<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 Solution {
public:

    vector<int>parent;
    vector<int>rank;

    int find(int x)
    {
        if(x == parent[x] )
            return x;
        return parent[x]=find(parent[x]);
    }

    void unionset(int x,int y)
    {
        int x_parent=find(x);
        int y_parent=find(y);

        if(x_parent == y_parent )
            return ;

        if(rank[x_parent] > rank[y_parent])
        {
            parent[y_parent] = x_parent;
        }
        else if(rank[x_parent] < rank[y_parent])
        {
           parent[x_parent] =y_parent;

        }
        else
        {
            parent[x_parent] =y_parent;
            rank[y_parent]++;
        }
    }
    long long countPairs(int n, vector<vector<int>>& edges) {
        
        parent.resize(n);
        rank.resize(n,0);
        for(int i=0;i<n;i++)
            parent[i]=i;


        for(auto &vec : edges)
        {
            int u=vec[0];
            int v=vec[1];

            unionset(u,v);
        }

        unordered_map<int,int>mp;

        for(int i=0;i<n;i++)
        {
            int papaji=find(i);
            mp[papaji]++;
        }


        long long int remain =n;
        long long int result=0;

        for( auto & x :  mp)
        {
            int size=x.second;

            result+=size * (remain-size);
            remain-=size;
            
        }
        return result;
    }
};
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