Maximum Repeating Substring

Maximum Repeating Substring - Problem

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.

Input & Output

Example 1 — Basic Repeating Pattern

$ Input: sequence = "ababc", word = "ab"

› Output: 2

💡 Note: "ab" appears once, "abab" (ab repeated 2 times) appears as substring in "ababc", but "ababab" (3 times) does not. Maximum k-repeating value is 2.

Example 2 — No Repetition

$ Input: sequence = "ababc", word = "ba"

› Output: 1

💡 Note: "ba" appears once in "ababc", but "baba" (repeated 2 times) does not appear. Maximum k-repeating value is 1.

Example 3 — Word Not Found

$ Input: sequence = "ababc", word = "ac"

› Output: 0

💡 Note: "ac" does not appear as substring in "ababc". Maximum k-repeating value is 0.

Constraints

1 ≤ sequence.length ≤ 100
1 ≤ word.length ≤ 100
sequence and word consists of lowercase English letters.

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is to incrementally build repeated patterns and check if they exist as substrings. Start with the word itself and keep concatenating until no match is found. Best approach is the optimized incremental building with Time: O(n × k), Space: O(m × k).

Common Approaches

✓ Brute Force Check All Positions

⏱️ Time: O(n × m × k) Space: O(m × k)

For each possible k from 1 to max possible, create word repeated k times and check if it's a substring of sequence. Continue until no match is found.

Optimized Single Pass

⏱️ Time: O(n × k) Space: O(m × k)

Start with k=1 and incrementally check if word repeated k times exists as substring. Use efficient string operations to avoid creating large temporary strings.

Brute Force Check All Positions — Algorithm Steps

Step 1: Start with k=1 and keep incrementing
Step 2: For each k, create word repeated k times
Step 3: Check if this repeated string is substring of sequence
Step 4: Return the maximum k where match was found

Visualization

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Step-by-Step Walkthrough

Try k=1

Check if 'word' exists in sequence

Try k=2

Check if 'wordword' exists in sequence

Continue

Keep trying until no match found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

int solution(char* sequence, char* word) {
    if (strstr(sequence, word) == NULL) {
        return 0;
    }
    
    int maxK = 0;
    int k = 1;
    int wordLen = strlen(word);
    int seqLen = strlen(sequence);
    
    while (true) {
        char* repeatedWord = malloc((k * wordLen + 1) * sizeof(char));
        repeatedWord[0] = '\0';
        
        for (int i = 0; i < k; i++) {
            strcat(repeatedWord, word);
        }
        
        if (strstr(sequence, repeatedWord) != NULL) {
            maxK = k;
            k++;
        } else {
            free(repeatedWord);
            break;
        }
        
        free(repeatedWord);
    }
    
    return maxK;
}

int main() {
    char sequence[1001], word[1001];
    
    fgets(sequence, sizeof(sequence), stdin);
    sequence[strcspn(sequence, "\n")] = 0;
    
    fgets(word, sizeof(word), stdin);
    word[strcspn(word, "\n")] = 0;
    
    int result = solution(sequence, word);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m × k)

For each k value, we create string of length k×m and search in string of length n

✓ Linear Growth

Space Complexity

O(m × k)

Store the repeated word string of maximum length m×k

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Check All Positions — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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