1152EasyArraySortingEnumeration

Cutting the Conveyor Into Three Stretches

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Trains the technique from

LeetCode 3010Divide an Array Into Subarrays With Minimum Cost I

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 holds crates whose weights are weights. Cut it into exactly three stretches of neighbouring crates, each stretch holding at least one crate.

A stretch is charged the weight of its own first crate, and the total charge is the three added up.

Return the smallest total charge.

Examples

Example 1

Input: weights = [1, 2, 3, 4, 5]
Output: 6

The first stretch has to begin at the first crate, so its 1 is unavoidable. The cheapest two crates left to begin stretches at are the 2 and the 3.

Example 2

Input: weights = [5, 4, 3, 2, 1]
Output: 8

The 5 is forced. Beginning the other two stretches at the 1 and the 2 gives 8, and no other pair is cheaper.

Example 3

Input: weights = [1, 1, 1]
Output: 3

Every crate weighs 1 and three stretches have to be charged.

Constraints

3 <= weights.length <= 50
1 <= weights[i] <= 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 minimum_cost(weights: list[int]) -> int:

Java

public int minimumCost(int[] weights)

The intended approach

Time O(n log n) for ordering the crates past the first.Space O(n) for that ordered copy.

Cutting into three stretches means picking two cut points, and the charge only ever depends on which crates end up first in a stretch. The first stretch has to begin at the conveyor's own first crate, so that weight is paid whatever happens. The other two stretches begin at the two cut points, and any two different crates from the second onwards can serve as those points: put the cuts just before them. So the charge is the first crate's weight plus any two distinct weights chosen from the rest, and taking the two smallest of those is best. That makes the whole thing an ordering of everything past the first crate, with no need to try the cut pairs one at a time.

Pythonentry point: minimum_cost

def minimum_cost(weights):
    rest = sorted(weights[1:])
    return weights[0] + rest[0] + rest[1]

The obvious version

Try every pair of cut points outright, adding the first crate's weight to the weights of the two crates the cuts land on, and keep the smallest total. Costs the crate count squared.

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_minimum_cost(weights):
    n = len(weights)
    best = None
    for a in range(1, n - 1):
        for b in range(a + 1, n):
            total = weights[0] + weights[a] + weights[b]
            if best is None or total < best:
                best = total
    return best

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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