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2316. 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.

Solutions (Python)

1. Solution

class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        parent = list(range(n))
        count = [0] * n

        for a, b in edges:
            while parent[a] != parent[parent[a]]:
                parent[a] = parent[parent[a]]
            while parent[b] != parent[parent[b]]:
                parent[b] = parent[parent[b]]
            if parent[a] < parent[b]:
                parent[parent[b]] = parent[a]
            else:
                parent[parent[a]] = parent[b]

        for i in range(n):
            while parent[i] != parent[parent[i]]:
                parent[i] = parent[parent[i]]
            count[parent[i]] += 1

        return sum(c * (n - c) for c in count) // 2