0188MediumArrayDynamic Programming

Restocking a Ring of Kiosks

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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.

Vending kiosks stand in a circle around a plaza. Kiosk i will take crates[i] crates of stock tonight, and the array closes on itself: the last kiosk and the first kiosk stand side by side.

One van does the round, and it cannot serve two kiosks that stand next to each other, because the plaza only opens one bay at a time and neighbouring bays share a shutter. Serving no kiosk at all is allowed.

Return the largest number of crates the van can deliver in a single night.

Examples

Example 1

Input: crates = [7, 4, 9, 5]
Output: 16

Serving the first and third kiosk delivers 7 plus 9 crates; the other legal pair, the second and fourth, only manages 9 in total.

Example 2

Input: crates = [6, 6, 6]
Output: 6

On a ring of three every kiosk touches both others, so the van can serve only one of them.

Example 3

Input: crates = [0, 0]
Output: 0

Neither kiosk needs stock, so the van delivers nothing.

Constraints

1 <= crates.length <= 100
0 <= crates[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 max_ring_restock(crates: list[int]) -> int:

Java

public int maxRingRestock(int[] crates)

The intended approach

Time O(n)Space O(1)

The only thing the ring adds over a straight row of kiosks is that position one and position n clash. Since at most one of them can be served, every legal plan fits in one of two straight rows: positions one through n minus one, or positions two through n. Each row is solved by carrying two running totals, the best that serves the current kiosk and the best that leaves it alone, and the answer is the larger of the two rows. A ring of one kiosk is its own case, since neither row would include it.

Pythonentry point: max_ring_restock

def max_ring_restock(crates):
    # The first and last kiosk cannot both be served, so one of them is always
    # left out. Solve the straight line that omits the last kiosk and the one
    # that omits the first, and keep the better total. Each line is a two-value
    # rolling recurrence: serve this kiosk on top of the best that skipped the
    # previous one, or skip it and carry the best so far.
    # Time O(n). Space O(1).
    n = len(crates)
    if n == 1:
        return crates[0]

    def best_line(lo, hi):
        served = 0
        skipped = 0
        for i in range(lo, hi + 1):
            served, skipped = skipped + crates[i], max(skipped, served)
        return max(served, skipped)

    return max(best_line(0, n - 2), best_line(1, n - 1))

The obvious version

Enumerate every subset of kiosks as a bitmask, throw away any subset holding two positions that are neighbours around the ring, and keep the largest surviving total.

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_ring_restock(crates):
    n = len(crates)
    best = 0
    for mask in range(1 << n):
        legal = True
        if n > 1:
            for i in range(n):
                if (mask >> i) & 1 and (mask >> ((i + 1) % n)) & 1:
                    legal = False
                    break
        if not legal:
            continue
        total = 0
        for i in range(n):
            if (mask >> i) & 1:
                total += crates[i]
        if total > best:
            best = total
    return best

How this problem was checked

Reference solution executed against 12 test cases — 12 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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