<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:
    vector<pair<int, int>> parents;
    int parentCount;

    pair<int, int>  findParent(int a) {
        if (parents[a].first == -1 || parents[a].first == a) return {a, parents[a].second};
        parents[a] = findParent(parents[a].first);
        return parents[a];
    }

    void merge (int a, int b) {
        auto p1 = findParent(a);
        auto pb = findParent (b);
        if (p1.first != pb.first) {
            --parentCount;
            parents[p1.first].first = pb.first;
            parents[pb.first].second += parents[p1.first].second;
        }
    }

    long long countPairs(int n, vector<vector<int>>& edges) {

        parents.resize(n, {-1, 1});
        parentCount = n;
        if (edges.size() == 0) {
            long long a = n;
            long long b = n-1;
            return (a*b)/2;
        }

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

        if (parentCount == 1) return 0;
        vector<bool> visited(n, false);
        vector<int> temp;
        
        for (int i = 0; i < n; ++i) {
            auto p = findParent(i);
            if (!visited[p.first]) {
                visited[p.first] = true;
                temp.push_back(p.second);
            }
        }

        unsigned long long ans = 0;
        for (int i = 0; i < temp.size(); ++i) {
            unsigned long long a = temp[i];
            ans += a*(n-a);
        }
        return ans/2;
        /*


        map<int, int> parentToChild;
        for (int i = 0; i < n; ++i) {
            int p = findParent(i);
            if (parentToChild.find(p) == parentToChild.end()) {
                parentToChild.insert({p, 0});
            }
            parentToChild[p]++;
        }
        long long ans = 0;
        //vector<int> ptemp (parentCount, 0);
        for (auto it1 : parentToChild) {
            for (auto it2 : parentToChild) {
                if (it1.first != it2.first) {
                    ans += (it1.second * it2.second);
                }
            }
        }
        return ans/2;
        */

        
    }
};
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