To solve the Longest Common Subpath problem, apply binary search over the possible subpath lengths and verify commonality using rolling hashes. The approach ensures efficient checking and avoids brute-force comparisons across all paths. This method works well within the problem's constraints, as binary search helps limit the possible answer space, and rolling hashes facilitate quick validation of subpath matches.
Problem Statement
You are given a country with n cities connected by roads. Each city is numbered from 0 to n - 1. m friends are traveling through this country, and each one follows a path consisting of a sequence of cities. The paths are represented as a 2D integer array, where paths[i] is the path taken by the ith friend, consisting of a series of cities visited in order. Some cities may be visited more than once, but no city appears consecutively in the path.
Your task is to find the length of the longest common subpath shared by all friends' paths, or return 0 if there is no common subpath. A subpath is defined as a contiguous subsequence of cities, and it must appear in the same order in every friend's path.
Examples
Example 1
Input: n = 5, paths = [[0,1,2,3,4], [2,3,4], [4,0,1,2,3]]
Output: 2
The longest common subpath is [2,3].
Example 2
Input: n = 3, paths = [[0],[1],[2]]
Output: 0
There is no common subpath shared by the three paths.
Example 3
Input: n = 5, paths = [[0,1,2,3,4], [4,3,2,1,0]]
Output: 1
The possible longest common subpaths are [0], [1], [2], [3], and [4]. All have a length of 1.
Constraints
- 1 <= n <= 105
- m == paths.length
- 2 <= m <= 105
- sum(paths[i].length) <= 105
- 0 <= paths[i][j] < n
- The same city is not listed multiple times consecutively in paths[i].
Solution Approach
Binary Search for Subpath Lengths
Start by performing a binary search over the length of the subpaths. The range of possible subpath lengths is from 0 to the length of the shortest path. This binary search efficiently narrows down the length of the longest common subpath by progressively verifying smaller or larger lengths.
Rolling Hash for Subpath Verification
For each candidate subpath length determined by the binary search, use a rolling hash to check whether there exists a common subpath of that length. The rolling hash technique allows for efficient comparison of subpaths by hashing them and checking for matching hash values across all paths.
Optimizing with Set Intersection
To confirm the existence of a common subpath, use set intersections of hashed values from different paths. If any subpath's hash is shared by all paths, then that subpath is a common subpath. The binary search narrows the search space, while the set intersection confirms the validity of the subpath.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity is driven by the binary search, which takes log(min_length) time, where min_length is the length of the shortest path. For each length tested, the rolling hash requires O(n) time for each path to compute and compare hashes, resulting in an overall time complexity of O(m * n * log(min_length)), where m is the number of paths and n is the average path length. The space complexity is O(m * n) due to the storage required for the rolling hash values of each path.
What Interviewers Usually Probe
- Ability to implement efficient hashing techniques such as rolling hashes.
- Understanding of binary search over a valid answer space to optimize problem-solving.
- Skill in managing multiple paths and confirming common subpaths through set intersection.
Common Pitfalls or Variants
Common pitfalls
- Failing to optimize the subpath verification process, leading to excessive brute-force comparisons.
- Misunderstanding the relationship between binary search and subpath verification, resulting in inefficient solution design.
- Overlooking edge cases, such as paths that are entirely distinct or paths with minimal overlap.
Follow-up variants
- What if paths can contain repeated cities in non-consecutive order?
- Can this solution be optimized further for large n and m values?
- How would the solution change if we needed to find the longest common subpath across only a subset of the paths?
How GhostInterview Helps
- GhostInterview offers an approach to solve this problem by applying binary search and rolling hash, efficiently handling path comparisons.
- The platform guides you through the optimal solution, ensuring you understand the key techniques like binary search, rolling hashes, and set intersection.
- You’ll get personalized hints on potential pitfalls, making sure you avoid common mistakes such as inefficient hashing or improper binary search implementation.
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 pattern to solve the Longest Common Subpath problem?
The problem can be efficiently solved using binary search over subpath lengths combined with rolling hashes to verify the presence of common subpaths.
How do I handle large inputs in the Longest Common Subpath problem?
For large inputs, it's important to optimize the subpath verification process using efficient algorithms like rolling hashes and minimize the number of comparisons with binary search.
What if there is no common subpath in the Longest Common Subpath problem?
If no common subpath exists, the solution will return 0, indicating that the friends' paths do not share any common subpath.
How do rolling hashes work in the Longest Common Subpath problem?
Rolling hashes allow you to compute hashes for subpaths in constant time as you slide over the paths, making it efficient to compare subpaths across different paths.
How does binary search help in the Longest Common Subpath problem?
Binary search is used to efficiently narrow down the possible lengths of the longest common subpath, reducing the number of candidate lengths to check.
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