0499MediumArrayHash TableBinary SearchSliding WindowPrefix Sum

Loading Crates From Both Ends

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Trains the technique from

LeetCode 1658Minimum Operations to Reduce X to Zero

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.

Crates stand in a single row on a loading dock. Crate i weighs weights[i] kilograms.

A picker can reach only the two ends of the row: one lift takes either the crate at the far left or the crate at the far right and puts it in the van. Once a crate is lifted the row closes up, so its neighbour becomes the new end of the row.

The van has to leave carrying exactly capacity kilograms. Return the smallest number of lifts that gets the loaded weight to exactly capacity, or -1 if no sequence of lifts reaches that weight. A lift count is never negative, so -1 can only mean the load is unreachable.

Examples

Example 1

Input: weights = [3, 2, 20, 1, 4], capacity = 7
Output: 2

Lift the 3 kg crate from the left end and the 4 kg crate from the right end. The van carries 3 + 4 = 7 kilograms after two lifts.

Example 2

Input: weights = [1, 1, 1, 9, 5], capacity = 12
Output: 4

Lift the three 1 kg crates off the left end, which brings the 9 kg crate to that end, then lift it as well. Four lifts carry 12 kilograms.

Example 3

Input: weights = [2, 4, 6], capacity = 5
Output: -1

No run of lifts off the two ends adds up to 5 kilograms, so the van cannot leave loaded as asked and the answer is -1.

Example 4

Input: weights = [4, 4, 4, 4], capacity = 16
Output: 4

Lifting all four crates loads 16 kilograms, which is the requested weight, so four lifts are used.

Constraints

1 <= weights.length <= 10^5
1 <= weights[i] <= 10^4
1 <= capacity <= 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 fewest_crates(weights: list[int], capacity: int) -> int:

Java

public int fewestCrates(int[] weights, int capacity)

The intended approach

Time O(n): the row total is one pass, and each end index advances at most n times.Space O(1): only the running window weight and two indices are kept.

Lifts always come off the two ends, so whatever stays on the dock is one unbroken stretch of the row, and its weight has to be the row total minus `capacity`. Fewest lifts therefore means widest such stretch. Since every crate weighs at least one kilogram, a window walked across the row grows on the right and is trimmed from the left whenever it holds too much, so each right end is checked against the heaviest window that still fits; the widest window whose weight matches is subtracted from the row length. A negative leftover, or no matching window at all, reports -1. An empty window counts, which covers loading the whole row.

Pythonentry point: fewest_crates

def fewest_crates(weights, capacity):
    # the crates left behind are always one unbroken stretch in the middle
    n = len(weights)
    stays = sum(weights) - capacity
    if stays < 0:
        return -1
    widest = -1
    held = 0
    left = 0
    for right in range(n):
        held += weights[right]
        while held > stays:
            held -= weights[left]
            left += 1
        if held == stays:
            width = right - left + 1
            if width > widest:
                widest = width
    if widest < 0:
        return -1
    return n - widest

The obvious version

Try every split directly: for each count of crates taken from the left and each count taken from the right that still fit in the row, add the two weights up and keep the smallest total count that hits the capacity.

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_fewest_crates(weights, capacity):
    n = len(weights)
    best = -1
    for head in range(n + 1):
        for tail in range(n - head + 1):
            lifted = sum(weights[:head]) + sum(weights[n - tail:]) if tail else sum(weights[:head])
            if lifted == capacity and (best < 0 or head + tail < best):
                best = head + tail
    return best

How this problem was checked

Reference solution executed against 18 test cases — 18 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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