<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 DisjointSet{
public:
    vector<int> parent,size;
    DisjointSet(int n){
        parent.resize(n+1,0);
        size.resize(n+1,0);
        for(int i=0;i<=n;i++){
            parent[i]=i;
        }
    }

    int findUPar(int node){
        if(node==parent[node]){
            return node;
        }

        return parent[node]=findUPar(parent[node]);
    }

    void UnionBySize(int u, int v){
        int ulp_u=findUPar(u);
        int ulp_v=findUPar(v);
        if(ulp_u==ulp_v) return;
        if(size[ulp_u]<size[ulp_v]){
            parent[ulp_u]=ulp_v;
            size[ulp_v]+=size[ulp_u];
        }
        else{
            parent[ulp_v]=ulp_u;
            size[ulp_u]+=size[ulp_v];
        }
    }
};

class Solution {
public:
    long long countPairs(int n, vector<vector<int>>& edges) {
        if(edges.size()==0){
            return (long long int)n*(n-1)/2;
        }

        // vector<int> adj[n];
        // for(auto it:edges){
        //     int u=it[0];
        //     int v=it[1];
        //     adj[u].push_back(v);
        //     adj[v].push_back(u);
        // }

        DisjointSet ds(n);

        for(auto it:edges){
            int u=it[0];
            int v=it[1];
            if(ds.findUPar(u)!=ds.findUPar(v)){
                ds.UnionBySize(u,v);
            }
        }

        unordered_map<int,int> umap;
        for(int i=0;i<n;i++){
            umap[ds.findUPar(i)]++;
        }

        vector<int> v;
        for(auto it:umap){
            cout<<it.first<<" - "<<it.second<<endl;
            v.push_back(it.second);
        }

        long long ans=0;
        for(int i=0;i<v.size();i++){
            for(int j=i+1;j<v.size();j++){
                // cout<<v[i]<<" : "<<v[j]<<endl;
                ans += (long long)v[i]*v[j];
            }
        }
        if(v.size()==1) return 0;
        return ans;
    }
};
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