<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>

class Solution {
public:
   bool dfs(int node, int col, int color[], vector<int> adj[]) {
        color[node] = col; 
        
        // traverse adjacent nodes
        for(auto it : adj[node]) {
            // if uncoloured
            if(color[it] == -1) {
                if(dfs(it, !col, color, adj) == false) return false; 
            }
            // if previously coloured and have the same colour
            else if(color[it] == col) {
                return false; 
            }
        }
        
        return true; 
    }
    bool possibleBipartition(int V, vector<vector<int>>& dislikes) {
        vector<int> adj[V + 1];
        for (auto it : dislikes) {
            adj[it[0]].push_back(it[1]);
            adj[it[1]].push_back(it[0]);
        }
       int color[20000];
	    for(int i = 0;i<=V;i++) color[i] = -1; 
	    
	    // for connected components
	    for(int i = 0;i<V;i++) {
	        if(color[i] == -1) {
	            if(dfs(i, 0, color, adj) == false) 
	                return false; 
	        }
	    }
	    return true; 
    }
};
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