<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>

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    #define ll long long
    const int N = 1e5 + 10;
    vector<ll> parent = vector<ll>(N), size = vector<ll>(N);

    void make(ll v) {
        parent[v] = v;
        size[v] = 1;
    }

    ll find(ll v) {
        if (v == parent[v]) return v;
        return parent[v] = find(parent[v]);
    }

    void unite(ll a, ll b) {
        a = find(a);
        b = find(b);
        if (a != b) {
            if (size[a] < size[b]) swap(a, b);
            parent[b] = a;
            size[a] += size[b];
        }
    }

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

        for (auto v : edges) unite(v[0], v[1]);
        vector<ll> val;

        for (int i = 0; i < n; i++) {
            if (i == find(i)) {
                val.push_back(size[i]);
            }
        }

        ll sum = accumulate(val.begin(), val.end(), 0LL);
        ll ans = 0;

        for (auto v : val) {
            ll rem = sum - v;
            ans += v * rem;
        }

        return ans / 2; // Divide by 2 to avoid double-counting pairs
    }
};

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