0122EasyArraySimulation

Two Cycle Playout

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Trains the technique from

LeetCode 1929Concatenation of Array

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 stadium ribbon screen fills each break by running its clip roster through twice, back to back. The roster arrives as the array clips, holding clip ids in the order the operator queued them.

Return the full playout order for one break. If the roster holds n clips, the answer has length 2 * n: slot i plays clips[i], and slot i + n plays clips[i] again. The roster is queued as-is, so a clip listed twice plays in both of its slots on each cycle, and the second cycle runs forward in the same direction as the first.

Examples

Example 1

Input: clips = [7, 4, 9]
Output: [7, 4, 9, 7, 4, 9]

Three clips fill six slots. The second cycle repeats the roster in its original direction, so slot 3 replays the clip from slot 0.

Example 2

Input: clips = [12, 12]
Output: [12, 12, 12, 12]

The operator queued the same clip twice, and both entries survive into both cycles, giving four slots.

Example 3

Input: clips = [5]
Output: [5, 5]

A one-clip roster fills a two-slot break.

Constraints

n == clips.length
1 <= n <= 1000
1 <= clips[i] <= 1000

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 two_cycle_playout(clips: list[int]) -> list[int]:

Java

public int[] twoCyclePlayout(int[] clips)

The intended approach

Time O(n)Space O(n) for the returned playout, no extra scratch

Reserve all `2 * n` slots up front, then make a single pass over the roster writing each clip into both of the slots it owns, `i` and `i + n`. One traversal of the input fills the whole playout with no reallocation.

Pythonentry point: two_cycle_playout

def two_cycle_playout(clips):
    n = len(clips)
    playout = [0] * (2 * n)
    for i, clip in enumerate(clips):
        playout[i] = clip
        playout[i + n] = clip
    return playout

The obvious version

Walk the output slots instead of the roster: for every slot from 0 to 2n-1, wrap the slot number back into the roster with a remainder and append whatever clip sits there.

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_two_cycle_playout(clips):
    n = len(clips)
    out = []
    for slot in range(2 * n):
        out.append(clips[slot % n])
    return out

How this problem was checked

Reference solution executed against 8 test cases — 8 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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