What is the time complexity of Maximum Repeating Substring?

The optimal solution has a time complexity of O(n*m^2).

What data structures are used to solve Maximum Repeating Substring?

This problem uses String, Dynamic Programming, String Matching. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Maximum Repeating Substring?

The problem can be solved iteratively by repeatedly concatenating `word` to itself and checking if the resulting string is a substring of `sequence`. The `contains()` or `find()` method of strings can be efficiently used to check for substring presence.

Which companies ask LeetCode 1668. Maximum Repeating Substring in interviews?

This problem is frequently asked at Asana, Pure Storage, Turing.

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Maximum Repeating Substring - LeetCode 1668 Solution

Maximum Repeating Substring - Complete Solution Guide

Maximum Repeating Substring is LeetCode problem 1668, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

For a string sequence , a string word is k -repeating if word concatenated k times is a substring of sequence . The word 's maximum k -repeating value is the highest value k where word is k -repeating in sequence . If word is not a substring of sequence , word 's maximum k -repeating value is 0 . Given strings sequence and word , return the maximum k -repeating value of word in sequence . Example 1: Input: sequence = "ababc", word = "ab" Output: 2 Explanation: "abab" is a substring in " abab c".

Detailed Explanation

The problem asks to find the maximum number of times a given word can be repeatedly concatenated to form a substring within a larger sequence string. The input consists of two strings: `sequence` (the larger string) and `word` (the smaller string to be repeated). The output is an integer representing the maximum number of times `word` can be repeated to create a substring of `sequence`. If `word` is not found in `sequence`, the output is 0. For example, if `sequence = "ababc"` and `word = "ab"`, the output is 2 because "abab" is a substring of "ababc".

Solution Approach

The provided solutions employ a simple iterative approach. They start with the `word` itself and repeatedly append `word` to a `repeatedWord` string. In each iteration, they check if `repeatedWord` is a substring of `sequence`. If it is, they increment a counter `k` and continue; otherwise, they stop and return the current value of `k`. This efficiently finds the maximum number of repetitions.

Step-by-Step Algorithm

Initialize a counter `k` to 0.
Initialize a string `repeatedWord` to the input `word`.
Enter a `while` loop that continues as long as `repeatedWord` is a substring of `sequence`.
Inside the loop, increment `k`.
Append `word` to `repeatedWord`.
After the loop terminates, return `k`.

Key Insights

The problem can be solved iteratively by repeatedly concatenating `word` to itself and checking if the resulting string is a substring of `sequence`.
The `contains()` or `find()` method of strings can be efficiently used to check for substring presence.
The loop continues until the concatenated string is no longer a substring of the sequence, indicating that we've found the maximum repetition.

Complexity Analysis

Time Complexity: O(n*m^2)

Space Complexity: O(m)

Topics

This problem involves: String, Dynamic Programming, String Matching.

Companies

Asked at: Asana, Pure Storage, Turing.

Maximum Repeating Substring | LeetCode 1668

Easy

String Dynamic Programming String Matching

Time: O(n*m^2)Space: O(m)

Solution Code

Available in:

class Solution {
    public int maxRepeating(String sequence, String word) {
        int k = 0;
        String repeatedWord = word;
        while (sequence.contains(repeatedWord)) {
            k++;
            repeatedWord += word;
        }
        return k;
    }
}

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Problem Info

Description

Problem 1668: Maximum Repeating Substring - Detailed explanation and solution approaches.

Topics

String Dynamic Programming String Matching

Companies

AsanaPure StorageTuring

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:1668

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

6 steps

1
Initialize a counter `k` to 0.
2
Initialize a string `repeatedWord` to the input `word`.
3
Enter a `while` loop that continues as long as `repeatedWord` is a substring of `sequence`.
4
Inside the loop, increment `k`.
5
Append `word` to `repeatedWord`.
6
After the loop terminates, return `k`.

💡

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(n*m^2)

💾

Space Complexity

O(m)

📝

Why?

The `contains` or `find` operation in strings takes O(m*n) in the worst case, where n is the length of `sequence` and m is the length of the repeatedly concatenated word. The space complexity is due to the `repeatedWord` string, whose length grows linearly with k (the maximum number of repetitions). In the worst case, k can be approximately n/m. Therefore, the space complexity is O(m * n/m) which simplifies to O(n). However, since k is at most n/m, the space complexity can also be expressed as O(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

String Matching Algorithms (e.g., Knuth-Morris-Pratt)

More sophisticated string matching algorithms like KMP could be used to find all occurrences of `word` in `sequence` more efficiently. This would reduce the time complexity of the repeated substring search.

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

Trade-offs: Using KMP improves the time complexity significantly, especially for large strings, but adds implementation complexity. For this problem's constraints, the simple iterative approach is sufficiently efficient.

✨

Key Insights

1The problem can be solved iteratively by repeatedly concatenating `word` to itself and checking if the resulting string is a substring of `sequence`.
2The `contains()` or `find()` method of strings can be efficiently used to check for substring presence.
3The loop continues until the concatenated string is no longer a substring of the sequence, indicating that we've found the maximum repetition.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

If `word` is not found in `sequence`, the loop will terminate immediately, and the function correctly returns 0.
If `word` is an empty string, the loop will run indefinitely. However, this edge case is handled implicitly since an empty string is always a substring; it will quickly lead to a string that exceeds the length of the sequence causing the loop to terminate naturally, returning 0.

⚠️

Common Mistakes to Avoid

✗Incorrectly handling the case where `word` is not found in `sequence`. A simple check before the loop is needed to handle this.

✗Not considering the potential for excessive memory consumption if `word` is very long and `k` becomes large. This can cause a StackOverflowError in recursive solutions or OutOfMemoryError in iterative solutions.

✅

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?

💼Detecting repeated patterns in DNA sequences (bioinformatics).

🏦Finding repeated code sections in software (code analysis).

🎮Analyzing text for plagiarism or copyright infringement (text processing).

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: July 19, 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

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 Repeating Substring"?

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

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