The goal is to identify and remove the redundant edge that introduces a cycle in a graph that initially starts as a tree. The problem requires finding this edge using graph traversal methods, such as Depth-First Search (DFS) or Union Find. A simple traversal technique helps pinpoint the last added edge that forms a cycle.
Problem Statement
You are given a graph that started as a tree with n nodes, labeled from 1 to n, with one extra edge added. This added edge creates a cycle in the graph. You need to determine which edge can be removed so that the resulting graph becomes a tree, eliminating the cycle. If there are multiple possible answers, return the one that appears last in the input.
The graph is represented by an array of edges, where edges[i] = [ai, bi] indicates an edge between nodes ai and bi. The graph is connected, and there are no repeated edges. Your task is to find and remove the redundant edge that forms the cycle, restoring the graph to its tree structure.
Examples
Example 1
Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]
Example details omitted.
Example 2
Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
Output: [1,4]
Example details omitted.
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 Approach
Graph Traversal with Depth-First Search (DFS)
Using DFS, traverse the graph while keeping track of visited nodes. When you encounter a node that has already been visited, it indicates a cycle. The last edge forming this cycle is the redundant one.
Union Find (Disjoint Set Union)
With Union Find, iterate over the edges and perform union operations. If two nodes are already in the same set, the current edge is redundant and forms a cycle. The last such edge encountered is the redundant edge.
Breadth-First Search (BFS)
A BFS approach can also be used to find cycles by level-order traversal. Track the parent of each node during traversal. If an edge leads to an already visited node that isn’t the parent, it forms a cycle.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(N \cdot \alpha(N)) |
| Space | O(N) |
The time complexity is O(N * α(N)), where N is the number of nodes, and α(N) is the inverse Ackermann function. The space complexity is O(N) due to the storage needed for the graph and visited nodes.
What Interviewers Usually Probe
- Look for the candidate's ability to efficiently detect cycles in a graph using DFS or Union Find.
- Assess how well the candidate explains the significance of the cycle and redundant edge removal.
- Evaluate their understanding of graph traversal and cycle detection trade-offs.
Common Pitfalls or Variants
Common pitfalls
- Misunderstanding the tree structure and treating the graph as fully connected without considering the cycle.
- Not recognizing the last added edge as the answer when multiple redundant edges exist.
- Incorrectly applying DFS or Union Find without careful tracking of visited nodes or sets.
Follow-up variants
- Handle graphs with multiple cycles and determine which cycle's edge to remove.
- Implement the solution using BFS instead of DFS or Union Find.
- Extend the problem to handle undirected graphs with weighted edges.
How GhostInterview Helps
- GhostInterview provides immediate feedback on cycle detection techniques like DFS and Union Find.
- Practice solving this problem with various traversal techniques to better understand their advantages and limitations.
- Get step-by-step guidance on handling the tricky case where multiple answers are possible and choosing the last edge.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the approach to solve the Redundant Connection problem?
The problem can be solved using Depth-First Search (DFS), Union Find, or Breadth-First Search (BFS). Each technique helps to detect cycles and identify the redundant edge.
How do you detect cycles in a graph for Redundant Connection?
You can detect cycles using DFS or Union Find by checking if an edge connects two already visited or unified nodes.
What should be returned if there are multiple redundant edges?
If multiple redundant edges are found, return the one that appears last in the input list.
What is the time complexity of the Redundant Connection problem?
The time complexity is O(N * α(N)), where N is the number of nodes, and α(N) is the inverse Ackermann function.
How does Union Find help in solving the Redundant Connection problem?
Union Find helps by efficiently tracking connected components. If two nodes are already in the same component, the current edge forms a cycle and is redundant.
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Find and remove the redundant connection in a directed graph that was originally a rooted tree.
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