Repeated Substring Pattern

Repeated Substring Pattern - Problem

Given a string s, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.

In other words, determine if the string is made up of a repeating pattern where the same substring is repeated 2 or more times consecutively to form the entire string.

Example: The string "abcabcabc" can be constructed by repeating the substring "abc" three times, so it returns true. However, "abcabcab" cannot be formed by repeating any substring, so it returns false.

Input & Output

Example 1 — Simple Repeat

$ Input: s = "abab"

› Output: true

💡 Note: The string can be constructed by repeating "ab" twice: "ab" + "ab" = "abab"

Example 2 — Triple Repeat

$ Input: s = "aba"

› Output: false

💡 Note: No substring can be repeated to form "aba". Single characters don't count as patterns.

Example 3 — Long Pattern

$ Input: s = "abcabcabcabc"

› Output: true

💡 Note: The pattern "abc" is repeated 4 times to form the string

Constraints

1 ≤ s.length ≤ 10⁴
s consists of lowercase English letters

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 12 f Facebook 8

The key insight is that if a string has a repeating pattern, then when doubled and trimmed, the original string will appear within it. The optimal approach uses string concatenation trick: create (s + s)[1:-1] and check if original string appears inside. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ String Concatenation Trick	O(n)	O(n)	Concatenate string with itself and check if original appears in the middle
Brute Force - Try All Substrings	O(n²)	O(n)	Try every possible substring length and check if it repeats to form the entire string

String Concatenation Trick — Algorithm Steps

Step 1: Create doubled string s + s
Step 2: Remove first and last characters to avoid trivial matches
Step 3: Check if original string appears in this modified doubled string

Visualization

Tap to expand

Step-by-Step Walkthrough

Double String

Create s + s

Trim Ends

Remove first and last chars

Check if original appears inside

Code -

solution.c — C

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

bool solution(char* s) {
    int n = strlen(s);
    if (n <= 1) return false;
    
    // Create s + s
    char* doubled = malloc(2 * n + 1);
    strcpy(doubled, s);
    strcat(doubled, s);
    
    // Remove first and last character
    char* trimmed = malloc(2 * n - 1);
    strncpy(trimmed, doubled + 1, 2 * n - 2);
    trimmed[2 * n - 2] = '\0';
    
    // Check if s appears in trimmed
    bool result = strstr(trimmed, s) != NULL;
    
    free(doubled);
    free(trimmed);
    return result;
}

int main() {
    char s[10001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;  // Remove newline
    
    bool result = solution(s);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

String concatenation and substring search both take linear time

✓ Linear Growth

Space Complexity

O(n)

Creating the doubled string requires space proportional to input length

⚡ Linearithmic Space

125.0K Views

Medium Frequency

~15 min Avg. Time

3.2K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

String Concatenation Trick — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler