<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 Disjoint{
    public:
    vector<int> parent , size ; 

    Disjoint(int n ){
        parent.resize(n+1,0); 
        size.resize(n+1, 1); 

        for(int i =0 ;i<=n ; i++){
            parent[i]=i; 
        }
    }

    int findupar(int node){
        if(parent[node]==node){
            return node; 
        }
        return parent[node] = findupar(parent[node]);
    }

    void union_size(int u , int v){
        int paru = findupar(u); 
        int parv = findupar(v); 

        if(paru==parv){
            return ; 
        }

        if(size[paru]>size[parv]){
            parent[parv] = paru ; 
            size[paru] = size[paru]+size[parv];
        }
        else{
            
            parent[paru] = parv ; 
            size[parv] = size[parv]+size[paru];
        
        }
    }
};

class Solution {
public:

    long long countPairs(int n, vector<vector<int>>& edges) {
        long long sum=0; 
        Disjoint ds(n); 
        for(auto it:edges){

            ds.union_size(it[0], it[1]); 
        }

         vector<long long> component_sizes;
        for (int i = 0; i < n; i++) {
            if (ds.findupar(i) == i) {  // Root node of the component
                component_sizes.push_back(ds.size[i]);
            }
        }

        // for(int i =0 ; i<component_sizes.size() ; i++){
        //     for(int j =i+1 ; j<component_sizes.size(); j++){
        //         sum = sum+ component_sizes[i]*component_sizes[j];
        //     }
        // }


       // long long sum =0 ; 
  // Count valid pairs
        long long total_pairs = (1LL * n * (n - 1)) / 2;  // Total possible pairs
        long long connected_pairs = 0;

        for (long long size : component_sizes) {
            connected_pairs += (size * (size - 1)) / 2;
        }

        // Subtract connected pairs from total pairs to get the result
        return total_pairs - connected_pairs;
      
        //return sum ; 


    }
};
163103