## DEV Community

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]

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:

``````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]
``````