<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:
    long long findpar(long long node, vector<long long>& parent) {
        if (node == parent[node]) {
            return node;
        }
        return parent[node] = findpar(parent[node], parent);
    }

    void uni(long long u, long long v, vector<long long>& parent, vector<long long>& rank) {
        long long pu = findpar(u, parent);
        long long pv = findpar(v, parent);
        if (pu != pv) {
            if (rank[pu] > rank[pv]) {
                parent[pv] = pu;
            } else if (rank[pu] < rank[pv]) {
                parent[pu] = pv;
            } else {
                parent[pv] = pu;
                rank[pu]++;
            }
        }
    }

    long long countPairs(int n, vector<vector<int>>& connections) {
        vector<long long> parent(n);
        vector<long long> rank(n, 0);
        for (long long i = 0; i < n; ++i) {
            parent[i] = i;
        }

        for (const auto& conn : connections) {
            long long u = conn[0];
            long long v = conn[1];
            uni(u, v, parent, rank);
        }

        map<long long,long long> uniqueParents;
        for (long long i = 0; i < n; ++i) {
            uniqueParents[findpar(i, parent)]++;
        }
        long long ans=0;
        for(auto i:uniqueParents){
            // for(auto j:uniqueParents){
            //     if(i.first!=j.first){
            //         ans+=(i.second)*(j.second);
            //     }
            // }
            ans+=(i.second)*(n-i.second);
            // cout<<i.first<<" "<<i.second<<endl;
        }

        // long long numComponents = uniqueParents.size();
        

        // long long required = numComponents - 1;

        return ans/2;
    }
};
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