0327MediumArrayMathDynamic ProgrammingMinimaxPrefix SumGame TheoryZero-Sum Game

Front Loaded Tray Haul

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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 belt presents piles trays in a fixed order, piles[0] nearest the loading bay. Tray i holds piles[i] blanks. Two crews, Rhea first and then Tomas, take turns clearing trays off the front of the belt.

An allowance m starts at 1. On a turn the crew takes the first x trays that are still on the belt for any x with 1 <= x <= 2 * m, and x may not exceed the number of trays remaining. Every blank on a taken tray goes to that crew. Once the turn is over the allowance becomes max(m, x), and it carries over to the other crew as well. Play stops when the belt is empty.

Both crews want as many blanks as they can get for themselves. Return the number of blanks Rhea ends up with when neither crew ever plays worse than it has to.

Examples

Example 1

Input: piles = [3, 9, 1, 2]
Output: 12

The reported figure is the total on Rhea's side of the table at the end of a run in which each crew always chooses the batch size that serves it best.

Example 2

Input: piles = [8]
Output: 8

One tray is on the belt and the opening allowance already covers it, so Rhea clears the belt on the first turn.

Example 3

Input: piles = [5, 5]
Output: 10

The opening allowance permits up to two trays, so Rhea can clear both and Tomas never gets a turn.

Example 4

Input: piles = [1, 1, 1, 1, 1, 1, 1]
Output: 4

Seven trays each hold one blank and the figure returned is Rhea's share of the seven.

Constraints

1 <= piles.length <= 100
1 <= piles[i] <= 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 best_first_haul(trays: list[int]) -> int:

Java

public int bestFirstHaul(int[] trays)

The intended approach

Time O(n^3)Space O(n^2)

Suffix sums give the blanks still on the belt from any position. Fill a table over (position, allowance) from the back of the belt forwards: whichever batch size the crew on turn picks, the remaining belt is handed to the other crew in a state the table already holds, so the mover's take is the whole remaining total minus the other crew's best from the new state. The answer is the entry for position 0 with allowance 1.

Pythonentry point: best_first_haul

def best_first_haul(trays):
    # Bottom-up over (start index, current allowance). suffix[i] is everything
    # still on the belt, so whatever the mover leaves behind is what the other
    # side then wins from the same table.
    # Time O(n^3). Space O(n^2).
    n = len(trays)
    suffix = [0] * (n + 1)
    for i in range(n - 1, -1, -1):
        suffix[i] = suffix[i + 1] + trays[i]
    table = [[0] * (n + 2) for _ in range(n + 1)]
    for i in range(n - 1, -1, -1):
        for m in range(1, n + 1):
            best = 0
            x = 1
            cap = 2 * m
            while x <= cap and i + x <= n:
                nm = m if m > x else x
                got = suffix[i] - table[i + x][nm]
                if got > best:
                    best = got
                x += 1
            table[i][m] = best
    return table[0][1]

The obvious version

Plain recursion with no table, re-exploring every batch size at every reachable state and re-adding the remaining trays each time.

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_best_first_haul(trays):
    n = len(trays)

    def rec(i, m):
        if i >= n:
            return 0
        left = sum(trays[i:])
        best = 0
        x = 1
        while x <= 2 * m and i + x <= n:
            best = max(best, left - rec(i + x, max(m, x)))
            x += 1
        return best

    return rec(0, 1)

How this problem was checked

Reference solution executed against 8 test cases — 8 passed
4 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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