<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, size):
        self.roots = [node for node in range(size)]
        self.rank = [1 for _ in range(size)]
        self.size = [1 for _ in range(size)]
    
    def find(self, node):
        if node != self.roots[node]:
            self.roots[node] = self.find(self.roots[node])
        return self.roots[node]
    
    def union(self, u, v):
        root_u, root_v = self.find(u), self.find(v)
        if root_u != root_v:
            if self.rank[root_u] < self.rank[root_v]:
                self.roots[root_u] = root_v
                self.size[root_v] += self.size[root_u]
            elif self.rank[root_v] < self.rank[root_u]:
                self.roots[root_v] = root_u
                self.size[root_u] += self.size[root_v]
            else:
                self.roots[root_u] = root_v
                self.rank[root_v] += 1
                self.size[root_v] += self.size[root_u]
    
    def is_connected(self, u, v):
        return self.find(u) == self.find(v)

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

        ans = 0
        nodes = 0
        seen = set()
        for node in uf.roots:
            root = uf.find(node) 
            if root not in seen:
                ans += nodes*uf.size[root]
                nodes += uf.size[root]
                seen.add(root)

        return ans

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