This problem involves finding the maximum length of a subarray that appears in both arrays. The solution uses dynamic programming, where dp[i][j] represents the longest common prefix of subarrays starting at indices i and j of nums1 and nums2 respectively.
Problem Statement
Given two integer arrays nums1 and nums2, your task is to find the maximum length of a subarray that appears in both arrays. A subarray is a contiguous part of an array.
For example, consider nums1 = [1,2,3,2,1] and nums2 = [3,2,1,4,7]. The longest repeated subarray is [3,2,1], which has length 3. Your solution should efficiently solve this problem by leveraging dynamic programming and optimization techniques.
Examples
Example 1
Input: nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7]
Output: 3
The repeated subarray with maximum length is [3,2,1].
Example 2
Input: nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0]
Output: 5
The repeated subarray with maximum length is [0,0,0,0,0].
Constraints
- 1 <= nums1.length, nums2.length <= 1000
- 0 <= nums1[i], nums2[i] <= 100
Solution Approach
State Transition Dynamic Programming
We use dynamic programming to track the longest common subarray. Let dp[i][j] represent the length of the longest common subarray ending at nums1[i] and nums2[j]. If nums1[i] equals nums2[j], then dp[i][j] is dp[i-1][j-1] + 1; otherwise, dp[i][j] is 0.
Binary Search Optimization
We can optimize the solution using binary search to find the maximum subarray length efficiently. This reduces the time complexity by narrowing down the search space for the longest common subarray.
Space Optimization with Rolling Hash
Using a rolling hash technique can help reduce the space complexity by storing only necessary values. This allows us to avoid storing the entire dp table and thus makes the algorithm more memory efficient.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O((M+N) * \log{(\min(M, N))}) |
| Space | O(M) |
The time complexity is O((M+N) * log(min(M, N))) due to binary search and dynamic programming operations. The space complexity is O(M), where M is the length of nums1, because we optimize the space by using only the current row of the dp table.
What Interviewers Usually Probe
- Look for an understanding of state transition dynamic programming.
- The candidate should propose optimizations such as binary search or rolling hash.
- Pay attention to how well they can balance time and space complexity in their approach.
Common Pitfalls or Variants
Common pitfalls
- Incorrectly handling indices while accessing elements in the dp table.
- Over-complicating the solution when a simple dynamic programming approach suffices.
- Failing to optimize for space when the problem constraints are large.
Follow-up variants
- What if the arrays are very large (greater than 1000 elements)?
- Can this problem be solved in linear time?
- What if the arrays contain non-integer values or negative numbers?
How GhostInterview Helps
- GhostInterview will guide you through the state transition dynamic programming approach, ensuring you understand how to track subarray matches.
- You'll learn how to apply binary search to reduce time complexity, improving your solution's efficiency.
- You'll get step-by-step hints on optimizing space usage by leveraging rolling hash techniques.
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 primary approach to solving the Maximum Length of Repeated Subarray problem?
The problem is typically solved using state transition dynamic programming, where dp[i][j] represents the longest common prefix of subarrays starting at nums1[i:] and nums2[j:].
How does dynamic programming help solve the problem?
Dynamic programming is used to efficiently compute the longest common subarray by storing results of overlapping subproblems, reducing redundant calculations.
Can binary search optimize the Maximum Length of Repeated Subarray solution?
Yes, binary search can be used to narrow down the length of the longest common subarray, improving the time complexity of the solution.
What is the time complexity of the Maximum Length of Repeated Subarray problem?
The time complexity is O((M+N) * log(min(M, N))) due to the binary search and dynamic programming steps.
How can space complexity be optimized in this problem?
Space complexity can be optimized by using a rolling hash or by keeping only the current row of the dp table, thus reducing the memory usage.
Need direct help with Maximum Length of Repeated Subarray instead of spending more time grinding it?
Download GhostInterview when you want a LeetCode solver, not another long practice loop. Capture Maximum Length of Repeated Subarray from a screenshot, get the answer path and complexity, and use supported stealth workflows that stay outside captured layers.
Capture the prompt fast instead of rewriting the problem by hand.
Get the solution path, trade-offs, and complexity summary in one pass.
Stay outside captured layers on supported screen-share workflows.
Stay in the same pattern family
Find two non-overlapping sub-arrays with a given target sum and return the minimal total length efficiently using array scanning and hash lookup.
Open problem page#713 Subarray Product Less Than KCount subarrays with a product strictly less than a given value k using efficient algorithms like binary search and sliding window.
Open problem page#689 Maximum Sum of 3 Non-Overlapping SubarraysMaximize the sum of three non-overlapping subarrays with length k in an integer array using dynamic programming.
Open problem page