0501MediumArrayMathGreedySortingNumber Theory

Balancing the Carousel Sweeps

Tracked in this browser only

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 carousel carries n trays bolted in a ring and numbered 0 through n - 1, where n is the length of loads. Tray i holds loads[i] grams of grain, and tray n - 1 sits immediately before tray 0, so the numbering closes up into a circle.

A sweep is span trays taken one after another around the ring: it begins at some tray, steps to the next tray each time, and carries on past tray n - 1 onto tray 0 if it has not finished yet. There are n sweeps, one beginning at each tray.

The carousel is balanced when all n sweeps hold the same weight of grain. One adjustment either adds a gram to a single tray or takes a gram off a single tray. Return the fewest adjustments that leave the carousel balanced. A tray may be brought to any whole number of grams, including zero.

Examples

Example 1

Input: loads = [5, 1, 2, 8, 4, 3], span = 4
Output: 10

Bringing the trays to 4, 3, 4, 3, 4, 3 grams takes 1 + 2 + 2 + 5 + 0 + 0 = 10 adjustments, and every sweep of four trays then holds 14 grams.

Example 2

Input: loads = [4, 1, 9, 4], span = 2
Output: 8

Bringing the trays to 9, 4, 9, 4 grams takes 5 + 3 + 0 + 0 = 8 adjustments, and each of the four sweeps of two trays then holds 13 grams.

Example 3

Input: loads = [6, 6, 6, 6], span = 2
Output: 0

Every sweep of two trays already holds 12 grams, so nothing has to be adjusted.

Constraints

1 <= span <= loads.length <= 10^5
1 <= loads[i] <= 10^9

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 min_adjustments(loads: list[int], span: int) -> int:

Java

public long minAdjustments(int[] loads, int span)

The intended approach

Time O(n log n): every tray lands in exactly one family and each family is sorted.Space O(n) for the strided slices taken off the input.

A sweep and the sweep beginning one tray later share all but two trays, so balance forces the tray a sweep drops and the tray it picks up to hold the same weight. Chaining that around the ring, trays whose numbers differ by a multiple of the greatest common divisor of n and span must all finish equal, and trays in different families never constrain each other. Take each family in turn, sort it, and charge the distance from every member to the middle member, which is the cheapest weight to settle a family on. The families are the strided slices of the input, so the sorting adds up to one sort of n values.

Pythonentry point: min_adjustments

def min_adjustments(loads, span):
    from math import gcd

    n = len(loads)
    stride = gcd(n, span)      # trays this far apart must finish on the same weight
    total = 0
    for start in range(stride):
        family = loads[start::stride]
        family.sort()
        aim = family[len(family) // 2]
        for grams in family:
            total += abs(grams - aim)
    return total

The obvious version

Follow the ring in steps of span from each unvisited tray to collect a chain of trays that have to agree, then try every whole weight between the chain's lightest and heaviest tray, cost the chain against each one, and keep the cheapest.

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_min_adjustments(loads, span):
    n = len(loads)
    placed = [False] * n
    total = 0
    for start in range(n):
        if placed[start]:
            continue
        chain = []
        here = start
        while not placed[here]:
            placed[here] = True
            chain.append(loads[here])
            here = (here + span) % n
        best = None
        for aim in range(min(chain), max(chain) + 1):
            cost = 0
            for grams in chain:
                cost += abs(grams - aim)
            if best is None or cost < best:
                best = cost
        total += best
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in