<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.parent = {i:i  for i in range(n)}
        self.size = {i:1 for i in range(n)}

    def find(self, x): 
        if x == self.parent[x]: 
            return x 
        self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, u,v): 
        root1 = self.find(u)
        root2 = self.find(v)

        if root1 != root2: 
            if self.size[root1] > self.size[root2]: 
                self.parent[root2] = root1 
                self.size[root1] += 1 
            else: 
                self.parent[root1] = root2 
                self.size[root2]  += 1 


class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        for u, v in edges: 
            uf.union(u,v)

        groups = defaultdict(int)
        
        #Getting number of nodes per group
        for i in range(n):
            groups[uf.find(i)] += 1
        
        answer = (n*(n-1))//2 

        for group_node, counts in groups.items():
            answer -= (counts*(counts-1))//2 

        return answer
1742