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Is Graph Bipartite?

Updated: Mar 25, 2021

There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties:

  • There are no self-edges (graph[u] does not contain u).

  • There are no parallel edges (graph[u] does not contain duplicate values).

  • If v is in graph[u], then u is in graph[v] (the graph is undirected).

  • The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them.

A graph is bipartite if the nodes can be partitioned into two independent sets A and B such that every edge in the graph connects a node in set A and a node in set B.

Return true if and only if it is bipartite.


Example 1:

Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
Output: false
Explanation: There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other.

Example 2:

Input: graph = [[1,3],[0,2],[1,3],[0,2]]
Output: true
Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}.

Constraints:

  • graph.length == n

  • 1 <= n <= 100

  • 0 <= graph[u].length < n

  • 0 <= graph[u][i] <= n - 1

  • graph[u] does not contain u.

  • All the values of graph[u] are unique.

  • If graph[u] contains v, then graph[v] contains u.

Solution:


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

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