1007MediumArrayDynamic Programming

The Biggest Swing in a Run of Adjustments

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Trains the technique from

LeetCode 1749Maximum Absolute Sum of Any Subarray

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 ledger of adjustments reads adjustments. A stretch is any run of neighbouring entries, and the empty stretch counts as one, with a total of nothing.

Return the largest size any stretch's total reaches, where size ignores whether the total came out above or below nothing.

Examples

Example 1

Input: adjustments = [4, -1, 2, 1]
Output: 6

The whole ledger totals six, and no shorter stretch swings further in either direction.

Example 2

Input: adjustments = [-1, -2, -3]
Output: 6

Every entry pulls downward, so the whole ledger's total of minus six is the furthest swing and its size is six.

Example 3

Input: adjustments = [10000, -10000, 10000]
Output: 10000

Either of the two rises on its own swings ten thousand. Taking more entries only cancels part of the swing away.

Constraints

1 <= adjustments.length <= 10^5
-10^4 <= adjustments[i] <= 10^4

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_absolute_sum(adjustments: list[int]) -> int:

Java

public int maxAbsoluteSum(int[] adjustments)

The intended approach

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

A total's size, sign ignored, is the larger of that total and its opposite. So the answer is the larger of two things: the biggest stretch total there is, and the biggest stretch total after every entry has its sign flipped. Both come out of a single pass. For the upward side, carry the best total of any stretch ending right here: at each entry either extend what came before or drop it and start over, whichever is larger, never letting it fall below nothing, which is what keeps the empty stretch available. Carry the mirror of that for the downward side, never letting it rise above nothing. The best either side ever shows is the answer. There is a second way to see the same number. A stretch's total is the difference between two running totals of the whole ledger, counting an empty running total of nothing at the start, so the answer is simply the spread between the largest and smallest of those running totals. Either reading costs one pass and nothing held but a few numbers.

Pythonentry point: max_absolute_sum

def max_absolute_sum(adjustments):
    up = 0
    down = 0
    best = 0
    for value in adjustments:
        up += value
        if up < 0:
            up = 0
        down += value
        if down > 0:
            down = 0
        if up > best:
            best = up
        if -down > best:
            best = -down
    return best

The obvious version

Take every starting place and run forward from it, keeping the total as it grows and watching its size against the best seen. Costs every stretch rather than one pass with two carried totals.

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_absolute_sum(adjustments):
    n = len(adjustments)
    best = 0
    for start in range(n):
        total = 0
        for end in range(start, n):
            total += adjustments[end]
            size = total if total >= 0 else -total
            if size > best:
                best = size
    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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