Redundant Connection
In an undirected graph of n nodes (1 to n), one extra edge creates a cycle. Given the edge list, return the redundant edge that can be removed to make the graph a tree. If multiple answers exist, return the last one in the input.
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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. Return an edge that can be removed so that the resulting graph is a tree of n nodes.
Input
A list of `edges` where `edges[i] = [a, b]`. The graph has n nodes.
Output
The redundant edge that can be removed. If multiple, return the last one.
Examples
Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]
[2,3] is the last edge that creates a cycle when added.
Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
Output: [1,4]
The brute-force approach
Try removing each edge (starting from the last). Check if the remaining graph is still connected. Return the first removal that keeps the graph a valid tree.
for i in range(len(edges)-1, -1, -1):
remaining = edges[:i] + edges[i+1:]
if is_valid_tree(n, remaining):
return edges[i]O(E × (V+E)) — for each edge, check connectivity. Very slow.
Spotting the pattern
This is a Union-Find problem. The key question to ask yourself:
If you process edges in order and union their endpoints, when would you first detect a cycle?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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