What is the time complexity of Maximum Length of Repeated Subarray?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Maximum Length of Repeated Subarray?

The space complexity is O(m*n).

What data structures are used to solve Maximum Length of Repeated Subarray?

This problem uses Array, Binary Search, Dynamic Programming, Sliding Window, Rolling Hash, Hash Function. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Length of Repeated Subarray?

Insight 1: Dynamic programming is well-suited for this problem because it allows us to build up solutions for smaller subproblems (smaller subarrays) and use them to find the solution for larger subproblems. Insight 2: The core idea is to use a 2D DP table where `dp[i][j]` stores the length of the longest common subarray ending at index `i-1` in `nums1` and index `j-1` in `nums2`.

Which companies ask LeetCode 718. Maximum Length of Repeated Subarray in interviews?

This problem is frequently asked at Citadel.

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Maximum Length of Repeated Subarray - LeetCode 718 Solution

Q: How to solve Maximum Length of Repeated Subarray?

The solution uses dynamic programming to find the maximum length of the repeated subarray. A 2D array `dp` is created with dimensions `(m+1) x (n+1)`, where `m` and `n` are the lengths of `nums1` and `nums2`, respectively. `dp[i][j]` stores the length of the longest common subarray ending at indices `i-1` in `nums1` and `j-1` in `nums2`. The algorithm iterates through the `dp` array, and if `nums1[i-1]` is equal to `nums2[j-1]`, it means we have found a match, so we increment the length of the common subarray by 1, using the value from the previous diagonal element (`dp[i-1][j-1]`). The maximum length found during this process is then returned.

Maximum Length of Repeated Subarray - Complete Solution Guide

Maximum Length of Repeated Subarray is LeetCode problem 718, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two integer arrays nums1 and nums2 , return the maximum length of a subarray that appears in both arrays . Example 1: Input: nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7] Output: 3 Explanation: 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 Explanation: 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

Detailed Explanation

The problem asks us to find the length of the longest subarray that is common to two given integer arrays, `nums1` and `nums2`. A subarray is a contiguous sequence of elements within an array. The task is to identify the subarray that appears in both input arrays and has the maximum possible length, then return that length. For example, if `nums1` is `[1,2,3,2,1]` and `nums2` is `[3,2,1,4,7]`, the longest common subarray is `[3,2,1]` with a length of 3. The constraints are that the lengths of the input arrays are between 1 and 1000, and the elements within the arrays are between 0 and 100.

Solution Approach

The solution uses dynamic programming to find the maximum length of the repeated subarray. A 2D array `dp` is created with dimensions `(m+1) x (n+1)`, where `m` and `n` are the lengths of `nums1` and `nums2`, respectively. `dp[i][j]` stores the length of the longest common subarray ending at indices `i-1` in `nums1` and `j-1` in `nums2`. The algorithm iterates through the `dp` array, and if `nums1[i-1]` is equal to `nums2[j-1]`, it means we have found a match, so we increment the length of the common subarray by 1, using the value from the previous diagonal element (`dp[i-1][j-1]`). The maximum length found during this process is then returned.

Step-by-Step Algorithm

Step 1: Initialize a 2D DP table `dp` of size `(m+1) x (n+1)` with all elements set to 0. `m` and `n` are the lengths of `nums1` and `nums2` respectively.
Step 2: Initialize a variable `max_length` to 0. This variable will store the maximum length of the common subarray found so far.
Step 3: Iterate through the `dp` table, starting from index (1, 1) up to (m, n).
Step 4: For each cell `dp[i][j]`, check if `nums1[i-1]` is equal to `nums2[j-1]`.
Step 5: If `nums1[i-1] == nums2[j-1]`, then set `dp[i][j] = dp[i-1][j-1] + 1`. This means we extend the common subarray by 1.
Step 6: Update `max_length` to be the maximum of its current value and `dp[i][j]`.
Step 7: After iterating through the entire `dp` table, return `max_length`.

Key Insights

Insight 1: Dynamic programming is well-suited for this problem because it allows us to build up solutions for smaller subproblems (smaller subarrays) and use them to find the solution for larger subproblems.
Insight 2: The core idea is to use a 2D DP table where `dp[i][j]` stores the length of the longest common subarray ending at index `i-1` in `nums1` and index `j-1` in `nums2`.
Insight 3: If `nums1[i-1]` and `nums2[j-1]` are equal, then we extend the common subarray by 1. Otherwise, the length of the common subarray ending at `i-1` and `j-1` is 0.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Binary Search, Dynamic Programming, Sliding Window, Rolling Hash, Hash Function.

Companies

Asked at: Citadel.

Maximum Length of Repeated Subarray | LeetCode 718

Medium

Array Binary Search Dynamic Programming Sliding Window Rolling Hash Hash Function

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

class Solution {
    public int findLength(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;
        
        int[][] dp = new int[m + 1][n + 1];
        
        int max_length = 0;
        
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (nums1[i-1] == nums2[j-1]) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                    max_length = Math.max(max_length, dp[i][j]);
                }
            }
        }
        
        return max_length;
    }
}

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Problem Info

Description

Problem 718: Maximum Length of Repeated Subarray - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming Sliding Window Rolling Hash Hash Function

Companies

Citadel

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:718

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a 2D DP table `dp` of size `(m+1) x (n+1)` with all elements set to 0. `m` and `n` are the lengths of `nums1` and `nums2` respectively.
2
Step 2: Initialize a variable `max_length` to 0. This variable will store the maximum length of the common subarray found so far.
3
Step 3: Iterate through the `dp` table, starting from index (1, 1) up to (m, n).
4
Step 4: For each cell `dp[i][j]`, check if `nums1[i-1]` is equal to `nums2[j-1]`.
5
Step 5: If `nums1[i-1] == nums2[j-1]`, then set `dp[i][j] = dp[i-1][j-1] + 1`. This means we extend the common subarray by 1.
6
Step 6: Update `max_length` to be the maximum of its current value and `dp[i][j]`.
7
Step 7: After iterating through the entire `dp` table, return `max_length`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

The nested loops in the code iterate through all possible pairs of indices from `nums1` and `nums2`, resulting in a time complexity of O(m*n). The DP table `dp` of size `(m+1)*(n+1)` dominates the space complexity, thus it is O(m*n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search with Rolling Hash

Use binary search to find the length of the longest common subarray. For each potential length, use rolling hash to check if there exists a common subarray of that length in both arrays.

Time: O((m+n) * log(min(m, n))), where m and n are the lengths of the arrays.Space: O(m+n)

Trade-offs: While potentially faster in some cases due to logarithmic complexity, rolling hash can have collision issues, and the implementation is more complex than dynamic programming. Prefer DP for its simplicity and reliability.

Sliding Window

Uses a sliding window approach to compare all possible subarrays of different lengths. The window slides along both arrays to find common sequences.

Time: O(m*n*min(m,n))Space: O(1)

Trade-offs: Less efficient than DP or rolling hash/binary search due to higher time complexity, especially when the arrays have significant overlap.

✨

Key Insights

1Insight 1: Dynamic programming is well-suited for this problem because it allows us to build up solutions for smaller subproblems (smaller subarrays) and use them to find the solution for larger subproblems.
2Insight 2: The core idea is to use a 2D DP table where `dp[i][j]` stores the length of the longest common subarray ending at index `i-1` in `nums1` and index `j-1` in `nums2`.
3Insight 3: If `nums1[i-1]` and `nums2[j-1]` are equal, then we extend the common subarray by 1. Otherwise, the length of the common subarray ending at `i-1` and `j-1` is 0.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: One or both of the input arrays are empty. The code handles this gracefully because the outer loops won't execute, and the initial value of `max_length` (0) will be returned.
Edge case 2: The input arrays have no common subarrays. In this case, all values in the DP table will remain 0, and `max_length` will remain 0, which is the correct result.
Edge case 3: When nums1 and nums2 are exactly the same the dp table will be populated diagonally from dp[1][1] with increasing numbers until reaching the smallest of nums1.length and nums2.length. The max length will then be this value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Initializing the DP table incorrectly. Make sure the dimensions are `(m+1) x (n+1)` and that all values are initialized to 0. Forgetting to initialize to zero will lead to incorrect results.

✗Common mistake 2: Incorrectly updating `max_length`. Ensure that you update `max_length` *during* the inner loop to capture the maximum length found so far, and not just at the end.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Bioinformatics: Identifying similarities between DNA sequences (strings). Finding the longest common subsequence can help understand evolutionary relationships or identify conserved regions.

🏦Real-world scenario 2: Plagiarism detection: Finding the longest common string between two documents can indicate potential plagiarism. The longer the common string, the higher the likelihood of plagiarism.

🎮Real-world scenario 3: Data compression: Identifying repeating sequences in data can be used for compression algorithms.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Length of Repeated Subarray"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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