<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>

// simple problem of dsu, just create a dsu, and define different connected
// components, you have to focus on the no of elements in a connected component

class Solution {
public:
    int find(unordered_map<int, int>& parent, int val) {
        if (parent[val] != val) {
            parent[val] = find(parent, parent[val]); // Path compression
        }
        return parent[val];
        // this is path compressing, storing the values in the parent[val] in
        // returning
    }

    void Union(unordered_map<int, int>& parent, unordered_map<int, int>& rank, int a, int b) {
        int rootA = find(parent, a);
        int rootB = find(parent, b);

        if (rootA != rootB) {
            if (rank[rootA] < rank[rootB]) {
                parent[rootA] = rootB;
            } else if (rank[rootA] > rank[rootB]) {
                parent[rootB] = rootA;
            } else {
                parent[rootB] = rootA;
                rank[rootA]++;
            }
        }
    }
    long long countPairs(int n, vector<vector<int>>& edges) {
        // can also be solved using dfs, but as edges are given so it's really
        // easy to do it with dsu
        unordered_map<int,int> parent;
        unordered_map<int,int> rank;
        for(int i = 0; i< n; i++) {
            parent[i] = i;
            rank[i] = 0;
        }

        for(auto x : edges){
            int a = x[0];
            int b = x[1];
            Union(parent,rank,a,b);
        }
        //now our unions are made, our connected components are made

        vector<long long> mp(n,0);
        for(int i=0;i<n;i++){
            int root = find(parent,i);
            mp[root]++;
        }
        mp.erase(remove(mp.begin(),mp.end(),0),mp.end());
        //now we have size of each connected component too in our map
        long long ans = 0;
        vector<long long> helper = mp;
        for(int i = mp.size()-2;i>=0;i--){
            mp[i] += mp[i+1];
        }
        for(int i = helper.size()- 2; i>=0; i--){
            ans += helper[i]*mp[i+1];
        }
        if(ans== 1) return 0;
        return ans;
    }
};
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