DEV Community

Abhishek Chaudhary
Abhishek Chaudhary

Posted on

Redundant Connection

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]


  • 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.


class Solution:
    def DFS(self, node, graph, visited, src, dest):
        for i in graph.get(node, []):
            if i not in visited:
                if (node, i) != (src, dest) and (i, node) != (dest, src):
                    self.DFS(i, graph, visited, src, dest)

    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        n = len(edges)
        graph = {}
        numnodes = 0
        for a, b in edges:
            numnodes = max(numnodes, a, b)
            graph[a] = graph.get(a, set()).union({b})
            graph[b] = graph.get(b, set()).union({a})
        for k in range(n - 1, -1, -1):
            a, b = edges[k]
            visited = set()
            self.DFS(a, graph, visited, a, b)
            if len(visited) == numnodes:
                return edges[k]
Enter fullscreen mode Exit fullscreen mode

Top comments (0)