0112EasyTwo PointersStringGreedy

Mosaic Border Inspection

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 mosaic studio finishes decorative border strips before they ship. A strip arrives as the string strip, one lowercase letter per tile naming that tile's glaze, listed from the left end of the strip to the right end.

The studio accepts a strip when it shows the same glaze sequence read from either end, and the finisher is allowed to pull out at most one tile first. Pulling a tile out closes the gap, so the two tiles that flanked it become neighbours. Pulling nothing out is also allowed, so a strip that already matches end to end is accepted as it stands.

Return true when the strip can be accepted and false when no single removal is enough.

Examples

Example 1

Input: strip = "levels"
Output: true

Pull the trailing `s` and the remaining six-tile run `level` shows the same glazes from either end.

Example 2

Input: strip = "cobalt"
Output: false

Dropping the `c` still leaves `obalt` mismatched at its ends, and dropping the `t` leaves `cobal` mismatched too, so one removal cannot rescue it.

Example 3

Input: strip = "kayak"
Output: true

The strip already matches end to end, and removing nothing is permitted.

Constraints

1 <= strip.length <= 10^5
strip 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 inspect_border(strip: str) -> bool:

Java

public boolean inspectBorder(String strip)

The intended approach

Time O(n)Space O(1)

Close in from both ends while the glazes agree. The first disagreement must be resolved by removing one of the two tiles involved, because every tile outside that span is already paired, so it is enough to test whether the span with its left tile dropped matches end to end, or the span with its right tile dropped does.

Pythonentry point: inspect_border

def inspect_border(strip):
    def mirrors(lo, hi):
        while lo < hi:
            if strip[lo] != strip[hi]:
                return False
            lo += 1
            hi -= 1
        return True

    lo = 0
    hi = len(strip) - 1
    while lo < hi:
        if strip[lo] != strip[hi]:
            return mirrors(lo + 1, hi) or mirrors(lo, hi - 1)
        lo += 1
        hi -= 1
    return True

The obvious version

Test the untouched strip, then rebuild the strip with each single tile removed and test each rebuild, 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_inspect_border(strip):
    if strip == strip[::-1]:
        return True
    for i in range(len(strip)):
        trimmed = strip[:i] + strip[i + 1:]
        if trimmed == trimmed[::-1]:
            return True
    return False

How this problem was checked

Reference solution executed against 12 test cases — 12 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