Is Graph Bipartite?


#1

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:

  • 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.
  • The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

#2

The below code uses BFS to color all nodes into two different colors, if this is not possible, we return false.

class Solution {
    public boolean isBipartite(int[][] graph) {
       int len = graph.length;
       int[] colors = new int[len];
        
        for(int i=0;i<len;i++){
            if(colors[i]!=0){continue;}
            Queue<Integer> queue = new LinkedList<>();
            queue.offer(i);
            colors[i] = 1;
            while(!queue.isEmpty()){
                int cur = queue.poll();
                for(int next: graph[cur]){
                    if(colors[next]==0){
                        colors[next]=-colors[cur];
                        queue.offer(next);
                    } else if (colors[next] != -colors[cur]){
                        return false;
                    }
                }
            }
        }
        return true;
    }
}