<p>We want to split a group of <code>n</code> people (labeled from <code>1</code> to <code>n</code>) into two groups of <strong>any size</strong>. Each person may dislike some other people, and they should not go into the same group.</p>

<p>Given the integer <code>n</code> and the array <code>dislikes</code> where <code>dislikes[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that the person labeled <code>a<sub>i</sub></code> does not like the person labeled <code>b<sub>i</sub></code>, return <code>true</code> <em>if it is possible to split everyone into two groups in this way</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4, dislikes = [[1,2],[1,3],[2,4]]
<strong>Output:</strong> true
<strong>Explanation:</strong> The first group has [1,4], and the second group has [2,3].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, dislikes = [[1,2],[1,3],[2,3]]
<strong>Output:</strong> false
<strong>Explanation:</strong> We need at least 3 groups to divide them. We cannot put them in two groups.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 2000</code></li>
	<li><code>0 &lt;= dislikes.length &lt;= 10<sup>4</sup></code></li>
	<li><code>dislikes[i].length == 2</code></li>
	<li><code>1 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt;= n</code></li>
	<li>All the pairs of <code>dislikes</code> are <strong>unique</strong>.</li>
</ul>

#define WHITE 0
#define RED 1
#define BLUE 2

class Solution 
{
public:
    bool possibleBipartition(int N, vector<vector<int>> &edges) 
    {
        vector<vector<int>> adj(N + 1); // adjacency list for undirected graph
        vector<int> color(N + 1, WHITE); // color of each vertex in graph, initially WHITE
        vector<bool> explored(N + 1, false); // to check if each vertex has been explored exactly once
        
        // create adjacency list from given edges
        for (auto &edge: edges)
        {
            int u = edge[0];
            int v = edge[1];
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        
        // print adjacency list (comment out before submitting)
        for (int i = 0; i <= N; ++i)
        {
            cout << "adj[" << i << "]: ";
            for (int j = 0; j < adj[i].size(); ++j)
            {
                cout << adj[i][j] << " ";
            }
            cout << "\n";
        }
        
        // queue to perform BFS over each connected component in the graph
        // while performing BFS, we check if we encounter any conflicts while
        // coloring the vertices of the graph
        // conflicts indicate that bi-partition is not possible
        queue<int> q;
        
        for (int i = 1; i <= N; ++i)
        {
            if (!explored[i])
            {
                // this component has not been colored yet
				// we color the first vertex RED and push it into the queue
                color[i] = RED;
                q.push(i);
                
                // perform BFS from vertex i
                while (!q.empty())
                {
                    int u = q.front();
                    q.pop();
                    
                    // check if u is already explored 
                    if (explored[u])
                    {
                        continue;
                    }
                    
                    explored[u] = true;
                    
                    // for each neighbor of u, execute this loop
                    for (auto v: adj[u])
                    {
                        // v is u's neighboring vertex
                        
                        // checking if there's any conflict in coloring
                        if (color[v] == color[u])
                        {
							// conflict edge found, so we return false 
							// as bi-partition will not be possible
                            return false;
                        }
                        
                        // we color v with the opposite color of u
                        if (color[u] == RED)
                        {
                            color[v] = BLUE;
                        }
                        else 
                        {
                            color[v] = RED;
                        }
                        
                        q.push(v);
                    }
                }
            }
        }
        
        // if no conflicts encountered then graph must be bipartite
        // so we return true
        
        return true;
    }
};
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