Tags: "leetcode", "graph", "dfs", "coloring", access_time 2-min read

Is Graph Bipartite

Created: February 21, 2020 by [lek-tin]

Last updated: February 21, 2020

Given an undirected 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 subsets.

Note

1. graph will have length in range [1, 100].
2. graph[i] will contain integers in range [0, graph.length - 1].
3. graph[i] will not contain i or duplicate values.
4. The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

Solution

class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        color = {}

        for node in range(len(graph)):
            if node not in color:
                color[node] = 0
                stack = [node]

                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