<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>

//way 2-Using DSU
class disjoint_set{
public:
    vector<int> rank,parent;
    disjoint_set(int n)
    {
        rank.resize(n+1,0);
        parent.resize(n+1);
        for(int i=0;i<=n;i++) parent[i]=i;
    }

    int find_ult_par(int node)
    {
        if(node==parent[node]) return node;
        return parent[node]=find_ult_par(parent[node]);
    }

    void union_by_rank(int u,int v)
    {
        int pu=find_ult_par(u);
        int pv=find_ult_par(v);
        if(pu==pv) return ;
        if(rank[pu]<rank[pv]) parent[pu]=pv;
        else if(rank[pv]<rank[pu]) parent[pv]=pu;
        else {
            rank[pu]++;
            parent[pv]=pu;
        }
    }
};

class Solution {
public:
    bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
        //step 1-make an adjacency list
        vector<int> adj[n+1];
        for(auto x:dislikes)
        {
            int u=x[0],v=x[1];//u----v
            adj[u].push_back(v);
            adj[v].push_back(u);
        }

        //step 2-use disjoint set and join the adjacents of a node,if current node and its adjacent is on same connected component means we can't partition , so return false in that case,take examples and see using pen & paper
        disjoint_set ds(n);
        for(int i=1;i<=n;i++)
        {
            int u=i;
            vector<int> adj_list=adj[i];
            int pu= ds.find_ult_par(u);
            for(auto v:adj_list)
            {
                int pv= ds.find_ult_par(v);
                if(pu==pv) return false;//they should not be on same group
                ds.union_by_rank(adj_list[0],v);//join the adjacents of 'u' to each other
            }
        }

        return true;
    }
};
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