0081MediumArrayBacktracking

Calibration Run Orders

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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.

A calibration bench applies a set of signed offsets, given as offsets, to an instrument. Each offset is applied exactly once per run, and the instrument drifts differently depending on which offset comes first, so the lab wants to inspect every run order before choosing one.

Return every arrangement of offsets that uses each offset exactly once, so n offsets yield n! arrangements. Each arrangement lists the offsets in the sequence they would be applied. The arrangements themselves may be handed back in whatever sequence you like.

The offsets are pairwise distinct, so no two arrangements are identical.

Examples

Example 1

Input: offsets = [-10, 0, 10]
Output: [[-10, 0, 10], [-10, 10, 0], [0, -10, 10], [0, 10, -10], [10, -10, 0], [10, 0, -10]]

Three offsets admit six run orders, one for each choice of which offset is applied first paired with each ordering of the remaining two.

Example 2

Input: offsets = [1, 2]
Output: [[1, 2], [2, 1]]

With two offsets the bench can only apply them in one order or the other.

Example 3

Input: offsets = [5]
Output: [[5]]

A single offset leaves nothing to reorder, so exactly one run order exists.

Constraints

1 <= offsets.length <= 6
-10 <= offsets[i] <= 10
All of the offsets are distinct

The values you return may be in any order.

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 run_orders(offsets: list[int]) -> list[list[int]]:

Java

public List<List<Integer>> runOrders(int[] offsets)

The intended approach

Time O(n * n!)Space O(n) beyond the returned arrangements

Grow one arrangement position by position. At each depth try every offset not yet taken, mark it taken, recurse, then unmark it and drop it from the partial arrangement so the next branch starts clean. Recording a copy of the partial arrangement at full depth yields each of the n! orders exactly once.

Pythonentry point: run_orders

def run_orders(offsets):
    n = len(offsets)
    taken = [False] * n
    partial = []
    orders = []

    def extend():
        if len(partial) == n:
            orders.append(partial[:])
            return
        for i in range(n):
            if taken[i]:
                continue
            taken[i] = True
            partial.append(offsets[i])
            extend()
            partial.pop()
            taken[i] = False

    extend()
    return orders

The obvious version

Build up by insertion instead of by recursion: start from the single empty arrangement and, for each offset in turn, splice it into every slot of every arrangement built so far.

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_run_orders(offsets):
    orders = [[]]
    for value in offsets:
        grown = []
        for order in orders:
            for slot in range(len(order) + 1):
                grown.append(order[:slot] + [value] + order[slot:])
        orders = grown
    return orders

How this problem was checked

Reference solution executed against 6 test cases — 6 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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