<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:
        dsu = DSU(range(n))
        for u, v in edges: dsu.union(u, v)
        counts = Counter(map(dsu.find, range(n))).values()
        return sum(map(mul, accumulate(counts, add, initial=0), counts))


T = Hashable
class DSU:
    def __init__(self, xs: Iterable[T] = ()) -> None:
        self.parents: Mapping[T, T] = {x: x for x in xs}
        self.sizes: Mapping[T, int] = {x: 1 for x in xs}

    def find(self, u: T) -> T:
        self.parents[u] = u if self.parents[u] == u else self.find(self.parents[u])
        return self.parents[u]
    
    def union(self, u: T, v: T) -> None:
        ur, vr = self.find(u), self.find(v)
        if ur == vr: return
        low, high = (ur, vr) if self.sizes[ur] < self.sizes[vr] else (vr, ur)
        self.parents[low] = high
        self.sizes[high] += self.sizes[low]
    
    def is_connected(self, u: T, v: T) -> bool:
        return self.find(u) == self.find(v)

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