Minor Improvements.

orsenthil · orsenthil · commit a0de20fbb3ca · 2016-05-28T15:31:36.000-07:00
diff --git a/python/dynamic/bitonicsequence.py b/python/dynamic/bitonicsequence.py
@@ -35,6 +35,7 @@ def longest_bitonic(sequence):
                 decreasing_sequence[i] = max(decreasing_sequence[i], decreasing_sequence[j] + 1)
 
     max_value = 0
+
     for i in range(len(sequence)):
         bitonic_sequence_length = increasing_sequence[i] + decreasing_sequence[i] - 1
         max_value = max(max_value, bitonic_sequence_length)
diff --git a/python/dynamic/boxstacking.py b/python/dynamic/boxstacking.py
@@ -36,7 +36,10 @@
 
 def create_rotation(given_dimensions):
     """
-    A rotation is an order wherein length is greater than or equal to width.
+    A rotation is an order wherein length is greater than or equal to width. Having this constraint avoids the
+    repetition of same order, but with width and length switched.
+
+    For e.g (height=3, width=2, length=1) is same the same box for stacking as (height=3, width=1, length=2).
 
     :param given_dimensions: Original box dimensions
     :return: All the possible rotations of the boxes with the condition that length >= height.
@@ -64,8 +67,9 @@ def box_stack_max_height(dimensions):
     for i in range(1, num_boxes):
         for j in range(0, i):
             if can_stack(boxes[i], boxes[j]):
-                if (T[j] + boxes[i].height) > T[i]:
-                    T[i] = T[j] + boxes[i].height
+                stacked_height = T[j] + boxes[i].height
+                if stacked_height > T[i]:
+                    T[i] = stacked_height
                     R[i] = j
 
     max_height = max(T)
diff --git a/python/dynamic/breakword.py b/python/dynamic/breakword.py
@@ -1,12 +1,13 @@
 """
 Problem Statement
 =================
+
 Given a string and a dictionary, split the string in to multiple words so that each word belongs to the dictionary.
 
 Video
 -----
-* https://youtu.be/WepWFGxiwRs
 
+* https://youtu.be/WepWFGxiwRs
 
 Analysis
 --------
diff --git a/python/dynamic/coin_change_num_ways.py b/python/dynamic/coin_change_num_ways.py
@@ -27,7 +27,6 @@ def coin_changing_num_ways(coins, total):
     for i in range(rows):
         for j in range(cols):
             if (i - 1) < 0:
-                T[i][j] = 1
                 continue
             if j < coins[i]:
                 T[i][j] = T[i - 1][j]
diff --git a/python/dynamic/egg_drop.py b/python/dynamic/egg_drop.py
@@ -2,8 +2,8 @@
 Problem Statement
 =================
 
-Given a certain number of eggs and a certain number of floors, determine the minimum number of attempts
-required to find the egg breaking floor.
+Given a certain number of eggs and a certain number of floors, determine the minimum number of attempts required to find
+the egg breaking floor.
 
 Analysis
 --------
diff --git a/python/dynamic/longest_increasing_subsequence.py b/python/dynamic/longest_increasing_subsequence.py
@@ -3,7 +3,7 @@
 =================
 
 Find a subsequence in given array in which the subsequence's elements are in sorted order, lowest to highest, and in
-which the subsequence is as long as possible
+which the subsequence is as long as possible.
 
 Video
 -----
@@ -14,7 +14,7 @@
 
 Dynamic Programming is used to solve this question. DP equation is.::
 
-        if(arr[i] > arr[j]) { T[i] = max(T[i], T[j] + 1 }
+        if(arr[i] > arr[j]) { T[i] = max(T[i], T[j] + 1) }
 
 * Time complexity is O(n^2).
 * Space complexity is O(n)
diff --git a/python/graph/dijkstrashortestpath.py b/python/graph/dijkstrashortestpath.py
@@ -1,9 +1,12 @@
 #dijkstra's algorithm
+
 # java code https://github.com/mission-peace/interview/blob/master/src/com/interview/graph/DijkstraShortestPath.java
+
 from priorityqueue import PriorityQueue
 from graph import Graph
 import sys
 
+
 def shortest_path(graph, sourceVertex):
 
     min_heap = PriorityQueue(True)
@@ -36,6 +39,7 @@ def shortest_path(graph, sourceVertex):
 
     return distance
 
+
 def get_other_vertex_for_edge(vertex, edge):
     if edge.vertex1.id == vertex.id:
         return edge.vertex2
