Redundant Connection
Problem Description
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.
Solve it on LeetCode
Approaches
1. Union-Find (Disjoint Set)
Intuition:
The goal of this problem is to identify an edge that, when added, creates a cycle in a graph that otherwise forms a tree. Trees have N nodes with exactly N-1 edges and no cycles, whereas adding an additional edge creates exactly one cycle.
The Union-Find data structure is ideal for this type of problem because it helps efficiently manage and detect the connected components of a graph. The idea is to iterate over each edge and try to union its two nodes. If the two nodes are already in the same connected component (same parent/root in Union-Find terms), adding the edge would create a cycle, and that's the redundant connection.
The steps are:
- Initialize a Union-Find structure for
nelements. - Iterate through each edge.
- For each edge, check if the two endpoints are in the same component using
find. - If they are not, union them.
- If they are, this edge is the redundant one.
Code:
- Time Complexity: O(N), where N is the number of edges. Each
findandunionoperation runs in near constant time thanks to the path compression and union by rank heuristics. - Space Complexity: O(N), for storing parent and rank arrays in the DSU structure.