<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> size, parent;
    DSU(int n) {
        size.resize(n, 1);
        parent.resize(n);
        for (int i = 0; i < n; i++) parent[i] = i;
    }
    int find(int root) {
        if (root == parent[root]) return root;
        return parent[root] = find(parent[root]);
    }
    void Union(int a, int b) {
        a = find(a);
        b = find(b);
        if (a == b) return;
        if (size[a] < size[b]) {
            size[b] += size[a];
            parent[a] = b;
        } else {
            size[a] += size[b];
            parent[b] = a;
        }
    }
};
class Solution {
public:
    long long countPairs(int n, vector<vector<int>>& edges) {
        long long i, j, ans = 0, N=n;
        if(edges.size()==0) return N*(N-1)/2;
        vector<long long> a;
        DSU ds(n);
        for (auto x : edges) ds.Union(x[0], x[1]);
        for (i = 0; i < n; i++) {
            if (ds.parent[i] == i)
                a.push_back(ds.size[i]);
        }
        if(a.size()<=1) return 0;
        // for (i = 0; i < a.size(); i++) {
        //     for (j = i + 1; j < a.size(); j++) {
        //         ans += (a[i] * a[j]);
        //     }
        // }
        for(i=0;i<a.size();i++) {
            if(a.size()>0) {
                ans += ((a[i])*(N-a[i]));
            }
        }
        return ans/2;
    }
};
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