<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:
    class DSU {
    public:
        vector<int> parent;
        vector<int> size;
        int n;
        DSU(int n) {
            this->n = n;
            parent.resize(n, -1);
            size.resize(n, 1);
            make_set();
        }
        void make_set() {
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }
        int find_set(int x) {
            if (parent[x] == x) {
                return x;
            }
            return parent[x] = find_set(parent[x]);
        }
        void union_set(int a, int b) {
            a = find_set(a);
            b = find_set(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) {
        if(edges.size()==0){
            long long m = n;
            return ((m-1)*m)/2;
        }
        unordered_map<int, vector<int>> adj;
        for (auto i : edges) {
            adj[i[0]].push_back(i[1]);
            adj[i[1]].push_back(i[0]);
        }
        DSU* dsu = new DSU(n);
        for (int i = 0; i < n; i++) {
            for (auto child : adj[i]) {
                dsu->union_set(i, child);
            }
        }
        set<int> bosses;
        for (int i = 0; i < n; i++) {
            bosses.insert(dsu->find_set(i));
        }
        long long ans = 0;
        vector<long long> ansVec;
        for (auto i : bosses) {
            ansVec.push_back(dsu->size[i]);
        }
        for (int i = 0; i < ansVec.size(); i++) {
            for (int j = i + 1; j < ansVec.size(); j++) {
                ans = ans + (ansVec[i] * ansVec[j]);
            }
        }
        return ans;
    }
};
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