<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 DSU {
public:
    vector<int> parent, size;
    DSU(int n) {
        parent.resize(n);
        size.resize(n, 1);
        iota(parent.begin(), parent.end(), 0);
    }

    int find(int u) { return u==parent[u] ? u : parent[u]=find(parent[u]); }

    void merge(int u, int v) {
        int pu = find(u);
        int pv = find(v);
        if (pu == pv) return;
        if (size[pu]>size[pv]) {
            parent[pv] = pu;
            size[pu] += size[pv];
        } else {
            parent[pu] = pv;
            size[pv] += size[pu];
        }
    }
};

class Solution {
public:
    long long countPairs(int n, vector<vector<int>>& edges) {
        if (edges.size()==0) return (long long)n*(n-1)/2;
        DSU ds(n);
        vector<int> par;
        unsigned long long res = 0;

        for (auto e:edges) {
            ds.merge(e[0], e[1]);
        }

        for (int i=0;i<n;i++) {
            if (ds.parent[i] == i) par.push_back(i);
        }

        if (par.size() == 1) return 0;
        for (int i=0;i<par.size();i++) {
            for (int j=0;j<par.size();j++) {
                if (i!=j) {
                    res += ((unsigned long long)ds.size[par[i]] * (unsigned long long)ds.size[par[j]]);
                }
            }
        }

        return res/2;
    }
};
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