Is Graph Bipartition

Given an undirected graph, returntrueif 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 indexesjfor which the edge between nodesiandjexists. Each node is an integer between0andgraph.length - 1. There are no self edges or parallel edges:graph[i]does not containi, and it doesn't contain any element twice.

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].

Example

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}.

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. Assign RED color to the source vertex (putting into set U).

2. Color all the neighbors with BLUE color (putting into set V).

3. Color all neighbor’s neighbor with RED color (putting into set U).

4. This way, assign color to all vertices such that it satisfies all the constraints of m way coloring problem where m = 2.

5. While assigning colors, if we find a neighbor which is colored with same color as current vertex, then the graph cannot be colored with 2 vertices (or graph is not Bipartite)

Code

class Solution {
    public boolean isBipartite(int[][] graph) {
        if (graph == null || graph.length == 0) return false;
        int[] color = new int[graph.length];
        Queue<Integer> q = new LinkedList<>();
        //Divide the node in two sets such that no two nodes in same set are linked
        //[[1,2,3], [0,2], [0,1,3], [0,2]]
        for (int i = 0; i < graph.length; i++) {
            if (color[i] == 0) {
                //color and send in queue
                q.offer(i);
                color[i] = 1;
                while (!q.isEmpty()) {
                    int node = q.poll();
                    for (int child : graph[node]) {
                        if (color[child] == 0) {
                            //color distinct color
                            if (color[node] == 2) color[child] = 1;
                            else color[child] = 2;
                            q.offer(child);
                        } else if (color[child] == color[node]) {
                            return false;
                        }                  
                    }
                }
            }
        }
        return true;
    }
}