1085EasyArray

The Heaviest Rising Run

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Trains the technique from

LeetCode 1800Maximum Ascending Subarray 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.

A strip of positive readings reads readings. A run of neighbouring readings rises when each of its readings is strictly larger than the one before it; a run of a single reading rises by default.

Return the largest total any rising run adds up to.

Examples

Example 1

Input: readings = [1, 2, 3, 1]
Output: 6

The first three readings rise and add to six. The last reading breaks the run and stands alone at one.

Example 2

Input: readings = [7, 7, 7]
Output: 7

No reading is strictly larger than the one before, so every rising run holds a single reading.

Example 3

Input: readings = [9, 4, 5]
Output: 9

The 9 stands alone at nine, while the rising run 4 then 5 adds only to nine as well, so nine is the answer either way.

Constraints

1 <= readings.length <= 100
1 <= readings[i] <= 100

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_ascending_sum(readings: list[int]) -> int:

Java

public int maxAscendingSum(int[] readings)

The intended approach

Time O(n) over the strip's length.Space O(1).

Walk the strip once carrying the total of the rising run that ends here. Each reading either continues the run, because it is strictly larger than the one before, which adds it to that total, or it breaks the run, which starts a fresh total at that reading alone. Keep the largest total seen. Since every reading is positive, a longer rising run always beats any of its own pieces, so there is nothing to weigh up beyond continuing or restarting. One pass and two numbers.

Pythonentry point: max_ascending_sum

def max_ascending_sum(readings):
    best = readings[0]
    run = readings[0]
    for at in range(1, len(readings)):
        if readings[at] > readings[at - 1]:
            run += readings[at]
        else:
            run = readings[at]
        if run > best:
            best = run
    return best

The obvious version

Try every run of the strip: for each starting place walk forward while each reading beats the one before it, adding as you go, and keep the largest total any such walk reaches. Costs the strip's length squared rather than one pass.

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_ascending_sum(readings):
    n = len(readings)
    best = 0
    for start in range(n):
        total = readings[start]
        if total > best:
            best = total
        at = start + 1
        while at < n and readings[at] > readings[at - 1]:
            total += readings[at]
            if total > best:
                best = total
            at += 1
    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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