Count Unreachable Pairs of Nodes in an Undirected Graph

You are given an integer n. There is an undirected graph with n nodes, numbered from 0 to n - 1. You are given a 2D integer array edges where edges[i] = [ai, bi] denotes that there exists an undirected edge connecting nodes ai and bi.

Return the number of pairs of different nodes that are unreachable from each other.

 

Example 1:

Input: n = 3, edges = [[0,1],[0,2],[1,2]]
Output: 0
Explanation: There are no pairs of nodes that are unreachable from each other. Therefore, we return 0.

Example 2:

Input: n = 7, edges = [[0,2],[0,5],[2,4],[1,6],[5,4]]
Output: 14
Explanation: 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.

 

Constraints:

  • 1 <= n <= 105
  • 0 <= edges.length <= 2 * 105
  • edges[i].length == 2
  • 0 <= ai, bi < n
  • ai != bi
  • There are no repeated edges.
SOLUTION:
from collections import defaultdict

class Solution:
    def DFS(self, graph, i, visited):
        for j in graph[i]:
            if j not in visited:
                self.ctr += 1
                visited.add(j)
                self.DFS(graph, j, visited)
    
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        graph = defaultdict(list)
        for a, b in edges:
            graph[a].append(b)
            graph[b].append(a)
        visited = set()
        total = 0
        res = 0
        for i in range(n):
            if i not in visited:
                visited.add(i)
                self.ctr = 1
                self.DFS(graph, i, visited)
                res += self.ctr * total
                total += self.ctr
        return res

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