<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 p){
        if(p == parent[p]){
            return p;
        }
        return parent[p] = find(parent[p]);
    }
    void Union(int x , int y){
        int x_p = find(x);
        int y_p = find(y);
        if(x_p == y_p) return;
        else if(rank[x_p] > rank[y_p]) parent[y_p] = x_p;
        else if(rank[x_p]<rank[y_p]) parent[x_p] = y_p;
        else {
            parent[x_p] = y_p;
            rank[y_p]++;
        }
    }
    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;
        int comp = n;
        for(auto p : edges){
            if(find(p[0])!=find(p[1])){
                Union(p[0] , p[1]);
            }
        }
        map<int , int> mp;
        for(int i = 0 ; i < n ; i++){
            parent[i] = find(i);
            mp[parent[i]]++;
        }
        long long res = 0;
        long long prev = 0;
        for(auto ans : mp){
            res += ans.second * (n - ans.second - prev);
            prev += ans.second;
        }
        return res;
    }
};
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