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OneCodeMonkeyOneCodeMonkey
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‎codeForces/Codeforces_0732B_The_Best_Friend_Of_a_Man.java‎

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// Time: 358 ms
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// Memory: 600 KB
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// dp | constructive
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// T:O(n), S:O(n)
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//
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import java.util.Scanner;
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public class Codeforces_0732B_The_Best_Friend_Of_a_Man {
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public static void main(String[] args) {
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Scanner sc = new Scanner(System.in);
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int n = sc.nextInt(), k = sc.nextInt(), prev = -1, ret = 0;
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int[] arr = new int[n];
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for (int i = 0; i < n; i++) {
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int a = sc.nextInt();
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if (i == 0) {
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prev = a;
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arr[i] = a;
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continue;
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}
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if (a + prev < k) {
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ret += k - a - prev;
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arr[i] = k - prev;
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} else {
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arr[i] = a;
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}
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prev = arr[i];
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}
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System.out.println(ret);
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for (int i = 0; i < n; i++) {
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System.out.print(arr[i]);
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if (i != n - 1) {
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System.out.print(" ");
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} else {
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System.out.println();
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}
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}
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}
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}

‎codeForces/Codeforces_1925A_We_Got_Everything_Covered.java‎

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// AC: 280 ms
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// Memory: 200 KB
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// Greedy. To Cover all possible k-alphabet character formed strings, just repeat "abc..." substring with length k,
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// repeat N times, this string can meets the requirement.
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// T:O(sum(ni * ki)), S:O(max(ni * ki))
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//
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import java.util.Scanner;
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public class Codeforces_1925A_We_Got_Everything_Covered {
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public static void main(String[] args) {
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Scanner sc = new Scanner(System.in);
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int t = sc.nextInt();
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for (int i = 0; i < t; i++) {
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int n = sc.nextInt(), k = sc.nextInt();
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String alphabetStr = "abcdefghijklmnopqrstuvwxyz";
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System.out.println(alphabetStr.substring(0, k).repeat(n));
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}
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}
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}

‎codeForces/Codeforces_1926A_Vlad_and_the_Best_of_Five.java‎

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// AC: 217 ms
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// Memory: 0 KB
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// .
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// T:O(t), S:O(1)
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//
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import java.util.Scanner;
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public class Codeforces_1926A_Vlad_and_the_Best_of_Five {
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public static void main(String[] args) {
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Scanner sc = new Scanner(System.in);
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int t = sc.nextInt();
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for (int i = 0; i < t; i++) {
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String s = sc.next();
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int countA = 0;
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for (char c : s.toCharArray()) {
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if (c == 'A') {
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countA++;
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}
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}
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System.out.println(countA > s.length() - countA ? 'A' : 'B');
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}
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}
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}

‎codeForces/Codeforces_1926B_Vlad_and_Shapes.java‎

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// AC: 280 ms
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// Memory: 100 KB
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// .
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// T:O(sum(ni)), S:O(max(ni))
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//
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import java.util.Scanner;
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public class Codeforces_1926B_Vlad_and_Shapes {
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public static void main(String[] args) {
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Scanner sc = new Scanner(System.in);
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int t = sc.nextInt();
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for (int i = 0; i < t; i++) {
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int n = sc.nextInt(), firstRowLen = 0, ret = 0;
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for (int j = 0; j < n; j++) {
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String row = sc.next();
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if (ret == 0 && row.contains("1")) {
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int countChar = 0;
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for (char c : row.toCharArray()) {
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if (c == '1') {
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countChar++;
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}
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}
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if (firstRowLen == 0) {
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firstRowLen = countChar;
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} else {
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if (countChar != firstRowLen) {
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ret = 1;
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} else {
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ret = -1;
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}
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}
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}
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}
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System.out.println(ret == 1 ? "TRIANGLE" : "SQUARE");
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}
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}
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}

‎codeForces/Codeforces_1931B_Make_Equal.java‎

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// AC: 685 ms
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// Memory: 600 KB
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// .
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// T:O(sum(ni)), S:O(max(ni))
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//
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import java.util.Scanner;
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public class Codeforces_1931B_Make_Equal {
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public static void main(String[] args) {
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Scanner sc = new Scanner(System.in);
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int t = sc.nextInt();
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for (int i = 0; i < t; i++) {
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int n = sc.nextInt();
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long sum = 0, remain = 0;
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int[] arr = new int[n];
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for (int j = 0; j < n; j++) {
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arr[j] = sc.nextInt();
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sum += arr[j];
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}
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boolean flag = true;
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long avg = sum / n;
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for (int j = 0; j < n; j++) {
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if (arr[j] + remain < avg) {
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flag = false;
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break;
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}
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remain = arr[j] + remain - avg;
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}
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System.out.println(flag ? "YES" : "NO");
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}
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}
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}

‎leetcode_solved/leetcode_0100_Same_Tree.java‎

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// Runtime 0 ms Beats 100.00% of users with Java
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// Memory 36.22 MB Beats 100.00% of users with Java
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// Recursion.
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// T:O(n), S:O(1)
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//
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class Solution {
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public boolean isSameTree(TreeNode p, TreeNode q) {
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if (p == null && q == null) {
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return true;
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}
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if ((p != null && q == null) || (p == null && q != null)) {
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return false;
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}
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if (p.val != q.val) {
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return false;
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}
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return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
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}
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}

‎leetcode_solved/leetcode_0892_Surface_Area_of_3D_Shapes.java‎

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// Runtime 6 ms Beats 42.58% of users with Java
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// Memory 44.26 MB Beats 43.54% of users with Java
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// Judge from 6 surfaces with each cell, sum up the surface area.
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// T:O(n ^ 2), S:O(1)
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//
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class Solution {
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public int surfaceArea(int[][] grid) {
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int len = grid.length, ret = 0;
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for (int i = 0; i < len; i++) {
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for (int j = 0; j < len; j++) {
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// top,bottom
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if (grid[i][j] > 0) {
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ret += 2;
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}
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// up
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if (i == 0) {
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ret += grid[i][j];
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} else {
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if (grid[i - 1][j] < grid[i][j]) {
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ret += grid[i][j] - grid[i - 1][j];
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}
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}
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// bottom
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if (i == len - 1) {
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ret += grid[i][j];
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} else {
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if (grid[i + 1][j] < grid[i][j]) {
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ret += grid[i][j] - grid[i + 1][j];
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}
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}
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// left
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if (j == 0) {
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ret += grid[i][j];
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} else {
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if (grid[i][j - 1] < grid[i][j]) {
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ret += grid[i][j] - grid[i][j - 1];
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}
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}
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// right
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if (j == len - 1) {
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ret += grid[i][j];
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} else {
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if (grid[i][j + 1] < grid[i][j]) {
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ret += grid[i][j] - grid[i][j + 1];
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}
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}
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}
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}
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return ret;
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}
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}

‎leetcode_solved/leetcode_3046_Split_the_Array.java‎

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// Runtime 3 ms Beats 50.00% of users with Java
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// Memory 42.78 MB Beats 50.00% of users with Java
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// Hashmap. Just check every element's occurence <= 2, if not, the answer is false.
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// T:O(n), S:O(n)
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//
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class Solution {
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public boolean isPossibleToSplit(int[] nums) {
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HashMap<Integer, Integer> record = new HashMap<>();
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for (int num : nums) {
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record.merge(num, 1, Integer::sum);
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}
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boolean ret = true;
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if (record.size() >= nums.length / 2) {
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for (int num : record.keySet()) {
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if (record.get(num) > 2) {
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ret = false;
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break;
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}
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}
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} else {
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ret = false;
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}
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return ret;
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}
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}

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