Possible Bipartition.py

'''
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

 

Note:

    1 <= N <= 2000
    0 <= dislikes.length <= 10000
    1 <= dislikes[i][j] <= N
    dislikes[i][0] < dislikes[i][1]
    There does not exist i != j for which dislikes[i] == dislikes[j].


'''

class Solution(object):
    def possibleBipartition(self, N, dislikes):
        """
        :type N: int
        :type dislikes: List[List[int]]
        :rtype: bool
        """
        dic = {}
        for i in xrange(1, N+1):
            dic[i] = []
        for a, b, in dislikes:
            dic[a].append(b)
            dic[b].append(a)
        
        
        color = {1: 1}
        vec = [1]
        visited = set([1])
        while vec:
            next_vec = []
            for p in vec:
                for c in dic[p]:
                    if c in color:
                        if color[c] == color[p]:
                            return False
                    else:
                        color[c] = 1 - color[p]
                    
                    if c not in visited:
                        visited.add(c)
                        next_vec.append(c)
            if not next_vec:
                for idx in xrange(1, N+1):
                    if idx not in visited:
                        next_vec.append(idx)
                        visited.add(idx)
                        break
            vec = next_vec
        
        return True