<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>

from collections import Counter
class UnionFind:
    def __init__(self, n):
        self.n = n
        self.parents = list(range(n))
        self.ranks = [0] * n
    
    def find(self, root):
        if root == self.parents[root]:
            return root
        self.parents[root] = self.find(self.parents[root])
        return self.parents[root]
    
    def union(self, root1, root2):
        r1 = self.find(root1)
        r2 = self.find(root2)
        if r1 == r2:
            return
        if self.ranks[r1] > self.ranks[r2]:
            self.parents[r1] = r2
        elif self.ranks[r1] < self.ranks[r2]:
            self.parents[r2] = r1
        else:
            self.parents[r1] = r2
            self.ranks[r1] += 1
    
    def helper(self):
        cnter = Counter()
        for i in range(self.n):
            cnter[self.find(i)] += 1
        return cnter

# 4, 1, 2
# 4 * (1 + 2) + 1 * 2
class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        for edge in edges:
            uf.union(edge[0], edge[1])
        
        cnter = uf.helper()
        total = cnter.total()
        print(cnter)
        res = 0
        for _, count in cnter.items():
            res += count * (total - count)
            total -= count
        
        return res
1732