1013MediumMathCombinatoricsEnumeration

Splitting a Batch Across Three Bins

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Trains the technique from

LeetCode 2929Distribute Candies Among Children II

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 batch of n identical parts is to be split across three bins, taken in order, and no bin may hold more than cap parts. A bin may be left empty.

Two splits count as different when some bin's count differs between them. Return how many splits there are.

Examples

Example 1

Input: n = 5, cap = 2
Output: 3

Five parts across three bins holding at most two each. The counts have to be some ordering of two, two and one, and there are three places the single part can go.

Example 2

Input: n = 3, cap = 3
Output: 10

Three parts with a cap of three means the cap never bites, so every way of writing three as an ordered triple counts: all in one bin, three ways; two and one, six ways; and one each, one way.

Example 3

Input: n = 10, cap = 2
Output: 0

Three bins of at most two hold six parts between them, which is short of ten, so no split works.

Constraints

1 <= n <= 10^6
1 <= cap <= 10^6

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 distribute_candies(n: int, cap: int) -> int:

Java

public long distributeCandies(int n, int cap)

The intended approach

Time O(1).Space O(1).

Ignore the cap for a moment. Splitting some batch of `m` parts across three ordered bins is the same as choosing where two dividers go in a line of `m` parts and two dividers, so there are as many splits as ways to pick two places out of `m + 2`, which comes to half of `m + 1` times `m + 2`. Now bring the cap back by counting what has to come off. Start from every uncapped split. For each of the three bins, subtract the splits where that bin overruns: setting `cap + 1` parts aside in it first and splitting the remainder freely counts exactly those, so each subtraction is the same uncapped count on a batch that much smaller. Splits with two bins overrunning were subtracted twice over, so add them back once per pair of bins, this time setting aside twice `cap + 1`. Splits with all three overrunning are now counted once too often, so subtract them once more. That is four evaluations of one small formula, with any term whose remainder falls below nothing contributing nothing since no such split exists.

Pythonentry point: distribute_candies

def distribute_candies(n, cap):
    def ways(spare):
        if spare < 0:
            return 0
        return (spare + 1) * (spare + 2) // 2
    step = cap + 1
    return (ways(n)
            - 3 * ways(n - step)
            + 3 * ways(n - 2 * step)
            - ways(n - 3 * step))

The obvious version

Try every allowed count for the first bin and every allowed count for the second, work out what the third would have to hold, and count the split when that lands inside the cap and not below nothing. Costs the square of the cap rather than a handful of arithmetic.

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_distribute_candies(n, cap):
    total = 0
    for first in range(cap + 1):
        if first > n:
            break
        for second in range(cap + 1):
            third = n - first - second
            if third < 0:
                break
            if third <= cap:
                total += 1
    return total

How this problem was checked

Reference solution executed against 16 test cases — 16 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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