0241MediumArrayDynamic Programming

Grade the Orchard Rows

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Trains the technique from

LeetCode 1043Partition Array for Maximum Sum

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.

An orchard is planted as one long line of rows. You are given yields, where yields[i] is the crop weight recorded for row i, and an integer span.

At grading time the manager cuts the line into consecutive blocks. Every row belongs to exactly one block, and a block holds at least one row and at most span rows. Each row is then written into the season report with the heaviest weight found anywhere in its own block, and the season figure is the sum of the written weights over all rows.

Return the largest season figure that some cut into blocks produces.

Examples

Example 1

Input: yields = [4, 1, 1, 12, 2, 6, 3], span = 3
Output: 58

Cutting into the blocks [4], [1, 1, 12] and [2, 6, 3] writes 4, then 12 three times, then 6 three times, giving 4 + 36 + 18 = 58.

Example 2

Input: yields = [1, 1, 9], span = 2
Output: 19

Cutting into the blocks [1] and [1, 9] writes 1, then 9 twice, giving 1 + 18 = 19.

Example 3

Input: yields = [6, 2, 5, 1, 8, 3, 4], span = 2
Output: 42

Cutting into the blocks [6, 2], [5, 1], [8, 3] and [4] writes 12 + 10 + 16 + 4 = 42.

Constraints

1 <= yields.length <= 500
0 <= yields[i] <= 10^9
1 <= span <= yields.length

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 max_graded_total(yields: list[int], span: int) -> int:

Java

public int maxGradedTotal(int[] yields, int span)

The intended approach

Time O(n * span)Space O(n) for the table of prefix answers

Let `best_to[end]` be the largest season figure for the first `end` rows. Whatever the cut looks like, the last block ends at row `end - 1` and covers between 1 and `span` rows, so `best_to[end]` is the best of `best_to[end - length] + top * length` over those lengths, where `top` is the heaviest weight in the last block. The code grows that block leftwards one row at a time and carries `top` along with it, so the block maximum never has to be rescanned. Every entry is filled in increasing `end`, and `best_to[n]` is the answer.

Pythonentry point: max_graded_total

def max_graded_total(yields, span):
    n = len(yields)
    best_to = [0] * (n + 1)
    for end in range(1, n + 1):
        best = 0
        top = 0
        length = 1
        while length <= span and length <= end:
            value = yields[end - length]
            if value > top:
                top = value
            candidate = best_to[end - length] + top * length
            if candidate > best:
                best = candidate
            length += 1
        best_to[end] = best
    return best_to[n]

The obvious version

Enumerate every legal cut with an explicit stack of (next row, running figure) states, scoring each block by scanning it, and keep the largest complete figure. Exponential in the number of rows.

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_max_graded_total(yields, span):
    n = len(yields)
    best = 0
    stack = [(0, 0)]
    while stack:
        start, total = stack.pop()
        if start == n:
            if total > best:
                best = total
            continue
        longest = span if span < n - start else n - start
        for length in range(1, longest + 1):
            block = yields[start:start + length]
            stack.append((start + length, total + max(block) * length))
    return best

How this problem was checked

Reference solution executed against 12 test cases — 12 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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