<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 DSU:
    def __init__(self, n):
        self.parents = {x:x for x in range(n)}
        self.ranks = [1]*n

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

    def union(self,x ,y):

        root_x, root_y = self.find(x), self.find(y)
        if root_x != root_y:
            if self.ranks[root_y] > self.ranks[root_x]:
                root_x, root_y = root_y, root_x
            self.parents[root_y] = root_x
            self.ranks[root_x] += self.ranks[root_y]
            self.ranks[root_y] = 0

class Solution:
    def countPairsOne(self, n: int, edges: List[List[int]]) -> int:

        graph = {x:set() for x in range(n)}
        for parent, child in edges:
            graph[parent].add(child)
            graph[child].add(parent)

        ans = 0
        for i in range(n):
            all_nodes = set()

            queue = [i]
            while queue:
                node = queue.pop(0)
                all_nodes.add(node)
                children = graph[node]

                for child in children:

                    if child not in all_nodes:
                        queue.append(child)

            ans += n - len(all_nodes)

        return ans // 2

    def countPairs(self, n: int, edges: List[List[int]]) -> int:

        finder = DSU(n)
        for a, b in edges:
            finder.union(a, b)
        
        finder.ranks.sort(reverse = False)
        answer = 0
        multiplicand = 0
        for x in finder.ranks:
            if x == 0:
                continue
            else:
                if multiplicand == 0:
                    multiplicand = x
                    continue
                answer += multiplicand * x
                multiplicand +=x
        return answer



                
        


        
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