How do you solve Maximum Repeating Substring on LeetCode?

Maximum Repeating Substring (LeetCode #1668) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding how to efficiently check for substrings can significantly reduce time complexity.

What is the brute force solution for Maximum Repeating Substring?

The brute force approach for Maximum Repeating Substring is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each possible k value starting from 1 and constructs the k-repeated word to see if it exists in the sequence. This is straightforward but can be inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Repeating Substring?

Maximum Repeating Substring uses the following concepts: String, Dynamic Programming, String Matching. The recommended patterns to study are: String Matching, Greedy Algorithms.

What are the interview tips for Maximum Repeating Substring?

Always start with a brute force solution to demonstrate your thought process. Explain your reasoning clearly as you optimize the solution, showing your understanding of the problem. Be prepared to discuss edge cases, such as when the word is longer than the sequence.

What are common mistakes when solving Maximum Repeating Substring?

Students often forget to check if the substring length is valid before accessing it, leading to index errors. Some may try to construct strings repeatedly without realizing the inefficiency.

#1668

Maximum Repeating Substring

Easy

StringDynamic ProgrammingString MatchingString MatchingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of constructing new strings, we can count how many times the word appears consecutively in the sequence. This reduces unnecessary string operations and leads to a linear time solution.

⚙️

Algorithm

4 steps

1Step 1: Initialize count to 0 and a variable to track the current position in the sequence.
2Step 2: While there are enough characters left in the sequence to match the word, check if the substring matches the word.
3Step 3: If it matches, increment count and move the position forward by the length of the word. Otherwise, break the loop.
4Step 4: Return count as the maximum k-repeating value.

solution.py11 lines

1def maxRepeating(sequence, word):
2    count = 0
3    word_length = len(word)
4    i = 0
5    while i + word_length <= len(sequence):
6        if sequence[i:i + word_length] == word:
7            count += 1
8            i += word_length
9        else:
10            break
11    return count

ℹ

Complexity note: This complexity is linear because we traverse the sequence only once, checking for matches with the word without constructing new strings.

1Understanding how to efficiently check for substrings can significantly reduce time complexity.
2Recognizing when to use string operations versus counting can lead to optimal solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.