<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;
    map<int,int> compCount;
    int find(int x){
        if(x == parent[x]){
            return x;
        }
        return parent[x]=find(parent[x]);
    }
    void Union(int x,int y){
        int parent_x= find(x);
        int parent_y= find(y);
        if(parent_x == parent_y){
            return;
        }
        if(parent_x != parent_y){
            if(rank[parent_x] > rank[parent_y]){
                parent[parent_y]= parent_x;
                if(compCount[parent_y]) compCount[parent_x] += (compCount[parent_y]+1);
                else compCount[parent_x]++;
            }
            else if(rank[parent_y] > rank[parent_x]){
                parent[parent_x]= parent_y;
                if(compCount[parent_x]) compCount[parent_y] += compCount[parent_x]+1;
                else compCount[parent_y]++;
            }
            else{
                parent[parent_x]= parent_y;
                rank[parent_y]++;
                if(compCount[parent_x]) compCount[parent_y] += compCount[parent_x]+1;
                else compCount[parent_y]++;
            }
        }
    }
    long long countPairs(int n, vector<vector<int>>& edges) {
        if(edges.size()==0){
            return ((long long)n*(n-1))/2;
        }
        // sort(edges.begin(),edges.end());
        parent.resize(n);
        rank.resize(n,0);
        for(int i=0;i<n;i++){
            parent[i]= i;
        }
        for(auto x:edges){
            int parent_x= find(x[0]);
            int parent_y= find(x[1]);
            Union(parent_x,parent_y);
        }
        vector<long long> comp;
        for(int i=0;i<n;i++){
            if(parent[i]==i){
                comp.push_back(compCount[i]+1);
            }
        }
        sort(comp.begin(),comp.end());
        long long s=0;
        for(int i=0;i<comp.size();i++){
            for(int j=(i+1);j<comp.size();j++){
                s += comp[i]*comp[j];
            }
        }
        return s;
    }
};
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