<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) {
        vector<int> p(n + 1, -1);
        unordered_map<int, vector<int>> ma;

        for (auto &x : dislikes) {
            ma[x[0]].push_back(x[1]);
            ma[x[1]].push_back(x[0]);
        }

        for (int i = 1; i <= n; i++) {
            queue<int> q;
            if (p[i] != -1)
                continue;
            q.push(i);
            p[i]=0;
            while (!q.empty()) {
                auto x = q.front();
                q.pop();
               
                for (auto &ne : ma[x]) {
                    if (p[ne] == p[x])
                        return false;
                    if (p[ne] == -1) {
                        p[ne] = 1 - p[x];
                        q.push(ne);
                    }
                    
                }
            }
        }
        return true;
    }
};
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