0165MediumLinked ListRecursion

Swap Neighbouring Clips

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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 editing bench holds a reel: a run of film clips spliced end to end, each clip carrying its own frame count. Splices cannot be handed across, so the reel reaches you as a plain JSON array of frame counts in projection order, the clip nearest the head of the reel first. Hand your answer back in the same form. An array with nothing in it is a reel carrying no clips at all.

Work along the reel two clips at a time and swap each couple over: the first clip and the second clip trade places, then the third and the fourth trade places, and so on to the end. When the reel carries an odd number of clips, the last one has no partner and must be left sitting where it is.

Re-cut the splices, not the film: every clip keeps the frame count it arrived with, and only its position along the run may change. Make a single pass over the reel, in O(n) steps, holding only a constant amount of bookkeeping beyond the reel you return.

Examples

Example 1

Input: reel = [4, 9, 2, 7]
Output: [9, 4, 7, 2]

The first couple trades to put 9 ahead of 4, and the second couple trades to put 7 ahead of 2.

Example 2

Input: reel = [5, 8, 3]
Output: [8, 5, 3]

The leading couple trades over, and the clip holding 3 has nobody to pair with, so it keeps the tail position.

Example 3

Input: reel = [12]
Output: [12]

A lone clip has no partner anywhere on the reel, so the reel is returned exactly as it came in.

Constraints

0 <= reel.length <= 100
0 <= reel[i] <= 100

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 swap_clips(reel: list[int]) -> list[int]:

Java

public int[] swapClips(int[] reel)

The intended approach

Time O(n)Space O(1) beyond the returned reel

The couples never overlap, so the reel can be re-cut left to right with no lookahead: stand at the head of a couple, trade that clip with the one behind it, then step forward by two to reach the head of the next couple. The loop guard stops as soon as there is no partner left, which leaves an odd reel's final clip untouched without a special case afterwards.

Pythonentry point: swap_clips

def swap_clips(reel):
    # Time O(n): each clip is touched once as part of exactly one couple.
    # Space O(1): the re-cut happens in the run itself, with no second reel.
    head = 0
    while head + 1 < len(reel):
        reel[head], reel[head + 1] = reel[head + 1], reel[head]
        head += 2
    return reel

The obvious version

Peel the leading couple off the front, trade the two clips, and glue the result in front of the reel that remains, recursing on that shorter reel each time.

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_swap_clips(reel):
    if len(reel) < 2:
        return list(reel)
    return [reel[1], reel[0]] + bf_swap_clips(reel[2:])

How this problem was checked

Reference solution executed against 10 test cases — 10 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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