0854MediumArrayGreedySorting

Crates Filled to the Brim

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Trains the technique from

LeetCode 2279Maximum Bags With Full Capacity of Rocks

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 yard holds n crates. Crate i takes capacity[i] bricks in total and already holds rocks[i] of them. A delivery brings additionalRocks more bricks, and each may be dropped into any crate that is not yet full.

Return the largest number of crates that can be filled to the brim.

Examples

Example 1

Input: capacity = [3, 6, 7], rocks = [1, 6, 2], additionalRocks = 5
Output: 2

The second crate is already full. Dropping 2 bricks into the first fills it, and the 3 bricks left do not cover the 5 the third crate still needs.

Example 2

Input: capacity = [4, 3, 2, 1], rocks = [0, 0, 0, 0], additionalRocks = 6
Output: 3

Filling the crates needing 1, 2 and 3 bricks uses all six bricks and fills three crates.

Example 3

Input: capacity = [5, 5, 5], rocks = [5, 5, 5], additionalRocks = 1
Output: 3

Every crate is already filled to the brim, so all three count and the delivery is not needed.

Constraints

1 <= capacity.length <= 50000
capacity.length == rocks.length
1 <= capacity[i] <= 10^9
0 <= rocks[i] <= 10^9
rocks[i] <= capacity[i] for every crate
1 <= additionalRocks <= 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 crates_filled(capacity: list[int], rocks: list[int], additionalRocks: int) -> int:

Java

public int cratesFilled(int[] capacity, int[] rocks, int additionalRocks)

The intended approach

Time O(n log n) for n crates, dominated by sorting the gaps.Space O(n) for the gaps.

Filling a crate costs the gap between its capacity and what it holds, and every crate filled counts the same, so spending on the smallest gaps first fills the most crates. Work out each gap, sort them, and take them in order while the delivery still covers the next one. A crate that is already full has a gap of zero and is taken for free.

Pythonentry point: crates_filled

def crates_filled(capacity, rocks, additionalRocks):
    gaps = sorted(capacity[i] - rocks[i] for i in range(len(capacity)))
    left = additionalRocks
    filled = 0
    for gap in gaps:
        if gap > left:
            break
        left -= gap
        filled += 1
    return filled

The obvious version

Repeatedly walk every crate that is not yet full, find the one needing the fewest bricks, and fill it if the delivery still stretches that far. Quadratic, but it re-examines the whole yard each time rather than trusting a sorted order.

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_crates_filled(capacity, rocks, additionalRocks):
    n = len(capacity)
    held = list(rocks)
    left = additionalRocks
    filled = sum(1 for i in range(n) if held[i] == capacity[i])
    while True:
        pick = -1
        for i in range(n):
            if held[i] == capacity[i]:
                continue
            need = capacity[i] - held[i]
            if pick < 0 or need < capacity[pick] - held[pick]:
                pick = i
        if pick < 0:
            break
        need = capacity[pick] - held[pick]
        if need > left:
            break
        left -= need
        held[pick] = capacity[pick]
        filled += 1
    return filled

How this problem was checked

Reference solution executed against 11 test cases — 11 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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