<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:

    int bfs(vector<int> adj[], int src, vector<int>& vis){

        queue<int> q;
        q.push(src);
        unordered_set<int> st;

        while(!q.empty()){
            int node= q.front();
            q.pop();
            vis[node]=1;
            st.insert(node);

            for(auto it: adj[node]){
                if(vis[it]==-1){
                    q.push(it);
                }
            }
        }

        return st.size();
    }
    long long countPairs(int n, vector<vector<int>>& edges) {
        
          if (edges.empty()) {
            return static_cast<long long>(n) * (n - 1) / 2;
        }
        vector<int> adj[n];

        for(auto it: edges){
            adj[it[0]].push_back(it[1]);
            adj[it[1]].push_back(it[0]);
        }
        vector<int> vis(n, -1);
        vector<int> count;

        for(int i=0; i<n; i++){
            if(vis[i]==-1){
                int x=bfs(adj, i, vis);
                count.push_back(x);
            }
        }

        long long ans=0;
        for(int i=0; i<count.size()-1; i++){
            for(int j=i+1; j<count.size(); j++){
                ans= ans + ((long long)count[i]* (long long)count[j]);
            }
        }

        return ans;
    }
};
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