<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 {
    int find_par(vector<int>&par,int node){
        if(par[node]==node)return node;
        return find_par(par,par[node]);
    }
public:
    bool possibleBipartition(int n, vector<vector<int>>& d) {
        vector<int>par(n+1);
        for(int i=0;i<n+1;i++)par[i]=i;
        unordered_map<int,int>mp;
        for(auto a:d){
            int one=find_par(par,a[0]);
            int two=find_par(par,a[1]);
            if(one==two)return false;
            if(mp.count(one)==0 && mp.count(two)==0){
                mp[one]=two;
                mp[two]=one;
                par[a[0]]=one;
                par[a[1]]=two;
            }
            else if(mp.count(one)!=0 && mp.count(two)==0){
                par[a[0]]=one;
                par[a[1]]=mp[one];
            }
            else if(mp.count(one)==0 && mp.count(two)!=0){
                par[a[0]]=mp[two];
                par[a[1]]=two;
            }
            else{
                if(mp[one]==two){
                    par[a[0]]=one;
                    par[a[1]]=two;
                }
                else{
                    par[a[0]]=one;
                    par[a[1]]=mp[one];
                    par[two]=mp[one];
                    par[mp[two]]=one;
                    mp.erase(two);
                }
            }
        }
        return true;
    }
};
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