<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>

#define ll long long
class Solution {
public:
    
    void dfs(vector<vector<int>>& adj, int curr, int currParent, int *chainPtr ,bool *visited, int chainNumber) {

        if(visited[curr]){
            return;
        }
        visited[curr] = true;
        chainPtr[curr] = chainNumber;

        for(int i=0;i<adj[curr].size();i++){
            int neighbour = adj[curr][i];
            if(!visited[neighbour]) {
                dfs(adj, neighbour, currParent, chainPtr, visited, chainNumber);
            }
        }
        return;
    }

    long long countPairs(int n, vector<vector<int>>& edges) {
        bool *visited = new bool[n+1];
        int *chainPtr = new int[n+1];
        vector<vector<int>> adj(n+1);
        for(int i=0;i<edges.size();i++){
            int start = edges[i][0];
            int end = edges[i][1];
            adj[start].push_back(end);
            adj[end].push_back(start);
        }


        for(int i=0;i<n+1;i++){
            visited[i] = false;
            chainPtr[i] = i;
        }
        int chainNumber = 0;
        for(int i=0;i<n;i++){
            if(!visited[i]){
                dfs(adj, i, i, chainPtr, visited, chainNumber);
                chainNumber++;
            }
        }

        for(int i=0;i<n;i++){
            cout<<chainPtr[i]<<",";
        }
        cout<<endl;

        unordered_map<int,int> chainMap;
        for(int i=0;i<n;i++){
            chainMap[chainPtr[i]] = chainMap[chainPtr[i]] + 1;
        }
        ll ans = 0 ;
        // for(auto it=chainMap.begin(); it != chainMap.end(); it++) {
        //     cout<<it->first<<"->"<<it->second<<"|";
        //     int diff = n - it->second; 
        //     // cout<<it->first<<","<<diff<<"|";
        //     ans += diff;
        // }
        for(int i=0;i<n;i++){
            int chainNumber = chainPtr[i];
            ans += (ll)(n - chainMap[chainNumber]);
        }

        
        adj.clear();
        chainMap.clear();
        delete[] chainPtr;
        return ans/2;
    }
};
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