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684. Redundant Connection

Difficulty Topics

Description

In this problem, a tree is an undirected graph that is connected and has no cycles.

You are given a graph that started as a tree with n nodes labeled from 1 to n, with one additional edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed. The graph is represented as an array edges of length n where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph.

Return an edge that can be removed so that the resulting graph is a tree of n nodes. If there are multiple answers, return the answer that occurs last in the input.

 

Example 1:

Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]

Example 2:

Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
Output: [1,4]

 

Constraints:

  • n == edges.length
  • 3 <= n <= 1000
  • edges[i].length == 2
  • 1 <= ai < bi <= edges.length
  • ai != bi
  • There are no repeated edges.
  • The given graph is connected.

Solution

redundant-connection.py
class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        n = len(edges)

        parent = [i for i in range(n)]

        def ufind(x):
            if parent[x] != x:
                parent[x] = ufind(parent[x])

            return parent[x]

        def uunion(x, y):
            px = ufind(x)
            py = ufind(y)

            parent[px] = py

        for a, b in edges:
            a -= 1; b -= 1

            if ufind(a) == ufind(b):
                return [a + 1, b + 1]

            uunion(a, b)