Adding Accepted Solution - Problem - 27 - Possible Bipartition

Author: Vanditg

Leetcode-May-Challenge/Day - 27/AcceptedSolution.py

@@ -0,0 +1,59 @@
+##==================================
+## Leetcode May Challenge 
+## Username: Vanditg
+## Year: 2020
+## Problem: 27
+## Problem Name: Possible Bipartition
+##===================================
+#
+#Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size.
+#
+#Each person may dislike some other people, and they should not go into the same group. 
+#
+#Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.
+#
+#Return true if and only if it is possible to split everyone into two groups in this way.
+#
+#Example 1:
+#
+#Input: N = 4, dislikes = [[1,2],[1,3],[2,4]]
+#Output: true
+#Explanation: group1 [1,4], group2 [2,3]
+#Example 2:
+#
+#Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
+#Output: false
+#Example 3:
+#
+#Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
+#Output: false
+class Solution:
+    def possibleBipartition(self, N, dislikes):
+        tmp = {}
+        for dislike in dislikes:
+            if dislike[0] in tmp:
+                tmp[dislike[0]].add(dislike[1])
+            else:
+                tmp[dislike[0]] = set([dislike[1]])
+            if dislike[1] in tmp:
+                tmp[dislike[1]].add(dislike[0])
+            else:
+                tmp[dislike[1]] = set([dislike[0]])
+        
+        lookup = {}
+        for i in range(1, N+1):
+            if i not in lookup:
+                lookup[i] = 0
+                stack = [i]
+                while stack:
+                    a = stack.pop()
+                    if a in tmp:
+                        for b in tmp[a]:
+                            if b in lookup:
+                                if lookup[a] == lookup[b]:
+                                    return False
+                            else:
+                                lookup[b] = (lookup[a]+1)%2
+                                stack.append(b)
+        return True
+