0216MediumArrayDivide and ConquerSortingHeap (Priority Queue)Merge SortBucket SortRadix SortCounting Sort

Tool Offset Ordering

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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 machine shop fills in a card for every finishing pass it runs. Each card carries one signed offset in microns: negative when the pass left the part under the nominal size, positive when it left the part over.

You are given offsets, the figures in the order the cards were filed. Return the same figures rearranged from lowest to highest. Every card keeps its own entry, so a figure filed three times has to appear three times in what you return.

Arrange the figures yourself: do not call a library sorting routine or ordering helper. Your routine must run in O(n log n) time and use as little extra room as you can manage.

Examples

Example 1

Input: offsets = [12, -4, 12, 0, -4]
Output: [-4, -4, 0, 12, 12]

Both -4 cards and both 12 cards survive, so the answer holds five figures rising from -4 to 12.

Example 2

Input: offsets = [7, 3, 9, 1, 8, 2, 6]
Output: [1, 2, 3, 6, 7, 8, 9]

The seven figures are all different, so each one lands in its own place in the rising run.

Example 3

Input: offsets = [42]
Output: [42]

A single card is already in order on its own.

Constraints

1 <= offsets.length <= 5 * 10^4
-5 * 10^4 <= offsets[i] <= 5 * 10^4

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

Java

public int[] toolOffsetOrdering(int[] offsets)

The intended approach

Time O(n log n)Space O(n)

Runs of length one are already ordered, so merge neighbouring runs into ordered runs of twice the width, doubling the width until one run covers everything. Each merge walks the two runs with one cursor apiece, always copying the smaller head and copying whatever is left over when one run empties. The passes are driven by a loop over the width rather than by recursion, and two arrays are swapped back and forth so only one extra array is ever allocated.

Pythonentry point: tool_offset_ordering

def tool_offset_ordering(offsets):
    n = len(offsets)
    cur = list(offsets)
    buf = [0] * n
    width = 1
    while width < n:
        start = 0
        while start < n:
            mid = min(start + width, n)
            end = min(start + 2 * width, n)
            i, j, k = start, mid, start
            while i < mid and j < end:
                if cur[i] <= cur[j]:
                    buf[k] = cur[i]
                    i += 1
                else:
                    buf[k] = cur[j]
                    j += 1
                k += 1
            while i < mid:
                buf[k] = cur[i]
                i += 1
                k += 1
            while j < end:
                buf[k] = cur[j]
                j += 1
                k += 1
            start += 2 * width
        cur, buf = buf, cur
        width *= 2
    return cur

The obvious version

Repeatedly scan what is left for the smallest figure and move it across, O(n^2).

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_tool_offset_ordering(offsets):
    remaining = list(offsets)
    out = []
    while remaining:
        best = 0
        for i in range(1, len(remaining)):
            if remaining[i] < remaining[best]:
                best = i
        out.append(remaining.pop(best))
    return out

How this problem was checked

Reference solution executed against 13 test cases — 13 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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