<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:
    //Solving Using Dsu
    vector<int>parent;
    vector<int>rank;
    int find(int a){
        if(parent[a] == a){
            return a;
        }
        return parent[a] = find(parent[a]);
    }
    void Union(int a,int b){
        int parentA = find(a);
        int parentB = find(b);
        if(parentA == parentB){
            return;
        }
        if(rank[parentA] > rank[parentB]){
            parent[parentB] = parentA;
        }
        else if(rank[parentA] < rank[parentB]){
            parent[parentA] = parentB;

        }
        else{
            parent[parentA] = parentB;
            rank[parentB]++;
        }


    }
    long long countPairs(int n, vector<vector<int>>& edges) {
        rank.resize(n,0);
        parent.resize(n);
        for(int  i = 0;i < n; i++){
            parent[i] = i;
        }

        for(auto edge : edges){
            Union(edge[0],edge[1]);
        }

        map<int,int>mp;
        for(int i = 0; i < n; i++){
            mp[find(i)]++;
        }
        long long ans = 0;
        for(auto it = mp.begin(); it != mp.end(); it++){
            ans += (long long)it -> second * (n - it -> second);
        }
        return ans/2;
    }
};
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