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Dividing a Conveyor Into k Parts

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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 conveyor is built from cars clipped one behind another. Each car carries a load and a clip fastened to the car behind it, and the car at the end has an empty clip. Nothing on the conveyor records how many cars it holds.

Because the harness passes plain JSON, the conveyor reaches you as chain, listing the loads in clip order starting at the front car. An empty list means there are no cars at all.

Divide the conveyor into k parts, keeping the cars in their order, so that no two parts differ in length by more than one car and no earlier part is shorter than a later one. Parts may come out empty.

Return the k parts in order, each as its own list of loads.

Examples

Example 1

Input: chain = [1, 2, 3], k = 5
Output: [[1], [2], [3], [], []]

Three cars into five parts gives a base length of nothing and a remainder of three, so the first three parts take one car each and the last two come out empty.

Example 2

Input: chain = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]

Ten cars into three parts gives a base length of three and a remainder of one, so the first part takes four cars and the other two take three.

Example 3

Input: chain = [4, 5], k = 2
Output: [[4], [5]]

Two cars split evenly, one to each part.

Constraints

0 <= chain.length <= 1000
0 <= chain[i] <= 1000
1 <= k <= 50

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 split_list_to_parts(chain: list[int], k: int) -> list[list[int]]:

Java

public List<List<Integer>> splitListToParts(int[] chain, int k)

The intended approach

Time O(n + k) over the cars and the parts.Space O(n) for the parts returned.

Divide the number of cars by the number of parts. The quotient is the length every part gets at least, and the remainder says how many parts get one car more. Those longer parts have to be the earliest ones, because no earlier part may be shorter than a later one, and no two parts may differ by more than a car, which rules out giving any part two extra. So the two rules together leave exactly one set of lengths. With the lengths fixed, one walk along the conveyor does the rest: cut off the base length plus one for the first however-many parts, then the base length for the remainder. When there are fewer cars than parts the base length is nothing, and the parts past the remainder come out empty on their own with no special handling.

Pythonentry point: split_list_to_parts

def split_list_to_parts(chain, k):
    n = len(chain)
    base = n // k
    extra = n % k
    out = []
    at = 0
    for part in range(k):
        take = base + 1 if part < extra else base
        out.append(chain[at:at + take])
        at += take
    return out

The obvious version

Work the parts out one at a time without dividing up front. Before each part, look at how many cars are still unassigned and how many parts are still to be filled, and give this part that many cars rounded upwards. Peel them off the front of what is left and move on. Recomputes a quota per part rather than settling every length from one division.

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_split_list_to_parts(chain, k):
    out = []
    left = len(chain)
    at = 0
    for part in range(k):
        parts_left = k - part
        take = left // parts_left
        if left % parts_left:
            take += 1
        piece = []
        for _ in range(take):
            piece.append(chain[at])
            at += 1
        out.append(piece)
        left -= take
    return out

How this problem was checked

Reference solution executed against 15 test cases — 15 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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