<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:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        
        par = dict()
        rank = dict()

        for i in range(0,n):
            rank[i] = 1 
            par[i] = i 

        def find(x): 
            if par[x] != x: 
                par[x] = find(par[x])
            return par[x]

        def union(x,y):
            x, y = find(x), find(y)
            if x == y: 
                return 
            if rank[x] > rank[y]: 
                rank[x] += rank[y]
                par[y] = x
                return 
            rank[y] += rank[x]
            par[x] = y
            return 
        
        for edge_1, edge_2 in edges: 
            union(edge_1, edge_2)
        
        master_map = dict()

        for i in range(0,n):
            f_i = find(i)
            if f_i in master_map: 
                continue 
            master_map[f_i] = rank[f_i]
        
        q = list(master_map.values())
        sum_count = 0 

        prefix_sum = [0]* len(q)
        for index in range(len(q)): 
            sum_count +=  q[index]
            prefix_sum[index] = sum_count
        # prefix sum? 
        res = 0 
        for index in range(1,len(q)):
            remain_sum = prefix_sum[len(q) - 1 ] - prefix_sum[index -1 ]
            mult_factor = q[index- 1 ]
            res += mult_factor * remain_sum

        return res 
1747