<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 {
private:
    int parent(vector<int>& unv, int a) {
        while(unv[a] != a) a = unv[a];
        return a;
    }

    void unionNode(vector<int>& unv, int a, int b) {
        int pa = parent(unv, a);
        int pb = parent(unv, b);

        if (pa != pb) unv[pb] = pa;
    }
public:
    long long countPairs(int n, vector<vector<int>>& edges) {
        vector<int> unv(n);
        for (int i = 0; i < n; i++) unv[i] = i;

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

        unordered_map<int, int> um;
        for (int num: unv) um[parent(unv, num)]++;

        unordered_map<int, int> umc;
        for (auto kvp: um) {
            int cnt = kvp.second;
            umc[cnt]++;
        }

        vector<vector<int>> gcv;
        for (auto kvp: umc) gcv.push_back({kvp.first, kvp.second});

        long long res = 0;
        for (int i = 0; i < gcv.size(); i++) {
            int cnti = gcv[i][0];
            int gp_cnti = gcv[i][1];

            long long lr = (long long)cnti * cnti * gp_cnti * (gp_cnti - 1) / 2;

            for (int j = i+1; j< gcv.size(); j++) {
                int cntj = gcv[j][0];
                int gp_cntj = gcv[j][1];

                lr += (long long)gp_cnti * cnti * cntj * gp_cntj;
            }

            res += lr;
        }

        return res;
    }
};
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