2: + if list1.val O(1). + +dp[i] := the cost to get to index i. +""" +class Solution: + def minCostClimbingStairs(self, cost: List[int]) -> int: + N = len(cost) + dp = [float('inf')]*(N+1) + dp[0] = 0 + dp[1] = 0 + + for i in range(2, len(dp)): + dp[i] = min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2]) + return dp[-1] \ No newline at end of file diff --git a/problems/python3/min-cost-to-connect-all-points.py b/problems/python3/min-cost-to-connect-all-points.py new file mode 100644 index 0000000..4cd6ada --- /dev/null +++ b/problems/python3/min-cost-to-connect-all-points.py @@ -0,0 +1,28 @@ +class Solution: + def minCostConnectPoints(self, points: List[List[int]]) -> int: + ans = 0 + visited = set() + adj = collections.defaultdict(list) + N = len(points) + + #build adjacency list + for i in range(N): + x0, y0 = points[i] + for j in range(i+1, N): + x1, y1 = points[j] + dis = abs(x0-x1)+abs(y0-y1) + adj[(x0, y0)].append((dis, x1, y1)) + adj[(x1, y1)].append((dis, x0, y0)) + + h = [(0, points[0][0], points[0][1])] #min heap + while len(visited) None: + self.stack.append(val) + self.minStack.append(val if (not self.minStack or val None: + self.minStack.pop() + return self.stack.pop() + + def top(self) -> int: + return self.stack[-1] + + def getMin(self) -> int: + return self.minStack[-1] \ No newline at end of file diff --git a/problems/python3/minimum-interval-to-include-each-query.py b/problems/python3/minimum-interval-to-include-each-query.py new file mode 100644 index 0000000..ca43de3 --- /dev/null +++ b/problems/python3/minimum-interval-to-include-each-query.py @@ -0,0 +1,23 @@ +class Solution: + def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]: + ans = [-1]*len(queries) + queries = sorted([(query, i) for i, query in enumerate(queries)]) + intervals.sort() + h = [] + + itervalIndex = 0 + for query, queryIndex in queries: + #push all intervals that include 'query' to the min heap + while itervalIndex str: + if len(t)>len(s): return "" + + ans = "" + counter1 = collections.Counter(t) + charSet = set(t) + counter2 = collections.Counter() #sliding window in string s, index between l and r + charSet2 = set(s) + matchCount = 0 #count of char in the sliding window that counts are larger than the char count in t. + + l = 0 + for r in range(len(s)): + counter2[s[r]] += 1 + + if s[r] in charSet and counter1[s[r]]==counter2[s[r]]: matchCount += 1 + + while lcounter1[s[l]]): + counter2[s[l]] -= 1 + l += 1 + + if matchCount==len(charSet) and (ans=="" or r-l+1 int: + N = len(nums) + ans = 0 + + for n in range(N+1): + ans ^= n + + for n in nums: + ans ^= n + + return ans \ No newline at end of file diff --git a/problems/python3/multiply-strings.py b/problems/python3/multiply-strings.py new file mode 100644 index 0000000..be9f675 --- /dev/null +++ b/problems/python3/multiply-strings.py @@ -0,0 +1,22 @@ +class Solution: + def multiply(self, num1: str, num2: str) -> str: + if num1=='0' or num2=='0': return '0' + M, N = len(num1), len(num2) + temp = [0]*(M+N+1) + + num1, num2 = num1[::-1], num2[::-1] + for i in range(M): + for j in range(N): + digits = int(num1[i])*int(num2[j]) + temp[i+j] += digits + temp[i+j+1] += temp[i+j]//10 + temp[i+j] = temp[i+j]%10 + + ans = '' + temp = temp[::-1] + isLeadingZero = True + for d in temp: + if d!=0 or not isLeadingZero: + isLeadingZero = False + ans += str(d) + return ans \ No newline at end of file diff --git a/problems/python3/n-queens.py b/problems/python3/n-queens.py new file mode 100644 index 0000000..2ab7335 --- /dev/null +++ b/problems/python3/n-queens.py @@ -0,0 +1,45 @@ +class Solution: + def solveNQueens(self, n: int) -> List[List[str]]: + def helper(row: int): + if row==n: + ans.append(convertFormat(queenCols)) + return + + for col in range(n): + posDiag = row+col + negDiag = row-col + + if col in colUsed or posDiag in posDiagUsed or negDiag in negDiagUsed: continue + + queenCols.append(col) + colUsed.add(col) + posDiagUsed.add(posDiag) + negDiagUsed.add(negDiag) + + helper(row+1) + + queenCols.pop() + colUsed.remove(col) + posDiagUsed.remove(posDiag) + negDiagUsed.remove(negDiag) + + def convertFormat(queenCols: List[int]) -> List[str]: + output = [] + for col in queenCols: + row = '' + for i in range(n): + if i==col: + row += 'Q' + else: + row += '.' + output.append(row) + return output + + ans = [] + queenCols = [] + colUsed = set() + posDiagUsed = set() + negDiagUsed = set() + + helper(0) + return ans \ No newline at end of file diff --git a/problems/python3/network-delay-time.py b/problems/python3/network-delay-time.py new file mode 100644 index 0000000..6da8613 --- /dev/null +++ b/problems/python3/network-delay-time.py @@ -0,0 +1,23 @@ +class Solution: + def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int: + ans = 0 + adj = collections.defaultdict(list) + h = [] + visited = set() + + for u, v, w in times: + adj[u].append((v, w)) + + heapq.heappush(h, (0, k)) + while h: + timeNeededToGetHere, node = heapq.heappop(h) + + if node in visited: continue + visited.add(node) + ans = max(ans, timeNeededToGetHere) + + for nei, time in adj[node]: + if nei in visited: continue + heapq.heappush(h, (time+timeNeededToGetHere, nei)) + + return ans if len(visited)==n else -1 \ No newline at end of file diff --git a/problems/python3/non-overlapping-intervals.py b/problems/python3/non-overlapping-intervals.py new file mode 100644 index 0000000..b3b6a0d --- /dev/null +++ b/problems/python3/non-overlapping-intervals.py @@ -0,0 +1,17 @@ +""" +Time: O(NLogN) for sorting. +""" +class Solution: + def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int: + intervals.sort() + + ans = 0 + prevEnd = intervals[0][1] + + for s, e in intervals[1:]: + if s>=prevEnd: + prevEnd = e + else: + ans += 1 + prevEnd = min(prevEnd, e) + return ans \ No newline at end of file diff --git a/problems/python3/number-of-1-bits.py b/problems/python3/number-of-1-bits.py new file mode 100644 index 0000000..2273859 --- /dev/null +++ b/problems/python3/number-of-1-bits.py @@ -0,0 +1,10 @@ +""" +n = n&(n-1) will turn the right most 1 to 0. +""" +class Solution: + def hammingWeight(self, n: int) -> int: + ans = 0 + while n>0: + n = n&(n-1) + ans += 1 + return ans \ No newline at end of file diff --git a/problems/python3/number-of-connected-components-in-an-undirected-graph.py b/problems/python3/number-of-connected-components-in-an-undirected-graph.py new file mode 100644 index 0000000..28571ea --- /dev/null +++ b/problems/python3/number-of-connected-components-in-an-undirected-graph.py @@ -0,0 +1,25 @@ +class Solution: + def countComponents(self, N: int, edges: List[List[int]]) -> int: + def union(n1, n2): + p1 = find(n1) + p2 = find(n2) + if p1==p2: + return + elif p1 int: + def dfs(i, j): + if i<0 or j<0 or i>=MAX_ROWS or j>=MAX_COLS: return + if grid[i][j]!='1': return + + grid[i][j] = '2' + dfs(i+1, j) + dfs(i-1, j) + dfs(i, j+1) + dfs(i, j-1) + + count = 0 + MAX_ROWS = len(grid) + MAX_COLS = len(grid[0]) + + for i in range(MAX_ROWS): + for j in range(MAX_COLS): + if grid[i][j]=='1': + dfs(i, j) + count += 1 + return count \ No newline at end of file diff --git a/problems/python3/pacific-atlantic-water-flow.py b/problems/python3/pacific-atlantic-water-flow.py new file mode 100644 index 0000000..05032b0 --- /dev/null +++ b/problems/python3/pacific-atlantic-water-flow.py @@ -0,0 +1,30 @@ +class Solution: + def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]: + def bfs(q, ocian): + while q: + i0, j0 = q.popleft() + if (i0, j0) in ocian: continue + ocian.add((i0, j0)) + + for i, j in ((i0+1, j0), (i0-1, j0), (i0, j0+1), (i0, j0-1)): + if i<0 or j<0 or i>=MAX_ROW or j>=MAX_COL: continue + if heights[i][j]>=heights[i0][j0]: q.append((i, j)) + + MAX_ROW = len(heights) + MAX_COL = len(heights[0]) + + #add the top and left to q1 + pacific = set() + q1 = collections.deque() + for j in range(MAX_COL): q1.append((0, j)) + for i in range(MAX_ROW): q1.append((i, 0)) + + #add botton and right to q2 + atlantic = set() + q2 = collections.deque() + for j in range(MAX_COL): q2.append((MAX_ROW-1, j)) + for i in range(MAX_ROW): q2.append((i, MAX_COL-1)) + + bfs(q1, pacific) + bfs(q2, atlantic) + return pacific.intersection(atlantic) \ No newline at end of file diff --git a/problems/python3/palindrome-partitioning.py b/problems/python3/palindrome-partitioning.py new file mode 100644 index 0000000..d02ef17 --- /dev/null +++ b/problems/python3/palindrome-partitioning.py @@ -0,0 +1,64 @@ +""" +`helper(i)` finds all the palindrome substring from i to j see if we can get an answer by recursively calling `helper(j+1)`. + +Time: O(2^N * N), for a string length S, there will be 2^N combination of substrings. For each substring, it will take O(N) to test if they are all palindrome. +Space: O(N). `partition` will takes O(N) and recursion stack will also take O(N). O(N + N) ~= O(N) +""" +class Solution: + def partition(self, s: str) -> List[List[str]]: + def helper(i): + if i>=len(s): + ans.append(partition.copy()) + return + + for j in range(i, len(s)): + if isPalindrome(i, j): + partition.append(s[i:j+1]) + helper(j+1) + partition.pop() + + def isPalindrome(i, j): + while i<=j: + if s[i]!=s[j]: return False + i += 1 + j -= 1 + return True + + ans = [] + partition = [] + helper(0) + return ans + +""" +If you look closely you can see that we execute `isPalindrome` on many repeated substrings. +We can optimize this by storing the result in `dp` and reuse it. +Since all the less difference i and j (shorter substring s[i:j+1]) will be process first in the `isPalindrome`. +When checking if s[i:j+1] is a palindrome or not, we can simply check the if s[i]==s[j] and the previous result (dp[i+1][j-1]) + +Time: O(2^N * N), The overall time complexity is the same, but isPalindrome is actually a lot faster. +Space: O(N^2) for `dp`. +""" +class Solution: + def partition(self, s: str) -> List[List[str]]: + def helper(i): + if i>=N: + ans.append(partition.copy()) + return + + for j in range(i, N): + if isPalindrome(i, j): + partition.append(s[i:j+1]) + helper(j+1) + partition.pop() + + def isPalindrome(i, j): + dp[i][j] = i==j or (j-i==1 and s[i]==s[j]) or (s[i]==s[j] and dp[i+1][j-1]) #len==1 palindrome or len==2 palindrome or len>=3 palindrome + return dp[i][j] + + N = len(s) + ans = [] + partition = [] + dp = [[False]*N for _ in range(N)] + + helper(0) + return ans \ No newline at end of file diff --git a/problems/python3/palindromic-substrings.py b/problems/python3/palindromic-substrings.py new file mode 100644 index 0000000..7ef7e96 --- /dev/null +++ b/problems/python3/palindromic-substrings.py @@ -0,0 +1,16 @@ +class Solution: + def countSubstrings(self, s: str) -> int: + def countPalindrome(l, r) -> int: + count = 0 + while l>=0 and r bool: + total = sum(nums) + if total%2!=0: return False + + target = total/2 + possibleSum = set() + possibleSum.add(0) + for num in nums: + temp = set() + for p in possibleSum: + if p==target or p+num==target: return True + temp.add(p) + temp.add(p+num) + possibleSum = temp + return False \ No newline at end of file diff --git a/problems/python3/partition-labels.py b/problems/python3/partition-labels.py new file mode 100644 index 0000000..0c12d00 --- /dev/null +++ b/problems/python3/partition-labels.py @@ -0,0 +1,26 @@ +""" +Time: O(N) +Space: O(N) + +For each char, store its max index in maxIndex. +Iterate through the string, update the "currMax" along the way. +currMax is the place we can partition unless it get updated again. +If the current index is the currMax, update "ans". +""" +class Solution: + def partitionLabels(self, s: str) -> List[int]: + ans = [] + maxIndex = {} + + for i, c in enumerate(s): + maxIndex[c] = i + + currMax = 0 + processedLength = 0 + for i, c in enumerate(s): + currMax = max(currMax, maxIndex[c]) + if i==currMax: + ans.append(i+1 - processedLength) + processedLength += ans[-1] + + return ans \ No newline at end of file diff --git a/problems/python3/permutation-in-string.py b/problems/python3/permutation-in-string.py new file mode 100644 index 0000000..cd22781 --- /dev/null +++ b/problems/python3/permutation-in-string.py @@ -0,0 +1,28 @@ +class Solution: + def checkInclusion(self, s1: str, s2: str) -> bool: + if len(s1)>len(s2): return False + + counter1 = collections.Counter(s1) + counter2 = collections.Counter(s2[:len(s1)]) + matches = 0 + for c in 'abcdefghijklmnopqrstuvwxyz': + if counter1[c]==counter2[c]: matches += 1 + if matches==26: return True + + l = 0 + for r in range(len(s1), len(s2)): + counter2[s2[r]] += 1 + if counter1[s2[r]]==counter2[s2[r]]: + matches += 1 + elif counter1[s2[r]]+1==counter2[s2[r]]: + matches -= 1 + + counter2[s2[l]] -= 1 + if counter1[s2[l]]==counter2[s2[l]]: + matches += 1 + elif counter1[s2[l]]-1==counter2[s2[l]]: + matches -= 1 + l += 1 + + if matches==26: return True + return False \ No newline at end of file diff --git a/problems/python3/permutations.py b/problems/python3/permutations.py new file mode 100644 index 0000000..98d536d --- /dev/null +++ b/problems/python3/permutations.py @@ -0,0 +1,28 @@ +""" +Time: O(N!), since we call helper() N! times. +Space: O(N) for recursion stacks. + +Whenever we call `helper()`, we pick a num that is not "used" and add it to the `permutation`. +Recursively call the `helper()` until we filled the `permutation`. +Resotre `permutation` and `used` and try another `num`. +""" +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + def helper(): + if len(permutation)==len(nums): + ans.append(permutation.copy()) + return + + for num in nums: + if num in used: continue + used.add(num) + permutation.append(num) + helper() + used.remove(num) + permutation.pop() + + ans = [] + permutation = [] + used = set() + helper() + return ans \ No newline at end of file diff --git a/problems/python3/plus-one.py b/problems/python3/plus-one.py new file mode 100644 index 0000000..8cfedb4 --- /dev/null +++ b/problems/python3/plus-one.py @@ -0,0 +1,16 @@ +class Solution: + def plusOne(self, digits: List[int]) -> List[int]: + i = len(digits)-1 + needAdditionDigit = True + + while i>=0 and needAdditionDigit: + if digits[i]==9: + digits[i] = 0 + i -= 1 + needAdditionDigit = True + else: + digits[i] += 1 + needAdditionDigit = False + if needAdditionDigit: digits.insert(0, 1) + return digits + \ No newline at end of file diff --git a/problems/python3/powx-n.py b/problems/python3/powx-n.py new file mode 100644 index 0000000..2b1b4fc --- /dev/null +++ b/problems/python3/powx-n.py @@ -0,0 +1,14 @@ +class Solution: + def myPow(self, x: float, k: int) -> float: + if k<0: return 1/self.myPow(x, -k) + + if k==0: + return 1 + elif k==1: + return x + elif k%2==0: + half = self.myPow(x, k//2) + return half * half + else: + half = self.myPow(x, (k-1)//2) + return half * half * x \ No newline at end of file diff --git a/problems/python3/product-of-array-except-self.py b/problems/python3/product-of-array-except-self.py new file mode 100644 index 0000000..2309896 --- /dev/null +++ b/problems/python3/product-of-array-except-self.py @@ -0,0 +1,45 @@ +class Solution: + def productExceptSelf(self, nums: List[int]) -> List[int]: + #left[i] := product of all nums left of nums[i] (not include nums[i]) + left = [1] + temp = 1 + for num in nums: + temp *= num + left.append(temp) + + #right[i] := product of all nums right of nums[i] (not include nums[i]) + right = [] + temp = 1 + for num in reversed(nums): + temp *= num + right.append(temp) + right.reverse() + right.append(1) + right = right[1:] + + ans = [] + for i in range(len(nums)): + ans.append(left[i]*right[i]) + return ans + +#In Place +class Solution: + def productExceptSelf(self, nums: List[int]) -> List[int]: + N = len(nums) + ans = [0]*N + + #generate "left" + ans[0] = 1 + for i in range(1, N): + ans[i] = ans[i-1]*nums[i-1] + + #generate "right" + temp = 1 + for i in range(N-2, -1, -1): + temp *= nums[i+1] + ans[i] *= temp + + return ans + + + \ No newline at end of file diff --git a/problems/python3/reconstruct-itinerary.py b/problems/python3/reconstruct-itinerary.py new file mode 100644 index 0000000..14c928f --- /dev/null +++ b/problems/python3/reconstruct-itinerary.py @@ -0,0 +1,27 @@ +""" +DFS with backtracking. +""" +class Solution: + def findItinerary(self, tickets: List[List[str]]) -> List[str]: + def dfs(start) -> bool: + if len(ans)==len(tickets)+1: return True + if start not in adj: return False + + temp = list(adj[start]) + for i, arr in enumerate(temp): + adj[start].pop(i) + ans.append(arr) + if dfs(arr): return True + adj[start].insert(i, arr) + ans.pop() + return False + + ans = ['JFK'] + adj = collections.defaultdict(list) + + tickets.sort() + for des, arr in tickets: + adj[des].append(arr) + + dfs('JFK') + return ans \ No newline at end of file diff --git a/problems/python3/redundant-connection.py b/problems/python3/redundant-connection.py new file mode 100644 index 0000000..47bda2e --- /dev/null +++ b/problems/python3/redundant-connection.py @@ -0,0 +1,32 @@ +class Solution: + def findRedundantConnection(self, edges: List[List[int]]) -> List[int]: + def union(n1, n2): + p1 = find(n1) + p2 = find(n2) + + if p1==p2: + return False #union failed, already united. + elif p1 bool: + def dfs(i, j): + if (i, j) in cache: return cache[(i, j)] + if i>=M and j>=N: return True + if j>=N: return False + + match = i Optional[ListNode]: + #count the length of the linked list + node = head + count = 0 + while node: + count += 1 + node = node.next + + + node = head + steps = count-n-1 + + #steps==-1 means that we need to remove the first node + if steps==-1: return head.next + + #traverse to the node before the node we wanted to remove + while steps>0: + node = node.next + steps -= 1 + + #remove "node.next" + node.next = node.next.next + + return head + + +#One pass. Fast pointer is ahead of slow pointer by n+1 +#So slow pointer will stop at the node before the node we wanted to remove +class Solution: + def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]: + dummy = ListNode() + dummy.next = head + + ahead = n+1 + fast = dummy + slow = dummy + + while fast: + fast = fast.next + ahead -= 1 + if ahead<0: slow = slow.next + + slow.next = slow.next.next + + return dummy.next \ No newline at end of file diff --git a/problems/python3/reorder-list.py b/problems/python3/reorder-list.py new file mode 100644 index 0000000..89c6ba2 --- /dev/null +++ b/problems/python3/reorder-list.py @@ -0,0 +1,37 @@ +class Solution: + def reorderList(self, head: Optional[ListNode]) -> None: + #find the middle point + slow = head + fast = head + while fast and fast.next: + slow = slow.next + fast = fast.next.next + middle = slow.next + + #reverse the linked list after the middle point + middle = self.reverseList(middle) + + #separate the linked list before the middle + slow.next = None + + #merge two linked list + node = head + while middle and node: + nextNode = node.next + node.next = middle + middle = middle.next + node.next.next = nextNode + node = nextNode + return head + + def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: + pre = None + node = head + + while node: + nextNode = node.next + node.next = pre + if not nextNode: return node + pre = node + node = nextNode + \ No newline at end of file diff --git a/problems/python3/reverse-bits.py b/problems/python3/reverse-bits.py new file mode 100644 index 0000000..dfc5191 --- /dev/null +++ b/problems/python3/reverse-bits.py @@ -0,0 +1,7 @@ +class Solution: + def reverseBits(self, n: int) -> int: + res = 0 + for i in range(32): + bit = (n >> i) & 1 + res = res | (bit << (31 - i)) + return res \ No newline at end of file diff --git a/problems/python3/reverse-integer.py b/problems/python3/reverse-integer.py new file mode 100644 index 0000000..76137ab --- /dev/null +++ b/problems/python3/reverse-integer.py @@ -0,0 +1,16 @@ +class Solution: + def reverse(self, x: int) -> int: + MAX = 2**31-1 + MIN = -2**31 + ans = 0 + + while x: + digit = int(math.fmod(x, 10)) + x = int(x/10) + + if ans>MAX//10 or (ans==MAX//10 and digit>MAX%10): return 0 + if ans Optional[ListNode]: + if not head or not head.next: return head + temp = self.reverseList(head.next) + head.next.next = head + head.next = None + return temp + +#Iterative +class Solution: + def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: + pre = None + node = head + + while node: + nextNode = node.next + node.next = pre + if not nextNode: return node + pre = node + node = nextNode \ No newline at end of file diff --git a/problems/python3/rotate-image.py b/problems/python3/rotate-image.py new file mode 100644 index 0000000..c8a872a --- /dev/null +++ b/problems/python3/rotate-image.py @@ -0,0 +1,16 @@ +class Solution: + def rotate(self, matrix: List[List[int]]) -> None: + l, r = 0, len(matrix[0])-1 + + while l int: + time = 0 + rotten = set() + aboutToRot = set() + fresh = set() + + for i in range(len(grid)): + for j in range((len(grid[0]))): + if grid[i][j]==2: + rotten.add((i, j)) + elif grid[i][j]==1: + fresh.add((i, j)) + + while rotten: + i0, j0 = rotten.pop() #randomly get one + grid[i0][j0] = 2 + + for i, j in ((i0+1, j0), (i0-1, j0), (i0, j0+1), (i0, j0-1)): + if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]): continue + if (i, j) in rotten or (i, j) in aboutToRot: continue + if (i, j) in fresh: + fresh.remove((i, j)) + aboutToRot.add((i, j)) + + if not rotten and not aboutToRot: break + + if not rotten: + time += 1 + rotten = aboutToRot + aboutToRot = set() + + if fresh: return -1 + + return time \ No newline at end of file diff --git a/problems/python3/same-tree.py b/problems/python3/same-tree.py new file mode 100755 index 0000000..6687f04 --- /dev/null +++ b/problems/python3/same-tree.py @@ -0,0 +1,16 @@ +""" +Time: O(N) +Space: O(LogN) if the tree is balanced. +""" +class Solution: + def isSameTree(self, node1: Optional[TreeNode], node2: Optional[TreeNode]) -> bool: + if not node1 and not node2: + return True + + if (not node1 and node2) or (node1 and not node2): + return False + + if node1.val!=node2.val: + return False + + return self.isSameTree(node1.left, node2.left) and self.isSameTree(node1.right, node2.right) \ No newline at end of file diff --git a/problems/python3/search-a-2d-matrix.py b/problems/python3/search-a-2d-matrix.py new file mode 100644 index 0000000..2afe301 --- /dev/null +++ b/problems/python3/search-a-2d-matrix.py @@ -0,0 +1,22 @@ +class Solution: + def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: + N = len(matrix) + M = len(matrix[0]) + + l = 0 + r = N*M-1 + + while l<=r: + m = l + int((r-l)/2) + + i = int(m/M) + j = m%M + + if matrix[i][j]target: + r = m-1 + else: + return True + + return False \ No newline at end of file diff --git a/problems/python3/serialize-and-deserialize-binary-tree.py b/problems/python3/serialize-and-deserialize-binary-tree.py new file mode 100644 index 0000000..1b057d0 --- /dev/null +++ b/problems/python3/serialize-and-deserialize-binary-tree.py @@ -0,0 +1,22 @@ +class Codec: + + def serialize(self, root): + if not root: return '#' + return str(root.val)+','+self.serialize(root.left)+','+self.serialize(root.right) + + + def deserialize(self, data): + def helper(): + if data[self.i]=='#': + self.i += 1 + return None + + node = TreeNode(int(data[self.i])) + self.i += 1 + node.left = helper() + node.right = helper() + return node + + data = data.split(",") + self.i = 0 + return helper() \ No newline at end of file diff --git a/problems/python3/set-matrix-zeroes.py b/problems/python3/set-matrix-zeroes.py new file mode 100644 index 0000000..dd836e5 --- /dev/null +++ b/problems/python3/set-matrix-zeroes.py @@ -0,0 +1,28 @@ +class Solution: + def setZeroes(self, matrix: List[List[int]]) -> None: + M = len(matrix) + N = len(matrix[0]) + firstRowZero = False + + for i in range(M): + for j in range(N): + if matrix[i][j]==0: + matrix[0][j] = 0 + if i==0: + firstRowZero = True + else: + matrix[i][0] = 0 + + for i in range(1, M): + if matrix[i][0]==0: + for j in range(N): + matrix[i][j] = 0 + + for j in range(N): + if matrix[0][j]==0: + for i in range(M): + matrix[i][j] = 0 + + if firstRowZero: + for j in range(N): + matrix[0][j] = 0 \ No newline at end of file diff --git a/problems/python3/single-number.py b/problems/python3/single-number.py new file mode 100644 index 0000000..daa3268 --- /dev/null +++ b/problems/python3/single-number.py @@ -0,0 +1,15 @@ +""" +^ (XOR) +The same will be 0 +0^0 = 0 +1^1 = 0 + +Different will be 1 +1^0 = 1 +0^1 = 1 +""" +class Solution: + def singleNumber(self, nums: List[int]) -> int: + ans = 0 + for num in nums: ans ^= num + return ans \ No newline at end of file diff --git a/problems/python3/sliding-window-maximum.py b/problems/python3/sliding-window-maximum.py new file mode 100644 index 0000000..a535c16 --- /dev/null +++ b/problems/python3/sliding-window-maximum.py @@ -0,0 +1,18 @@ +class Solution: + def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: + ans = [] + q = collections.deque() + l = r = 0 + + while r List[int]: + ans = [] + x0 = 0 + y0 = 0 + dx = len(matrix[0])-1 + dy = len(matrix)-1 + direction = 'right' + isFirst = True + + ans.append(matrix[x0][y0]) + while True: + if direction=='right': + for x in range(x0+1, x0+dx+1): + ans.append(matrix[y0][x]) + x0 += dx + direction = 'down' + + if isFirst: + isFirst = False + else: + dx -= 1 + + if dy==0: break + + elif direction=='left': + for x in range(x0-1, x0-dx-1, -1): + ans.append(matrix[y0][x]) + x0 -= dx + direction = 'up' + dx -= 1 + if dy==0: break + + elif direction=='down': + for y in range(y0+1, y0+dy+1): + ans.append(matrix[y][x0]) + y0 += dy + direction = 'left' + dy -= 1 + if dx==0: break + + elif direction=='up': + for y in range(y0-1, y0-dy-1, -1): + ans.append(matrix[y][x0]) + y0 -= dy + direction = 'right' + dy -= 1 + if dx==0: break + return ans + + +""" +Answer from Neetcode, more elegant. +left, right, top, bottom is the border (index is exclusive on the border. +In other words, for matrix[i][j] +i: top List[int]: + res = [] + left, right = 0, len(matrix[0]) + top, bottom = 0, len(matrix) + + while left < right and top < bottom: + # get every i in the top row + for i in range(left, right): + res.append(matrix[top][i]) + top += 1 + # get every i in the right col + for i in range(top, bottom): + res.append(matrix[i][right - 1]) + right -= 1 + if not (left < right and top < bottom): + break + # get every i in the bottom row + for i in range(right - 1, left - 1, -1): + res.append(matrix[bottom - 1][i]) + bottom -= 1 + # get every i in the left col + for i in range(bottom - 1, top - 1, -1): + res.append(matrix[i][left]) + left += 1 + + return res \ No newline at end of file diff --git a/problems/python3/subsets-ii.py b/problems/python3/subsets-ii.py new file mode 100644 index 0000000..be0b0b7 --- /dev/null +++ b/problems/python3/subsets-ii.py @@ -0,0 +1,23 @@ +""" +Time: O(N * 2^N) +Space: O(N) +""" +class Solution: + def subsetsWithDup(self, nums: List[int]) -> List[List[int]]: + def helper(i): + if i==len(nums): + ans.append(subset.copy()) + return + + subset.append(nums[i]) + helper(i+1) + subset.pop() + + while i+1 List[List[int]]: + def helper(i): + if not i List[int]: + N = len(s) + M = len(words) + W = len(words[0]) + wordSet = set(words) + ans = [] + counter = collections.Counter(words) + + for i in range(W): #[0] + windowCounter = collections.Counter() #counter for the word in words + notInWords = 0 #number of word not in the wordSet + theSame = 0 #number of word with the same count with "counter" + j = i + + while jcounter[word]: + theSame -= 1 + else: + notInWords += 1 + + popStart = j-M*W + if popStart>=0: + popWord = s[popStart:popStart+W] + if popWord in wordSet: + windowCounter[popWord] -= 1 + if windowCounter[popWord]==counter[popWord]: + theSame += 1 + elif windowCounter[popWord] bool: + if not root or not subRoot: return root==subRoot + if self.isSame(root, subRoot): return True + return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot) + + def isSame(self, p, q): + if not p or not q: return p==q + if p.val!=q.val: return False + return self.isSame(p.left, q.left) and self.isSame(p.right, q.right) \ No newline at end of file diff --git a/problems/python3/sum-of-two-integers.py b/problems/python3/sum-of-two-integers.py new file mode 100644 index 0000000..25f16f9 --- /dev/null +++ b/problems/python3/sum-of-two-integers.py @@ -0,0 +1,12 @@ +""" +Does not work with negative values yet. +""" +class Solution: + def getSum(self, a: int, b: int) -> int: + ans = a^b + carry = (a&b)<<1 + + while carry!=0: + ans, carry = ans^carry, (ans&carry)<<1 + + return ans \ No newline at end of file diff --git a/problems/python3/surrounded-regions.py b/problems/python3/surrounded-regions.py new file mode 100644 index 0000000..42aa4c0 --- /dev/null +++ b/problems/python3/surrounded-regions.py @@ -0,0 +1,34 @@ +""" +Time: O(N) +Space: O(N) +""" +class Solution: + def solve(self, board: List[List[str]]) -> None: + def dfs(i0, j0): + if i0<0 or j0<0 or i0>=MAX_ROW or j0>=MAX_COL: return + if board[i0][j0]!='O': return + if (i0, j0) in survived: return + + survived.add((i0, j0)) + dfs(i0+1, j0) + dfs(i0-1, j0) + dfs(i0, j0+1) + dfs(i0, j0-1) + + + MAX_ROW = len(board) + MAX_COL = len(board[0]) + survived = set() + + for i in range(MAX_ROW): + dfs(i, 0) + dfs(i, MAX_COL-1) + + for j in range(MAX_COL): + dfs(0, j) + dfs(MAX_ROW-1, j) + + for i in range(MAX_ROW): + for j in range(MAX_COL): + board[i][j] = 'O' if (i, j) in survived else 'X' + return board \ No newline at end of file diff --git a/problems/python3/swim-in-rising-water.py b/problems/python3/swim-in-rising-water.py new file mode 100644 index 0000000..dd5dbef --- /dev/null +++ b/problems/python3/swim-in-rising-water.py @@ -0,0 +1,23 @@ +""" +Time: O(N^2 * LogN^2) = O(N^2 * 2LogN) = O(N^2LogN), N is the number of elements in a row or column. +Space: O(N^2) +""" +class Solution: + def swimInWater(self, grid: List[List[int]]) -> int: + ROWS = len(grid) + COLS = len(grid[0]) + + visited = set() + h = [(grid[0][0], 0, 0)] + + while h: + t, r0, c0 = heapq.heappop(h) + + if (r0, c0) in visited: continue + visited.add((r0, c0)) + if r0==ROWS-1 and c0==COLS-1: return t + + for r, c in ((r0+1, c0), (r0-1, c0), (r0, c0+1), (r0, c0-1)): + if r<0 or c<0 or r>=ROWS or c>=COLS: continue + if (r, c) in visited: continue + heapq.heappush(h, (max(t, grid[r][c]), r, c)) \ No newline at end of file diff --git a/problems/python3/target-sum.py b/problems/python3/target-sum.py new file mode 100644 index 0000000..832fadd --- /dev/null +++ b/problems/python3/target-sum.py @@ -0,0 +1,25 @@ +""" +Time: O(NS), S is sum(nums), N is len(nums). This is the max possible number of element in "history". Which will be lesser than 2^N. +Space: O(NS) +""" +class Solution: + def findTargetSumWays(self, nums: List[int], target: int) -> int: + def dfs(i, curr): + #cache + if (i, curr) in history: + return history[(i, curr)] + + #ending condition + if i==len(nums): + if curr==target: + history[(i, curr)] = 1 + else: + history[(i, curr)] = 0 + return history[(i, curr)] + + history[(i, curr)] = dfs(i+1, curr+nums[i])+dfs(i+1, curr-nums[i]) + return history[(i, curr)] + + ans = 0 + history = {} + return dfs(0, 0) \ No newline at end of file diff --git a/problems/python3/task-scheduler.py b/problems/python3/task-scheduler.py new file mode 100644 index 0000000..ddc4625 --- /dev/null +++ b/problems/python3/task-scheduler.py @@ -0,0 +1,23 @@ +class Solution: + def leastInterval(self, tasks: List[str], n: int) -> int: + q = collections.deque() + h = [] + time = 0 + + counter = collections.Counter(tasks) + for task in counter: + heapq.heappush(h, (-counter[task], task)) + + while h or q: + if q and q[0][0]<=time: + _, count, task = q.popleft() + heapq.heappush(h, (-count, task)) + + if h: + count, task = heapq.heappop(h) + count*=-1 + count -= 1 + if count>0: q.append((time+n+1, count, task)) + time += 1 + + return time \ No newline at end of file diff --git a/problems/python3/time-based-key-value-store.py b/problems/python3/time-based-key-value-store.py new file mode 100644 index 0000000..f43facf --- /dev/null +++ b/problems/python3/time-based-key-value-store.py @@ -0,0 +1,28 @@ +class TimeMap: + + def __init__(self): + self.data = collections.defaultdict(list) + + def set(self, key: str, value: str, timestamp: int) -> None: + self.data[key].append((timestamp, value)) + + def get(self, key: str, timestamp: int) -> str: + dataList = self.data[key] + l = 0 + r = len(dataList)-1 + ans = '' + + while l<=r: + m = l + int((r-l)/2) + t = dataList[m][0] + + if ttimestamp: + r = m-1 + else: + ans = dataList[m][1] + break + + return ans \ No newline at end of file diff --git a/problems/python3/top-k-frequent-elements.py b/problems/python3/top-k-frequent-elements.py new file mode 100644 index 0000000..7d728cb --- /dev/null +++ b/problems/python3/top-k-frequent-elements.py @@ -0,0 +1,53 @@ +#Bucket Sort +class Solution: + def topKFrequent(self, nums: List[int], k: int) -> List[int]: + bucket = collections.defaultdict(list) + counter = collections.Counter(nums) + ans = [] + + for num in counter: + bucket[counter[num]].append(num) + + tempCount = len(nums) + while len(ans) List[int]: + def quickSelect(freqs, s, e, K): + i = s + t = s + j = e + pivot = freqs[(s+e)//2][1] + + while t<=j: + if freqs[t][1]pivot: + freqs[j], freqs[t] = freqs[t], freqs[j] + j -= 1 + else: + t += 1 + if e-j>=K: + return quickSelect(freqs, j+1, e, K) + elif e-(i-1)>=K: + return pivot + else: + return quickSelect(freqs, s, i-1, K-(e-i+1)) + + counts = collections.Counter(nums) + freqs = [(num, counts[num]) for num in counts] + ans = [] + + KthLargestFreq = quickSelect(freqs, 0, len(freqs)-1, K) + + for num, freq in freqs: + if freq>=KthLargestFreq: + ans.append(num) + return ans \ No newline at end of file diff --git a/problems/python3/trapping-rain-water.py b/problems/python3/trapping-rain-water.py new file mode 100644 index 0000000..fe01411 --- /dev/null +++ b/problems/python3/trapping-rain-water.py @@ -0,0 +1,35 @@ +""" +The water stored at i is `min(leftMax, rightMax) - height[i]` +leftMax: max height left of i +rightMax: max height right of i + +To use only constant space. +We can calculete the leftMax or rightMax and update ans along the way. +And we always go with the side where leftMax or rightMax is smaller. + +Time: O(N) +Space: O(1) +""" +class Solution: + def trap(self, height: List[int]) -> int: + ans = 0 + l = 0 + r = len(height)-1 + + leftMax = float('-inf') + rightMax = float('-inf') + + while l<=r: + if leftMax=leftMax: + leftMax = height[l] + else: + ans += leftMax-height[l] + l += 1 + else: + if height[r]>=rightMax: + rightMax = height[r] + else: + ans += rightMax-height[r] + r -= 1 + return ans \ No newline at end of file diff --git a/problems/python3/two-sum-ii-input-array-is-sorted.py b/problems/python3/two-sum-ii-input-array-is-sorted.py new file mode 100644 index 0000000..2890404 --- /dev/null +++ b/problems/python3/two-sum-ii-input-array-is-sorted.py @@ -0,0 +1,11 @@ +class Solution: + def twoSum(self, numbers: List[int], target: int) -> List[int]: + i = 0 + j = len(numbers)-1 + + while numbers[i]+numbers[j] != target: + if numbers[i]+numbers[j] > target: + j -= 1 + else: + i += 1 + return (i+1, j+1) \ No newline at end of file diff --git a/problems/python3/two-sum.py b/problems/python3/two-sum.py new file mode 100644 index 0000000..aa90d66 --- /dev/null +++ b/problems/python3/two-sum.py @@ -0,0 +1,8 @@ +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + #needed: {number required : counter part index} + needed = {} + for i, num in enumerate(nums): + if num in needed: + return (needed[num], i) + needed[target-num] = i \ No newline at end of file diff --git a/problems/python3/unique-paths.py b/problems/python3/unique-paths.py new file mode 100644 index 0000000..9c7ce40 --- /dev/null +++ b/problems/python3/unique-paths.py @@ -0,0 +1,11 @@ +class Solution: + def uniquePaths(self, m: int, n: int) -> int: + m = m-1 #number of steps need to move down + n = n-1 #number of steps need to move right + + #the total combination of m and n to sort will be (m+n)! + #since all "move down" are consider the same, we need to remove the repeatition of it sorting: m!. + #since all "move right" are consider the same, we need to remove the repeatition of it sorting: n!. + #(m+n)!/m!n! + + return math.factorial(m+n)//(math.factorial(m)*math.factorial(n)) \ No newline at end of file diff --git a/problems/python3/valid-anagram.py b/problems/python3/valid-anagram.py new file mode 100644 index 0000000..6c37cb0 --- /dev/null +++ b/problems/python3/valid-anagram.py @@ -0,0 +1,17 @@ +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + counter = collections.Counter() + + for c in s: + counter[c] += 1 + + for c in t: + if c not in counter: + return False + counter[c] -= 1 + + for c in counter: + if counter[c]>0: + return False + + return True \ No newline at end of file diff --git a/problems/python3/valid-palindrome.py b/problems/python3/valid-palindrome.py new file mode 100644 index 0000000..0069c83 --- /dev/null +++ b/problems/python3/valid-palindrome.py @@ -0,0 +1,15 @@ +class Solution: + def isPalindrome(self, s: str) -> bool: + i = 0 + j = len(s)-1 + + while i<=j: + while i bool: + mapping = {')': '(', ']': '[', '}':'{'} + stack = [] + + for c in s: + if c not in mapping: + #open parentheses + stack.append(c) + else: + #close parentheses + if stack and stack[-1]==mapping[c]: + stack.pop() + else: + return False + + return not stack \ No newline at end of file diff --git a/problems/python3/valid-parenthesis-string.py b/problems/python3/valid-parenthesis-string.py new file mode 100644 index 0000000..52ea446 --- /dev/null +++ b/problems/python3/valid-parenthesis-string.py @@ -0,0 +1,23 @@ +class Solution: + def checkValidString(self, s: str) -> bool: + leftMin = 0 + leftMax = 0 + + for c in s: + if c=='(': + leftMin += 1 + leftMax += 1 + elif c==')': + leftMin -= 1 + leftMax -= 1 + else: + leftMin -= 1 + leftMax += 1 + + if leftMax<0: + return False + + if leftMin<0: + leftMin = 0 + + return leftMin == 0 \ No newline at end of file diff --git a/problems/python3/valid-sudoku.py b/problems/python3/valid-sudoku.py new file mode 100644 index 0000000..b2e6fa7 --- /dev/null +++ b/problems/python3/valid-sudoku.py @@ -0,0 +1,37 @@ +class Solution: + def isValidSudoku(self, board: List[List[str]]) -> bool: + def rowIsValid(i): + used = set() + for j in range(N): + if board[i][j]=='.': continue + if board[i][j] in used: return False + used.add(board[i][j]) + return True + + def columnIsValid(i): + used = set() + for j in range(N): + if board[j][i]=='.': continue + if board[j][i] in used: return False + used.add(board[j][i]) + return True + + def isBoxValid(i1, i2, j1, j2): + used = set() + for i in range(i1, i2+1): + for j in range(j1, j2+1): + if board[i][j]=='.': continue + if board[i][j] in used: return False + used.add(board[i][j]) + return True + + N = 9 + for i in range(N): + if not rowIsValid(i): return False + if not columnIsValid(i): return False + + for i in range(0, N, 3): + for j in range(0, N, 3): + if not isBoxValid(i, i+2, j, j+2): return False + + return True \ No newline at end of file diff --git a/problems/python3/validate-binary-search-tree.py b/problems/python3/validate-binary-search-tree.py new file mode 100755 index 0000000..4185bc6 --- /dev/null +++ b/problems/python3/validate-binary-search-tree.py @@ -0,0 +1,24 @@ +""" +The inorder travsersal of a BST is always increasing. + +Time: O(N) +Space: O(N) +""" +class Solution: + def isValidBST(self, root: Optional[TreeNode]) -> bool: + lastVal = float('-inf') + stack = [] + + node = root + while stack or node: + while node: + stack.append(node) + node = node.left + + node = stack.pop() + + if not lastVal None: + def bfs(i0, j0) -> None: + q = collections.deque([(i0+1, j0, 1), (i0-1, j0, 1), (i0, j0+1, 1), (i0, j0-1, 1)]) + visited = set() + + while q: + i, j, dis = q.popleft() + + if i<0 or j<0 or i>=MAX_ROW or j>=MAX_COL: continue + if rooms[i][j]==0 or rooms[i][j]==-1: continue + if (i, j, dis) in visited: continue + visited.add((i, j, dis)) + + if dis bool: + def dfs(i): + if i==len(s): return True + if i in history and not history[i]: return False + + for word in wordDict: + if i+len(word)<=len(s) and s[i:i+len(word)]==word: + history[i] = True + if dfs(i+len(word)): return True + history[i] = False + return False + + history = {} + return dfs(0) \ No newline at end of file diff --git a/problems/python3/word-ladder.py b/problems/python3/word-ladder.py new file mode 100644 index 0000000..453fb4a --- /dev/null +++ b/problems/python3/word-ladder.py @@ -0,0 +1,40 @@ +""" +Time: O(NxM^2). N is the number of words. M is the length of the word. +Note that, getPatterns() takes O(M^2) since creating new string will also takes O(M) and for each word we do that O(M) times. + +Space: O(NxM^2) +""" +class Solution: + def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int: + def getPatterns(word) -> List[str]: + patterns = [] + for i in range(len(word)): + pattern = word[:i]+'*'+word[i+1:] + patterns.append(pattern) + return patterns + + + if endWord not in wordList: return 0 + wordList.append(beginWord) + + #build adjacency list + nei = collections.defaultdict(list) + for word in wordList: + for pattern in getPatterns(word): + nei[pattern].append(word) + + #BFS + q = collections.deque([(beginWord, 1)]) + visited = set() + while q: + word, steps = q.popleft() + + if word in visited: continue + visited.add(word) + if word==endWord: return steps + + for pattern in getPatterns(word): + for nextWord in nei[pattern]: + if nextWord in visited: continue + q.append((nextWord, steps+1)) + return 0 \ No newline at end of file diff --git a/problems/python3/word-search-ii.py b/problems/python3/word-search-ii.py new file mode 100644 index 0000000..a285655 --- /dev/null +++ b/problems/python3/word-search-ii.py @@ -0,0 +1,54 @@ +""" +Time: O(M * 4 * 3^(L-1)), M is board elements count, L is the average word length. +Space: O(N), N is the number of char in the words + +This solution will TLE. We need to further "prune" the trie once we reach the leaf node. +""" +class TrieNode(): + def __init__(self): + self.children = {} + self.isEnd = False + def addWord(self, word): + curr = self + for c in word: + if c not in curr.children: + curr.children[c] = TrieNode() + curr = curr.children[c] + curr.isEnd = True + +class Solution: + def findWords(self, board: List[List[str]], words: List[str]) -> List[str]: + def dfs(r, c, node, word): + if c<0 or r<0 or c==COLS or r==ROWS: return + if board[r][c] not in node.children: return + if (r, c) in visited: return + + visited.add((r, c)) + node = node.children[board[r][c]] + word += board[r][c] + if node.isEnd: + ans.add(word) + node.isEnd = False + + dfs(r+1, c, node, word) + dfs(r-1, c, node, word) + dfs(r, c+1, node, word) + dfs(r, c-1, node, word) + + visited.remove((r, c)) + + + root = TrieNode() + for word in words: + root.addWord(word) + + ROWS = len(board) + COLS = len(board[0]) + + visited = set() + ans = set() + node = root + for r in range(ROWS): + for c in range(COLS): + dfs(r, c, root, '') + return ans \ No newline at end of file diff --git a/problems/python3/word-search.py b/problems/python3/word-search.py new file mode 100644 index 0000000..b7ebb09 --- /dev/null +++ b/problems/python3/word-search.py @@ -0,0 +1,27 @@ +class Solution: + def exist(self, board: List[List[str]], word: str) -> bool: + def helper(x, i, j): + if i<0 or i>=N or j<0 or j>=M: return False + if (i, j) in usedWords: return False + usedWords.add((i, j)) + + if word[x]==board[i][j]: + if x==len(word)-1: + return True + else: + if helper(x+1, i+1, j): return True + if helper(x+1, i, j+1): return True + if helper(x+1, i-1, j): return True + if helper(x+1, i, j-1): return True + + + usedWords.remove((i, j)) + return False + + usedWords = set() + N = len(board) + M = len(board[0]) + for i in range(N): + for j in range(M): + if helper(0, i, j): return True + return False \ No newline at end of file diff --git a/problems/same-tree.py b/problems/same-tree.py deleted file mode 100644 index e17c40c..0000000 --- a/problems/same-tree.py +++ /dev/null @@ -1,23 +0,0 @@ -from collections import deque - -class Solution(object): - def isSameTree(self, p, q): - q1 = deque([p]) - q2 = deque([q]) - while q1 or q2: - if not q1 or not q2: return False - n1 = q1.popleft() - n2 = q2.popleft() - - if n1 and n2: - if n1.val!=n2.val: return False - q1.append(n1.left) - q1.append(n1.right) - q2.append(n2.left) - q2.append(n2.right) - elif n1 and not n2: - return False - elif not n1 and n2: - return False - return True - diff --git a/problems/search-in-a-binary-search-tree.py b/problems/search-in-a-binary-search-tree.py deleted file mode 100644 index 6fc233e..0000000 --- a/problems/search-in-a-binary-search-tree.py +++ /dev/null @@ -1,19 +0,0 @@ -class Solution(object): - def searchBST(self, root, val): - node = root - - while node: - if node.val==val: - return node - elif node.val>val: - node = node.left - else: - node = node.right - - return None - - -""" -Time complexity: O(LogN) -Space complexity: O(1) -""" \ No newline at end of file diff --git a/problems/unique-binary-search-trees.py b/problems/unique-binary-search-trees.py deleted file mode 100644 index 65dd64e..0000000 --- a/problems/unique-binary-search-trees.py +++ /dev/null @@ -1,14 +0,0 @@ - -# Implement from https://www.youtube.com/watch?v=GgP75HAvrlY -class Solution(object): - def numTrees(self, n): - dp = [1, 1, 2] - - if n<=2: return dp[n] - - for i in xrange(3, n+1): - dp.append(0) - for root in xrange(1, i+1): - dp[i] += dp[root-1]*dp[i-root] - - return dp[n] \ No newline at end of file diff --git a/problems/vertical-order-traversal-of-a-binary-tree.py b/problems/vertical-order-traversal-of-a-binary-tree.py deleted file mode 100644 index 9cc1849..0000000 --- a/problems/vertical-order-traversal-of-a-binary-tree.py +++ /dev/null @@ -1,28 +0,0 @@ -""" -The description wants us to report from left to right, top to bottom. -Thus, we keep add value into `temp` with `x_position` as the key and `(y_position, node.val)` as the value. -And maintain `min_x`, `max_x` at the same time. So we know how to iterate the `temp`. -""" -from collections import defaultdict - -class Solution(object): - def verticalTraversal(self, root): - temp = defaultdict(list) - min_x = 0 - max_x = 0 - - stack = [] - stack.append((root, 0, 0)) - while stack: - node, x, y = stack.pop() - temp[x].append((y, node.val)) #append the value of height, so we can sort by height later on - min_x = min(min_x, x) - max_x = max(max_x, x) - if node.left: stack.append((node.left, x-1, y+1)) - if node.right: stack.append((node.right, x+1, y+1)) - - opt = [] - for i in range(min_x, max_x+1): - opt.append([v for y, v in sorted(temp[i])]) #the temp[i] will be sorted by y_position then sorted by node.val - - return opt \ No newline at end of file diff --git a/problems/word-search-ii.py b/problems/word-search-ii.py deleted file mode 100644 index e7d7b91..0000000 --- a/problems/word-search-ii.py +++ /dev/null @@ -1,50 +0,0 @@ -# TLE -class Solution(object): - def findWords(self, board, words): - def getNeighbor(i, j): - opt = [] - if i+1=0: opt.append((i-1, j)) - if j-1>=0: opt.append((i, j-1)) - return opt - - def dfs(i, j, l, word): - if word in found: return - if board[i][j]!=word[l]: return - - if l==len(word)-1 and board[i][j]==word[l]: - opt.append(word) - found.add(word) - return - - char = board[i][j] - board[i][j] = '#' - - for ni, nj in getNeighbor(i, j): - dfs(ni, nj, l+1, word) - - board[i][j] = char - - opt = [] - M = len(board) - N = len(board[0]) - found = set() - - for i in xrange(M): - for j in xrange(N): - for word in words: - if word not in found: - dfs(i, j, 0, word) - return opt - - -board = [ - ['o','a','a','n'], - ['e','t','a','e'], - ['i','h','k','r'], - ['i','f','l','v'] -] -words = ["oath","pea","eat","rain"] - -print Solution().findWords(board, words)