# Time:  O(|V| + |E|)
# Space: O(|V|)

# Given a graph, return true if and only if it is bipartite.
#
# Recall that a graph is bipartite if we can split it's set of nodes into
# two independent subsets A and B such that every edge in the graph has
# one node in A and another node in B.
#
# The graph is given in the following form: graph[i] is a list of indexes j
# for which the edge between nodes i and j exists. 
# Each node is an integer between 0 and graph.length - 1.
# There are no self edges or parallel edges: graph[i] does not contain i,
# and it doesn't contain any element twice.
#
# Example 1:
# Input: [[1,3], [0,2], [1,3], [0,2]]
# Output: true
# Explanation: 
# The graph looks like this:
# 0----1
# |    |
# |    |
# 3----2
# We can divide the vertices into two groups: {0, 2} and {1, 3}.
#  
# Example 2:
# Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
# Output: false
# Explanation: 
# The graph looks like this:
# 0----1
# | \  |
# |  \ |
# 3----2
# We cannot find a way to divide the set of nodes into two independent ubsets.
#
# Note:
# - graph will have length in range [1, 100].
# - graph[i] will contain integers in range [0, graph.length - 1].
# - graph[i] will not contain i or duplicate values.

class Solution(object):
    def isBipartite(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: bool
        """
        color = {}
        for node in xrange(len(graph)):
            if node in color:
                continue
            stack = [node]
            color[node] = 0
            while stack:
                curr = stack.pop()
                for neighbor in graph[curr]:
                    if neighbor not in color:
                        stack.append(neighbor)
                        color[neighbor] = color[curr] ^ 1
                    elif color[neighbor] == color[curr]:
                        return False
        return True

Create is-graph-bipartite.py