<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 possibleBipartition(int n, vector<vector<int>>& dislikes) {
        unordered_map<int, vector<int>> mp;
        
        // Create adjacency list for the undirected graph
        for (int i = 0; i < dislikes.size(); i++) {
            mp[dislikes[i][0]].push_back(dislikes[i][1]);
            mp[dislikes[i][1]].push_back(dislikes[i][0]); // Add the reverse edge
        }
        
        vector<int> visited(n + 1, 0);
        vector<int> room(n + 1, -1); // -1 indicates uncolored
        
        for (int i = 1; i <= n; i++) {
            if (visited[i]) {
                continue;
            }
            
            queue<pair<int, int>> q;
            q.push({i, 0});
            
            while (!q.empty()) {
                int a = q.front().first;
                int ct = q.front().second;
                q.pop();
                if (room[a] != -1) {
                    if (room[a] != ct) {
                        return false;
                    }
                    continue;
                }
                
                room[a] = ct; 
                visited[a] = 1;
                
                for (int x : mp[a]) {
                    if (room[x] == ct) {
                        return false; 
                    }
                    if (!visited[x]) {
                        q.push({x, 1 - ct});
                    }
                }
            }
        }
        return true;
    }
};

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