<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:
    int par[(int)1e5+10];
    int size[(int)1e5+10];
    void make(int v){
        par[v] = v;
        size[v] = 1;
    }

    int Find(int v){
        if(par[v]==v) return v;
        return par[v] = Find(par[v]);
    }

    void Union(int a, int b){
        a = Find(a);
        b = Find(b);

        if(a!=b){
            if(size[a]>size[b]){
                swap(a,b);
            }
            par[a] = b;
            size[b] += size[a];
        }
    }

    long long countPairs(int n, vector<vector<int>>& edges) {
        vector<int> compsize(n);
        for(int i = 0; i<n; i++){
            make(i);
        }

        for(int i = 0; i<edges.size(); i++){
            int a = edges[i][0];
            int b = edges[i][1];
            Union(a, b);
        }

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

        long long pairs = 0;
        int rem = n;
        for(int i = 0; i<n; i++){
            if(compsize[i]==0) continue;

            int curr = compsize[i];
            rem-=curr;
            long long curP = curr*1LL*rem;
            pairs+=curP;
        }

        return pairs;
        
    }
};
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