0046HardArrayBinary SearchDynamic ProgrammingGreedyPrefix Sum

Balanced Truckloads

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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.

Parcels are queued along a loading bay in one fixed line, and their masses are the integer array weights. A fleet of trucks vehicles is waiting at the far end.

The crew works front to back and never reorders the queue, so each vehicle is handed one unbroken run of neighbouring parcels. Every parcel is loaded, no run overlaps another, and no vehicle may drive off empty. A vehicle's load is the total mass of the run it received.

The shift only ends when the most heavily laden vehicle has finished its route, so the crew wants the biggest of the trucks loads to come out as small as it possibly can. Report the mass of that heaviest load under the best possible cut.

Examples

Example 1

Input: weights = [4, 9, 3, 12, 6, 1], trucks = 3
Output: 15

Cutting the line into 4 9 | 3 12 | 6 1 gives loads of 13, 15 and 7, and no other set of two cuts pushes the biggest load below 15.

Example 2

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

There are exactly as many vehicles as parcels, so each vehicle carries a single parcel and the heaviest load is the heaviest parcel.

Example 3

Input: weights = [8, 1, 1, 1], trucks = 1
Output: 11

With one vehicle the whole queue travels together, so the answer is the total mass.

Constraints

1 <= weights.length <= 1000
0 <= weights[i] <= 10^6
1 <= trucks <= min(50, weights.length)
Each run of parcels must be contiguous and the queue order is fixed

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 min_heaviest_load(weights: list[int], trucks: int) -> int:

Java

public int minHeaviestLoad(int[] weights, int trucks)

The intended approach

Time O(n log S) where S is the total massSpace O(1)

Guess a ceiling for the heaviest load and ask whether the queue can be cut into at most `trucks` runs that all stay under it: walk the queue and open a new run only when the current one would overflow, which uses the fewest runs possible for that ceiling. Feasibility is monotone in the ceiling, so binary search the ceiling between the heaviest single parcel and the total mass.

Pythonentry point: min_heaviest_load

def min_heaviest_load(weights, trucks):
    low = max(weights)
    high = sum(weights)
    while low < high:
        mid = (low + high) // 2
        used = 1
        load = 0
        for w in weights:
            if load + w > mid:
                used += 1
                load = w
            else:
                load += w
        if used <= trucks:
            high = mid
        else:
            low = mid + 1
    return low

The obvious version

Exact table walk instead of a search on the answer: best[i][j] is the smallest possible heaviest load when the first i parcels are dealt out to j vehicles. Try every position for the last cut, using prefix masses for the trailing run. O(n^2 * trucks).

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_min_heaviest_load(weights, trucks):
    n = len(weights)
    prefix = [0] * (n + 1)
    for i in range(n):
        prefix[i + 1] = prefix[i] + weights[i]
    huge = prefix[n] + 1
    best = [[huge] * (trucks + 1) for _ in range(n + 1)]
    best[0][0] = 0
    for i in range(1, n + 1):
        for j in range(1, trucks + 1):
            for cut in range(j - 1, i):
                if best[cut][j - 1] == huge:
                    continue
                tail = prefix[i] - prefix[cut]
                worst = best[cut][j - 1] if best[cut][j - 1] > tail else tail
                if worst < best[i][j]:
                    best[i][j] = worst
    return best[n][trucks]

How this problem was checked

Reference solution executed against 10 test cases — 10 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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