0068EasyTwo PointersString

Flip the Stamp Strip

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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.

An engraving machine loaded a strip of stamps back to front. The strip is given as strip, a list whose entries are single-character strings, one printable ASCII character per stamp, listed from the front of the strip to the back.

Re-seat the stamps so the strip reads back to front: the stamp at the back moves to the front slot, the one before it moves to the second slot, and so on.

The machine has no spare tray, so you may not build a second list of stamps. Rearrange the entries inside strip itself, using only a constant amount of extra space on top of the input, then return that same list so the result can be inspected.

Examples

Example 1

Input: strip = ["s", "t", "a", "m", "p"]
Output: ["p", "m", "a", "t", "s"]

The five stamps end up in the opposite order, and the middle one never moves.

Example 2

Input: strip = [" ", "~", "0", "Z"]
Output: ["Z", "0", "~", " "]

Spaces, digits and punctuation are ordinary stamps and are re-seated like any letter.

Example 3

Input: strip = ["Q"]
Output: ["Q"]

A one-stamp strip already reads the same in both directions, so nothing moves.

Constraints

1 <= strip.length <= 10^5
strip[i] is a string of exactly one printable ASCII character
Only O(1) extra space is allowed; the stamps must be re-seated inside the given list

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 flip_strip(strip: list[str]) -> list[str]:

Java

public char[] flipStrip(char[] strip)

The intended approach

Time O(n)Space O(1)

Pair the front slot with the back slot, the second slot with the second-to-last, and so on: each pair simply trades stamps. Hold one cursor at each end, swap the two entries they point at, then step both cursors inward and stop once they meet or cross, which handles an odd-length strip by leaving its middle stamp alone. Nothing beyond the two cursors is stored, so the strip is rearranged in place.

Pythonentry point: flip_strip

def flip_strip(strip):
    left = 0
    right = len(strip) - 1
    while left < right:
        strip[left], strip[right] = strip[right], strip[left]
        left += 1
        right -= 1
    return strip

The obvious version

Read the slots from the back to the front and append each stamp to a fresh list, which is an obviously correct oracle even though it ignores the in-place rule.

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_flip_strip(strip):
    rebuilt = []
    for index in range(len(strip) - 1, -1, -1):
        rebuilt.append(strip[index])
    return rebuilt

How this problem was checked

Reference solution executed against 9 test cases — 9 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

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