<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<long long>parent;
     vector<long long>siz;


      void make_parent(vector<long long> & parent){
        for(int i = 0; i < parent.size() ; i++){
                   parent[i] = i;
        }
      }

      int find_par(int v){
           
        if(v == parent[v])
             return v;

        return parent[v] = find_par(parent[v]);
      }

      void join(int u , int v){

         int x = find_par(u);
         int y = find_par(v);

         if(x != y){

            if(siz[x] < siz[y]){
                swap(x , y);
            }  
            siz[x] += siz[y];
            
            parent[y] = x;
         }
      }
       
     


    long long countPairs(int n, vector<vector<int>>& edges) {
               long long  count = 0; 

                if(edges.size() == 0){
                  long long k = (long long)n * (n - 1) / 2;
                    return k;
                }


               parent.resize(n);
               siz.resize(n , 1);

               make_parent(parent);

            
            for(int i = 0; i < edges.size() ; i++){
                       int u = edges[i][0];
                       int v = edges[i][1];                
                       join(u , v);
            }

                set<long long>st;

            for(int i = 0; i < n ; i++){
                long long x = find_par(i);
                st.insert(x);
            }

            if(st.size() == 1)
                 return 0;

            vector<long long>ans;

            for(auto x : st){
                ans.push_back(x);
            }
            
            for(int i = 0; i < ans.size() ; i++){
                 for(int j = i+1 ; j < ans.size() ; j++){
                        count +=  siz[ans[i]]*siz[ans[j]];
                 }
            }

            return count;
    }
};
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