0073EasyStringString Matching

Carousel Alignment

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 baggage carousel is a closed loop of slots, each stencilled with one lowercase letter. A scanner sits at one point of the loop and reads the letters clockwise from wherever the loop currently rests, producing the string belt.

Nudging the carousel forward by one slot moves the letter that was under the scanner to the far end of the reading and shifts everything else one place earlier. So a nudge turns "cargo" into "argoc".

Return true when some number of nudges, possibly zero, makes the scanner's reading equal target, and false otherwise.

Examples

Example 1

Input: belt = "loops", target = "opslo"
Output: true

Two nudges send l and o to the back of the reading, which lands exactly on the target.

Example 2

Input: belt = "loops", target = "loosp"
Output: false

The target uses the same five letters, but swapping the last two is not something nudging the loop can do.

Example 3

Input: belt = "loops", target = "ops"
Output: false

Nudging never removes a slot, so a shorter target can never be reached.

Constraints

1 <= belt.length, target.length <= 100
belt and target contain lowercase 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 can_align_carousel(belt: str, target: str) -> bool:

Java

public boolean canAlignCarousel(String belt, String target)

The intended approach

Time O(n^2) in the worst case, from scanning for the target inside the doubled readingSpace O(n)

Every reading the carousel can ever show is a window of length n inside `belt` written out twice, since the doubled string contains each nudge of the loop as a contiguous slice. So check the lengths agree and then search for `target` inside `belt + belt`.

Pythonentry point: can_align_carousel

def can_align_carousel(belt, target):
    if len(belt) != len(target):
        return False
    return target in belt + belt

The obvious version

Apply one nudge at a time, rebuilding the reading and comparing it with the target after each of the n possible rests, O(n^2).

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_can_align_carousel(belt, target):
    if len(belt) != len(target):
        return False
    reading = belt
    for _ in range(len(belt)):
        if reading == target:
            return True
        reading = reading[1:] + reading[0]
    return False

How this problem was checked

Reference solution executed against 15 test cases — 15 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