<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 {
class Disjoint {
    public:
        vector<int> rank, parent, size;
        Disjoint(int n) {
            rank.resize(n + 1, 0);
            parent.resize(n + 1);
            size.resize(n + 1, 1);
            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 unionByRank(int u, int v){
            int ulp_u = findUPar(u);
            int ulp_v = findUPar(v);

            if(ulp_u == ulp_v) return;

            if(rank[ulp_u] > rank[ulp_v]){
                parent[ulp_v] = ulp_u;
            }
            else if(rank[ulp_u] < rank[ulp_v]){
                parent[ulp_u] = ulp_v;
            }
            else{
                parent[ulp_v] = ulp_u;
                rank[ulp_u]++;
            }
        }
        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_v] = ulp_u;
                size[ulp_u] += size[ulp_v];
                size[ulp_v] = 0;
            }
            else if(size[ulp_u] < size[ulp_v]){
                parent[ulp_u] = ulp_v;
                size[ulp_v] += size[ulp_u];
                size[ulp_u] = 0;
            }
            else{
                parent[ulp_v] = ulp_u;
                size[ulp_u] += size[ulp_v];
                size[ulp_v] = 0;
            }
        }
};
public:
    long long countPairs(int n, vector<vector<int>>& edges) {
        vector<vector<int>> adj(n); 
        Disjoint ds(n);
        vector<long long> ans;
        for(int i = 0; i < edges.size(); i++){ 
            int u = edges[i][0];
            int v = edges[i][1];

            ds.unionBySize(u, v);
        }

        for(int i = 0; i < n; i++){
            ans.push_back(ds.size[i]);
        }

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

        return total_cnt;
    }
};
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