0629EasyStringString MatchingZ AlgorithmKnuth–Morris–Pratt Algorithm

Ribbon Repeats a Motif

Tracked in this browser only

Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A printed ribbon is described by the lowercase string s, one letter per printed cell, read from the left end to the right end.

Call the ribbon patterned when there is a block of cells strictly shorter than the whole ribbon such that laying that block down twice or more, end to end and always in the same direction, reproduces s exactly. Return true when the ribbon is patterned and false when it is not.

Examples

Example 1

Input: s = "abcabcabc"
Output: true

The block abc is shorter than the ribbon, and laying it down three times spells out the ribbon exactly, so the answer is true.

Example 2

Input: s = "aab"
Output: false

A block would have to divide the length of 3, so the only candidate shorter than the ribbon has length 1. Neither a nor b repeated three times spells aab, so the answer is false.

Example 3

Input: s = "abaababaab"
Output: true

The block abaab laid down twice spells abaababaab, which is the ribbon, so the answer is true.

Constraints

1 <= s.length <= 10^4
s consists of lowercase English letters.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def repeated_substring_pattern(s: str) -> bool:

Java

public boolean repeatedSubstringPattern(String s)

The intended approach

Time O(n) — one pass builds the table, and the pointer only ever moves back to earlier table entries.Space O(n) for the failure table.

Build the failure table used by prefix matching: fail[i] is the length of the longest proper prefix of the first i+1 cells that also ends those cells. Read the last entry, which is the longest overlap between a prefix and a suffix of the whole ribbon. Subtracting that overlap from the length leaves the shift that maps the ribbon onto itself; the ribbon is patterned when the overlap is non-empty and that shift divides the length evenly.

Pythonentry point: repeated_substring_pattern

def repeated_substring_pattern(s):
    # O(n) time, O(n) space: build the failure table where fail[i] is the length
    # of the longest proper prefix of s[:i+1] that is also a suffix of it. The
    # ribbon tiles from a shorter block exactly when that final overlap is
    # non-empty and the length left over after removing it divides n.
    n = len(s)
    fail = [0] * n
    k = 0
    for i in range(1, n):
        while k > 0 and s[i] != s[k]:
            k = fail[k - 1]
        if s[i] == s[k]:
            k += 1
        fail[i] = k
    overlap = fail[n - 1]
    if overlap == 0:
        return False
    return n % (n - overlap) == 0

The obvious version

Try every block length that divides the ribbon length and is strictly smaller than it. For each, repeat the leading block enough times to reach the full length and compare the two strings directly.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_repeated_substring_pattern(s):
    n = len(s)
    for block in range(1, n):
        if n % block == 0 and s[:block] * (n // block) == s:
            return True
    return False

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in