0859MediumArrayHash TableGreedy

Fewest Crates to Sort the Bearings

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Trains the technique from

LeetCode 2910Minimum Number of Groups to Create a Valid Assignment

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 tray of bearings is given as balls, where balls[i] is the grade of bearing i.

Every bearing must go into a crate, and a crate may only hold bearings of a single grade. The packing is tidy when any two crates differ in count by at most one.

Return the fewest crates a tidy packing can use.

Examples

Example 1

Input: balls = [5, 5, 5, 5, 5, 6]
Output: 4

Grade 6 appears once, so every crate must hold one or two bearings. The five grade 5 bearings need three crates at that size, and the grade 6 bearing needs one of its own.

Example 2

Input: balls = [1, 1, 1, 2]
Output: 3

Grade 2 appears once, so every crate must hold one or two bearings. Three crates of one grade-1 bearing each, plus one for the grade 2, gives four.

Example 3

Input: balls = [7, 7, 7]
Output: 1

All three bearings share a grade, so a single crate holds them and no other crate exists to differ from.

Constraints

1 <= balls.length <= 10^5
1 <= balls[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 fewest_crates(balls: list[int]) -> int:

Java

public int fewestCrates(int[] balls)

The intended approach

Time O(g * t) for g distinct grades and the smallest tally t, bounded by the number of bearings.Space O(g) for the tallies.

Only the tally of each grade matters, since grades never share a crate. A tidy packing has some smaller crate size `size` with every crate holding `size` or `size + 1`, so a grade with tally `t` can be split into `c` crates exactly when `c * size <= t <= c * (size + 1)`, and the fewest such `c` is `t` divided by `size + 1`, rounded up. Try each candidate `size` from the smallest tally downwards, since a size above the smallest tally could never hold that grade, and for each check every grade fits and add up the crates. The first size that works from the largest downwards gives the fewest crates.

Pythonentry point: fewest_crates

def fewest_crates(balls):
    tally = {}
    for grade in balls:
        tally[grade] = tally.get(grade, 0) + 1
    counts = list(tally.values())
    low = min(counts)
    for size in range(low, 0, -1):
        crates = 0
        ok = True
        for t in counts:
            c = (t + size) // (size + 1)      # fewest crates of size or size + 1
            if c * size > t:
                ok = False
                break
            crates += c
        if ok:
            return crates
    return len(balls)

The obvious version

Try every crate size and, for each, try every crate count for every grade, checking directly that the grade's bearings can be split into that many crates each holding the size or one more. Keeps the smallest total that works anywhere.

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(balls):
    tally = {}
    for grade in balls:
        tally[grade] = tally.get(grade, 0) + 1
    counts = list(tally.values())
    best = None
    for size in range(1, max(counts) + 1):
        total = 0
        ok = True
        for t in counts:
            pick = None
            for c in range(1, t + 1):
                if c * size <= t <= c * (size + 1):
                    pick = c
                    break
            if pick is None:
                ok = False
                break
            total += pick
        if ok and (best is None or total < best):
            best = total
    return best

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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