<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 UnionFind:
    def __init__(self, n):
        self.par = [i for i in range(n)]
        self.rank = [1] * n
        self.N = n
        # self.count = defaultdict(int)
        # self.visit = set()

    def find(self, n):
        n = self.par[n]
        while self.par[n] != n:
            self.par[n] = self.par[self.par[n]]
            n = self.par[n]
        return n

    def unconnect(self, n):
        return self.N - self.rank[self.find(n)]

    def union(self, n1, n2):
        p1, p2 = self.find(n1), self.find(n2)
        if p1 == p2:
            return True
        small, big = sorted([p1, p2], key = lambda x: self.rank[x])
        self.rank[big] += self.rank[small]
        self.par[small] = big
    
class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        res = 0
        if not edges:
            tmp = n - 1
            for i in range(1, n):
                res += tmp
                tmp -= 1
            return res

      
        uf = UnionFind(n)
        for n1, n2 in edges:
            uf.union(n1,n2)
     
        for i in range(n):
            res += uf.unconnect(i)        
        return res // 2
                
        # key_count = list(uf.count.values())
        # unvisit = n - len(uf.visit)
        # for i in range(unvisit):
        #     key_count.append(1)
        # for i in range(len(key_count)):
        #     for j in range(i + 1, len(key_count)):
        #         res += key_count[i] * key_count[j]
        # return res

        
76000