<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 DisjointSet
{
public:
    vector<int> rank, parent,size;
    DisjointSet(int n)
    {
        rank.resize(n + 1, 0);
        parent.resize(n + 1);
        size.resize(n+1, 1);
        for (int i = 0; i <= n; i++)
            parent[i] = i;
    }
    int findPar(int node)
    {
        if (node == parent[node])
        {
            return node;
        }
        return parent[node] = findPar(parent[node]);
    }
    void unionByRank(int u, int v)
    {
        int ult_u = findPar(u);
        int ult_v = findPar(v);
        if (ult_u == ult_v)
            return;
        if (rank[ult_u] < rank[ult_v])
        {
            parent[ult_u] = ult_v;
        }
        else if (rank[ult_u] > rank[ult_v])
        {
            parent[ult_v] = ult_u;
        }
        else{
            parent[ult_v] = ult_u;
            rank[ult_u]++;
        }
    }
    
};
class Solution {
public:
    long long countPairs(int n, vector<vector<int>>& edges) {
        DisjointSet ds(n);
        for(auto it: edges){
            int f= it[0];
            int s = it[1];
            ds.unionByRank(f, s); 
        }

        vector<int> comp;
        unordered_map<int,int>mp;
        for(int i=0; i<n; i++){
            if(ds.findPar(i) == i){
                comp.push_back(i);
            }
            int parent = ds.findPar(i);
            mp[parent]++;
        }

        long long result=0;
        long long remainingnodes=n;


        for(auto it:mp){
            long long component=it.second;
            result += (component)*(remainingnodes-component);
            remainingnodes-=component;
        }
      
      return result;
    }
};
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