0693HardArrayDynamic ProgrammingPrefix Sum

Combining Grain Lots

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Trains the technique from

LeetCode 1000Minimum Cost to Merge Stones

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 a row of grain lots, and weights[i] is the weight of the lot in position i.

One combine takes exactly groupSize lots that are neighbours in the current row and replaces them with a single lot whose weight is their combined weight. The handling charge for that combine is the combined weight of the lots it took. Combines may be repeated, and each one works on the row as it stands at that moment.

Return the smallest total handling charge that leaves the conveyor holding one lot. If no sequence of combines can leave exactly one lot, return -1.

Examples

Example 1

Input: weights = [8, 3, 5], groupSize = 2
Output: 24

Combine the lots of weight 3 and 5 for a charge of 8, leaving lots of weight 8 and 8, then combine those for a charge of 16. The charges add up to 24.

Example 2

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

A combine turns three lots into one, so the row of four can only become a row of two, and a row of two admits no further combine. One lot is unreachable.

Example 3

Input: weights = [2, 9, 1, 8, 3], groupSize = 3
Output: 35

Combine the lots of weight 1, 8 and 3 for a charge of 12, leaving weights 2, 9 and 12, then combine all three for a charge of 23. The charges add up to 35.

Constraints

1 <= weights.length <= 30
1 <= weights[i] <= 100
2 <= groupSize <= 30

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 merge_lots(weights: list[int], groupSize: int) -> int:

Java

public int mergeLots(int[] weights, int groupSize)

The intended approach

Time O(n^3 / groupSize) for n lots: every stretch tries every legal split point.Space O(n^2) for the stretch table.

Every combine reduces the lot count by `groupSize - 1`, so the row can reach a single lot only when `n - 1` is a multiple of that step; otherwise the answer is the sentinel. Otherwise work over stretches of the original row in increasing length, storing for each stretch the least charge to combine it as far as it will go. A stretch is split into a left part that closes up into one lot, whose length must therefore be one more than a multiple of the step, and the remainder; when the whole stretch itself closes up, the final combine adds the stretch weight, read off a prefix-sum table.

Pythonentry point: merge_lots

def merge_lots(weights, groupSize):
    n = len(weights)
    step = groupSize - 1
    # each combine turns groupSize lots into one, so it drops the lot count by
    # step; the row can only ever reach a single lot when n - 1 is a multiple
    if (n - 1) % step != 0:
        return -1
    prefix = [0] * (n + 1)
    for i, w in enumerate(weights):
        prefix[i + 1] = prefix[i] + w
    # best[i][j] = least cost to combine the stretch i..j as far as it will go
    best = [[0] * n for _ in range(n)]
    for span in range(groupSize, n + 1):
        for i in range(n - span + 1):
            j = i + span - 1
            cheapest = None
            # the left part must itself be combinable to one lot, so its length
            # runs over 1, 1 + step, 1 + 2 * step, ...
            for mid in range(i, j, step):
                cand = best[i][mid] + best[mid + 1][j]
                if cheapest is None or cand < cheapest:
                    cheapest = cand
            best[i][j] = cheapest
            if (span - 1) % step == 0:
                # this stretch collapses all the way to one lot, and the final
                # combine costs the whole stretch weight
                best[i][j] += prefix[j + 1] - prefix[i]
    return best[0][n - 1]

The obvious version

Search the rows themselves: from a row, try each group of neighbouring lots that a combine could take, pay its charge and recurse on the shorter row, memoising on the row contents. A row that is stuck above one lot reports an unreachable cost. Explodes past about eight lots, but it makes no assumption about which lot counts are reachable.

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_merge_lots(weights, groupSize):
    INF = float('inf')
    memo = {}

    def solve(row):
        if len(row) == 1:
            return 0
        if len(row) < groupSize:
            return INF
        if row in memo:
            return memo[row]
        cheapest = INF
        for i in range(len(row) - groupSize + 1):
            total = 0
            for w in row[i:i + groupSize]:
                total += w
            rest = solve(row[:i] + (total,) + row[i + groupSize:])
            if rest + total < cheapest:
                cheapest = rest + total
        memo[row] = cheapest
        return cheapest

    got = solve(tuple(weights))
    return -1 if got == INF else got

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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