0400EasyMathStringEuclidean AlgorithmGreatest Common Divisor

Longest Shared Ribbon Tile

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Trains the technique from

LeetCode 1071Greatest Common Divisor of Strings

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 weaving shop records a ribbon as a string of uppercase letters, one letter per stitched panel.

A pattern x tiles a pattern y when laying out some whole number of copies of x back to back reproduces y letter for letter. So RG tiles RGRG and RGRGRG, and every pattern tiles itself.

You are given ribbons ribbonA and ribbonB. Return the longest pattern that tiles both of them. If no pattern tiles both, return the empty string. Whenever such a pattern exists there is exactly one longest one, so the answer is never ambiguous.

Examples

Example 1

Input: ribbonA = "RGBRGB", ribbonB = "RGBRGBRGB"
Output: "RGB"

Two copies of `RGB` give `ribbonA` and three copies give `ribbonB`. Laying `RGBRGB` or `RGBRGBRGB` end to end does not reproduce both ribbons.

Example 2

Input: ribbonA = "PQPQPQPQ", ribbonB = "PQPQ"
Output: "PQPQ"

Two copies of `PQPQ` give `ribbonA`, and one copy is `ribbonB` itself.

Example 3

Input: ribbonA = "MNM", ribbonB = "MN"
Output: ""

Laying `M`, `N`, `MN` or `MNM` end to end never reproduces both ribbons, so there is no shared tile and the answer is the empty string.

Example 4

Input: ribbonA = "TTTT", ribbonB = "TT"
Output: "TT"

Two copies of `TT` give `ribbonA` and one copy is `ribbonB`.

Example 5

Input: ribbonA = "XYXY", ribbonB = "XYX"
Output: ""

Laying `X`, `Y`, `XY`, `XYX` or `XYXY` end to end never reproduces both ribbons, so the answer is the empty string.

Constraints

1 <= ribbonA.length, ribbonB.length <= 1000
ribbonA and ribbonB consist of uppercase English letters only.

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 gcd_of_strings(ribbonA: str, ribbonB: str) -> str:

Java

public String gcdOfStrings(String ribbonA, String ribbonB)

The intended approach

Time O(n + m) where n and m are the ribbon lengthsSpace O(n + m) for the two joined patterns and the returned tile

If some pattern tiles both ribbons then laying ribbonA before ribbonB gives the same letters as laying ribbonB before ribbonA, and when that test fails no shared tile exists at all. Once it passes, a tile's panel count has to divide both lengths, and the prefix whose length is the greatest common divisor of the two lengths is the longest one that works, so run Euclid on the lengths and slice that prefix.

Pythonentry point: gcd_of_strings

def gcd_of_strings(ribbonA, ribbonB):
    # Time O(n + m): the two joins and the single comparison dominate, and the
    # length gcd loop is O(log min(n, m)).
    # Space O(n + m) for the two joined patterns plus the returned tile.
    if ribbonA + ribbonB != ribbonB + ribbonA:
        return ""
    a = len(ribbonA)
    b = len(ribbonB)
    while b:
        a, b = b, a % b
    return ribbonA[:a]

The obvious version

Take prefixes of the shorter ribbon from the full length downwards, and return the first one whose repeats rebuild both ribbons exactly.

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_gcd_of_strings(ribbonA, ribbonB):
    shortest = ribbonA if len(ribbonA) <= len(ribbonB) else ribbonB
    for size in range(len(shortest), 0, -1):
        if len(ribbonA) % size or len(ribbonB) % size:
            continue
        tile = shortest[:size]
        if tile * (len(ribbonA) // size) == ribbonA:
            if tile * (len(ribbonB) // size) == ribbonB:
                return tile
    return ""

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
5 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

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