Problem

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:

!https://assets.leetcode.com/uploads/2021/05/02/reduntant1-1-graph.jpg

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

Example 2:

!https://assets.leetcode.com/uploads/2021/05/02/reduntant1-2-graph.jpg

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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.

Solve

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

        n = len(edges)
        
        parent = [-1]*( len(edges) + 1 )

        def find(u):
            if parent[u] == -1:
                return u
            
            return find(parent[u])

        def union(u,v):

            u_p = find(u)
            v_p = find(v)

            if u_p != v_p:
                parent[v_p] = u_p

                return False
            else:

                return True

        for i in range(len(edges)):
            if  union( edges[i][0] , edges[i][1] )  :

                return [ edges[i][0] , edges[i][1]   ]