<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:
    map<int,int> UF, pre;
    int Find(int i){
        while(UF[i] != i){
            i = UF[i];
        }
        return i;
    }

    bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
        if(dislikes.size() == 0)return true;

        for(auto &dislike : dislikes){
            if(UF.find(dislike[0])==UF.end()){
                UF[dislike[0]] = dislike[0];
            }
            if(UF.find(dislike[1])==UF.end()){
                UF[dislike[1]] = dislike[1];
            }
           
            if(UF[dislike[0]] == dislike[0] && pre.find(dislike[1])!=pre.end()){
                UF[dislike[0]] = Find(pre[dislike[1]]);
            }
           
            if(UF[dislike[1]] == dislike[1] && pre.find(dislike[0])!=pre.end()){
                UF[dislike[1]] = Find(pre[dislike[0]]);
            }

            int root1 = Find(dislike[0]);
            int root2 = Find(dislike[1]);
            
            if(root1==root2){
                return false;
            }

            pre[dislike[0]] = dislike[1];
            pre[dislike[1]] = dislike[0];
        }
        return true;
    }
};
61403