Redundant Connection
The key idea
Add the edges one by one with a Disjoint Set Union. For each edge, if its two endpoints are already in the same set, that edge would create a cycle, so it is the redundant one. Because we scan in order, the first such edge we hit is also the last redundant edge the problem asks for.
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 given as an array edges of length n, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi. 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.
Constraints
n == edges.length3 <= n <= 1000edges[i].length == 21 <= ai < bi <= edges.lengthai != bi- There are no repeated edges.
- The given graph is connected.
Examples
Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]
Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
Output: [1,4]
Complexity
Time: O(n α(n)) Space: O(n)
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