<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> rank, parents;

    disjointset(int n) {
        rank.resize(n + 1, 0);
        parents.resize(n + 1);
        for (int i = 0; i <= n; i++) {
            parents[i] = i;
        }
    }

    int findupr(int node) {
        if (node == parents[node]) {
            return node;
        }
        return parents[node] = findupr(parents[node]); // Path compression
    }

    void unionbyrank(int u, int v) {
        int upl_u = findupr(u);
        int upl_v = findupr(v);
        if (upl_u == upl_v) return;
        if (rank[upl_u] < rank[upl_v]) {
            parents[upl_u] = upl_v;
        } else if (rank[upl_v] < rank[upl_u]) {
            parents[upl_v] = upl_u;
        } else {
            parents[upl_u] = upl_v;
            rank[upl_v]++;
        }
    }
};

class Solution {
public:
    long long countPairs(int n, vector<vector<int>>& edges)
     {
        disjointset ds(n);
vector<vector<int>>adj(n);
for(auto it:edges)
{
    adj[it[0]].push_back(it[1]);
     adj[it[1]].push_back(it[0]);
}
for(auto it:edges)
{
    if(ds.parents[it[0]]==ds.parents[it[1]])continue;
    ds.unionbyrank(it[0] ,it[1]);
     //ds.unionbyrank(it[1] ,it[0]);

}

unordered_map<int ,int>mp;

for(int i=0;i<n;i++)
{
    int x=ds.findupr(i);
    mp[x]++;
}


    vector<int>vec;
    for(auto it :mp)
    {
        vec.push_back(it.second);
        cout<<it.second<<" ";
    }
    if(vec.size()==1)return 0;


long long ans=(1LL*n*(n-1)/2);
for(int i=0;i<vec.size();i++)
{
    
    
        ans-=(1ll*vec[i]*(vec[i]-1)/2);
    
        
}


    return ans;    
    }
};
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