Repeated String Match

Repeated String Match - Problem

Given two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it.

If it is impossible for b to be a substring of a after repeating it, return -1.

Notice: string "abc" repeated 0 times is "", repeated 1 time is "abc" and repeated 2 times is "abcabc".

Input & Output

Example 1 — Basic Case

$ Input: a = "abcd", b = "cdabcdab"

› Output: 3

💡 Note: We need "abcdabcdabcd" (3 repetitions) to contain "cdabcdab" as substring

Example 2 — Single Repetition

$ Input: a = "abc", b = "cab"

› Output: 2

💡 Note: "abc" repeated once gives "abc", doesn't contain "cab". "abcabc" contains "cab"

Example 3 — Impossible Case

$ Input: a = "abc", b = "aabcd"

› Output: -1

💡 Note: Character 'd' in b doesn't exist in a, so impossible to form b from repetitions of a

Constraints

1 ≤ a.length, b.length ≤ 10⁴
a and b consist of lowercase English letters

Visualization

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Asked in

G Google 15 f Facebook 12 a Amazon 8

The key insight is that we need at most ceil(len(b)/len(a)) + 1 repetitions. Best approach validates character sets first, then tests only the minimum required bounds. Time: O(n+m), Space: O(1)

Common Approaches

✓ Brute Force Concatenation

⏱️ Time: O(n × m × k) Space: O(n × k)

We repeatedly concatenate string a to itself and check if b becomes a substring. We stop when either we find b or when the concatenated string becomes too long to ever contain b efficiently.

Character Set Validation + Bounded Search

⏱️ Time: O(n + m) Space: O(1)

We optimize by first validating that all characters in b exist in a (otherwise impossible). Then we calculate the minimum and maximum possible repetitions needed and search within those bounds only.

Brute Force Concatenation — Algorithm Steps

Start with empty result string
Keep adding string a to result
After each addition, check if b is substring of result
Return count when found, or -1 if length exceeds limit

Visualization

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Step-by-Step Walkthrough

Start

Begin with empty string

Add & Check

Add string a and check for substring b

Found

Return the count when b is found

Code -

solution.c — C

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

int solution(char* a, char* b) {
    if (strlen(b) == 0) return 0;
    if (strlen(a) == 0) return -1;
    
    int lenA = strlen(a);
    int lenB = strlen(b);
    
    // Maximum reasonable repetitions
    int maxReps = (lenB / lenA) + 2;
    
    // Create buffer for repeated string
    char* repeated = malloc((lenA * maxReps + 1) * sizeof(char));
    repeated[0] = '\0';
    
    for (int i = 1; i <= maxReps; i++) {
        strcat(repeated, a);
        if (strstr(repeated, b) != NULL) {
            free(repeated);
            return i;
        }
    }
    
    free(repeated);
    return -1;
}

int main() {
    char a[1001], b[1001];
    fgets(a, sizeof(a), stdin);
    fgets(b, sizeof(b), stdin);
    
    // Remove newlines
    a[strcspn(a, "\n")] = 0;
    b[strcspn(b, "\n")] = 0;
    
    int result = solution(a, b);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m × k)

n = len(a), m = len(b), k = repetitions needed. String concatenation and substring search for each repetition

✓ Linear Growth

Space Complexity

O(n × k)

Storing the concatenated string of length n × k

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Concatenation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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