0685EasyArraySortingHeap (Priority Queue)Simulation

Shelving Boxes in Turns

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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.

Two archivists, Ada and Ben, clear a pile of boxes. Each box carries a label, given as the list labels, and the pile holds an even number of boxes.

They repeat the following round until the pile is empty:

Ada lifts the box with the smallest label still in the pile.
Ben lifts the box with the smallest label still in the pile.
Ben places his box at the end of the shelf.
Ada places her box at the end of the shelf.

Labels may repeat; when two boxes carry the same label it makes no difference which of them is lifted, because the shelf reads the same either way.

Return the list of labels along the shelf, from the first box placed to the last.

Examples

Example 1

Input: labels = [12, 9, 4, 30]
Output: [9, 4, 30, 12]

Round one: Ada lifts `4`, Ben lifts `9`, so the shelf starts `9`, `4`. Round two: Ada lifts `12`, Ben lifts `30`, so the shelf continues `30`, `12`.

Example 2

Input: labels = [8, 6, 6, 8, 6, 8]
Output: [6, 6, 8, 6, 8, 8]

The first two rounds each lift a `6` and a `6`, and a `6` and an `8`; the last round lifts the remaining two `8`s. Placing each round's second box first gives the shelf shown.

Example 3

Input: labels = [100, 1, 2, 99]
Output: [2, 1, 100, 99]

Ada lifts `1` and Ben lifts `2` in the first round, then Ada lifts `99` and Ben lifts `100`. Ben places before Ada each round.

Constraints

2 <= labels.length <= 100
1 <= labels[i] <= 100
labels.length is even.

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 shelve_boxes(labels: list[int]) -> list[int]:

Java

public int[] shelveBoxes(int[] labels)

The intended approach

Time O(n log n) for n labels, dominated by the sort.Space O(n) for the sorted copy and the returned shelf.

Sorting the labels once gives the exact sequence in which the boxes leave the pile, since every round removes the two smallest labels left. Walk that sorted sequence two entries at a time and emit the second entry of each pair before the first, which is the order the two archivists place them on the shelf.

Pythonentry point: shelve_boxes

def shelve_boxes(labels):
    order = sorted(labels)                 # O(n log n)
    shelf = []
    # the pile is emptied two boxes at a time, and within each pair the second
    # box lifted is the first one shelved
    for i in range(0, len(order), 2):
        shelf.append(order[i + 1])
        shelf.append(order[i])
    return shelf

The obvious version

Simulate the rounds literally on a mutable copy of the pile: scan it for the smallest label and delete that entry, twice per round, then append the two labels in placement order. Quadratic because of the repeated scans, but it follows the wording step by step.

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_shelve_boxes(labels):
    pile = list(labels)
    shelf = []
    while pile:
        first = min(pile)
        pile.remove(first)
        second = min(pile)
        pile.remove(second)
        shelf.append(second)
        shelf.append(first)
    return shelf

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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