Problem Statement:

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:

The idea is to check if a two-color coloring can be done to the graph. Colors={1,-1} and 0 denotes uncolored. Start with node 0 and traverse by BFS. Then for any other connected component repeat. To do this, we can run for all i from 0 to n and only start doing BFS if it is uncolored.

Note that uncolored = unvisited.

 
class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) 
    {
        int V = graph.size();
        vector<int> colors = vector<int>(V,0);
        queue<int> Q;

        for (int i=0; i<V; i++)
        {
            if (colors[i]!=0) continue;
            colors[i] = 1;
            Q.push(i);
            while(!Q.empty())
            {
                int u = Q.front();
                Q.pop();
                for (int v: graph[u])
                {
                    if (colors[v]==0)
                    {
                        colors[v] = -colors[u];
                        Q.push(v);
                    }
                    else
                        if (colors[v]==colors[u])
                            return false;
                }
            }
        }
        return true;
    }
};