<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;
    vector<long long>size; 
    Dsu(int n)
    {
        parent.resize(n);
        size.resize(n);

        for(int i=0;i<n;i++) {parent[i]=i; size[i]=1;}
    }
    int Upar(int node)
    {
        if(parent[node]==node)
        return node;
        return parent[node]=Upar(parent[node]);
    }
    void Union(int u,int v)
    {
        int ulp_u=Upar(u);
        int ulp_v=Upar(v);

        if(ulp_u==ulp_v) return;

        if(size[ulp_u]<size[ulp_v])
        {
            parent[ulp_u]=ulp_v;
            size[ulp_v]+=size[ulp_u];
        }
        else
        {
            parent[ulp_v]=ulp_u;
            size[ulp_u]+=size[ulp_v];
        }
    }
};

class Solution {
public:
    long long countPairs(int n, vector<vector<int>>& edges) {
        
        Dsu ds(n);
        vector<long long>v;

        for(auto it:edges)
        {
            if(ds.Upar(it[0]) != ds.Upar(it[1]))
            ds.Union(it[0],it[1]);
        }

        for(int i=0;i<n;i++)
        {
            if(ds.Upar(i) == i)
            v.push_back(ds.size[i]);
        }
        sort(v.begin(),v.end());

        int m=v.size();

        if(m == 1) return 0;

        vector<long long>pf(m-1,0);
        pf[0] = v[0];
        for(int i=1;i<m-1;i++)
        {
            pf[i] = pf[i-1]+v[i];
        }
        long long ans = 0;

        for(int i=m-1;i>0;i--)
        {
            ans += v[i]*pf[i-1];
        }

        return ans;

    }
};
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