<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:
//toh dekh muje jitne bhi edges derakhi 
//mai phle unko pair krke union karlunga
//theeke aise mai poor agraph components mai bat jaayega
//aur unko respective aprent bhi ban jaayenge

//ab kux ni parent banane ke baad 
// i will create a map jisme mera har component ke nodes count hojaaega
//then i will count the no. of pairs of unconnected nodes

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

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

        return parent[x]=find(parent[x]);
    }

    void Union(int x,int y)
    {
        int x_par=find(x);
        int y_par=find(y);

        if(x_par==y_par)
        return;//merge krne ki jroort nhi aise mei

        if(rank[x_par]>rank[y_par])
        {
            parent[y_par]=x_par;
        }
        else if(rank[x_par]<rank[y_par])
        {
            parent[x_par]=y_par;
        }
        else
        {
            parent[x_par]=y_par;
            rank[y_par]++;
        }

    }
    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)
        {
            Union(vec[0],vec[1]);
        }

        unordered_map<int,int> mp;

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

        long long result=0;
        long long remaining_nodes=n;

        for(auto it:mp)
        {
            //it.first mei papa hoga

            int size=it.second;

            result+=size*(remaining_nodes-size);
            remaining_nodes-=size;
        }

        return result;
        
    }
};
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