<p>You are given an integer <code>n</code>. There is an <strong>undirected</strong> graph with <code>n</code> nodes, numbered from <code>0</code> to <code>n - 1</code>. You are given a 2D integer array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> denotes that there exists an <strong>undirected</strong> edge connecting nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>.</p>

<p>Return <em>the <strong>number of pairs</strong> of different nodes that are <strong>unreachable</strong> from each other</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/05/tc-3.png" style="width: 267px; height: 169px;" />
<pre>
<strong>Input:</strong> n = 3, edges = [[0,1],[0,2],[1,2]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no pairs of nodes that are unreachable from each other. Therefore, we return 0.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/05/tc-2.png" style="width: 295px; height: 269px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[0,2],[0,5],[2,4],[1,6],[5,4]]
<strong>Output:</strong> 14
<strong>Explanation:</strong> There are 14 pairs of nodes that are unreachable from each other:
[[0,1],[0,3],[0,6],[1,2],[1,3],[1,4],[1,5],[2,3],[2,6],[3,4],[3,5],[3,6],[4,6],[5,6]].
Therefore, we return 14.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= edges.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated edges.</li>
</ul>

class DSU
{
    public:
    vector<int>parent,size;
    DSU(int n)
    {
        for(int i=0;i<n;++i)
        {
            parent.push_back(i);
            size.push_back(1);
        }
    }
    
    int findparent(int node)
    {
        if(parent[node]==node)
            return node;
        return parent[node]=findparent(parent[node]);
    }
    
    void unionOf(int n1,int n2)
    {
        int up1=findparent(n1);
        int up2=findparent(n2);
        if(up1==up2)
            return;
        
        if(size[up1]<size[up2])
        {
            size[up2]+=size[up1];
            parent[up1]=up2;
        }
        else
        {
            size[up1]+=size[up2];
            parent[up2]=up1;   
        }
    }
};


class Solution {
public:
    long long countPairs(int n, vector<vector<int>>& edges) {
        
        DSU dsu(n);
    
        for(auto x:edges)
        {
            int u=x[0];
            int v=x[1];
            dsu.unionOf(u,v);
        }
        
        vector<int>sz;
        unordered_set<int>st;
        for(int i=0;i<n;++i)
        {
            int up1=dsu.findparent(i);
            if(st.find(up1)==st.end())
            {
                st.insert(up1);
                sz.push_back(dsu.size[up1]);
            }
        }
        
        
        int m=sz.size();
        vector<long long>prefix(m);
        prefix[m-1]=sz[m-1];
        for(int i=m-2;i>=0;--i)
            prefix[i]=prefix[i+1]+sz[i];
        
        long long int res=0;
        for(int i=0;i<m-1;i++)
            res+=(sz[i]*prefix[i+1]);
        
        
       
        return res;
        
    }
};










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