1149MediumArrayGreedySortingPrefix Sum

Ordering Entries to Keep the Totals Above Zero

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Trains the technique from

LeetCode 2587Rearrange Array to Maximize Prefix Score

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 list entries holds whole numbers, and they may be arranged in any order at all.

Once arranged, form the running totals: the first entry, then the first two added together, then the first three, and so on, giving one total per entry.

Return the largest number of those running totals that can be strictly above zero.

Examples

Example 1

Input: entries = [-6, -5, 4, 3, 2, 1]
Output: 5

Taking the four positive entries first gives totals of 4, 7, 9 and 10. Adding -5 leaves 5, still above zero, and adding -6 drops it to -1, so five totals stay above zero.

Example 2

Input: entries = [5, -5, 5, -5]
Output: 3

The two entries of 5 first give 5 and then 10, one -5 leaves 5, and the last brings the total exactly to zero, which does not count.

Example 3

Input: entries = [-1, -2, -3]
Output: 0

Every entry is below zero, so whatever the order the very first total is already negative.

Constraints

1 <= entries.length <= 10^5
-10^6 <= entries[i] <= 10^6

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_score(entries: list[int]) -> int:

Java

public int maxScore(int[] entries)

The intended approach

Time O(n log n) for the ordering.Space O(n) for the ordered copy.

Put the entries in falling order. That is best because a running total is only ever helped by taking the bigger entries earlier: swapping a smaller entry in front of a bigger one lowers every total between them and leaves the rest alone, so it can never turn a total from not-above-zero into above-zero. Then walk the running totals and count the ones above zero. A total of exactly zero does not count, which matters when the entries balance out or include zeros. Once a total falls to zero or below the walk can stop, and nothing is lost by stopping: for a later total to climb back above zero the entry added would have to be positive, but the entries are in falling order, so every entry before it would be positive too and the earlier total could never have fallen in the first place.

Pythonentry point: max_score

def max_score(entries):
    order = sorted(entries)
    order.reverse()
    total = 0
    count = 0
    for value in order:
        total += value
        if total > 0:
            count += 1
        else:
            break
    return count

The obvious version

Try every arrangement of the entries, form its running totals, count the ones above zero, and keep the largest count found. Costs the factorial of the entry count.

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_score(entries):
    from itertools import permutations
    best = 0
    for order in set(permutations(entries)):
        total = 0
        count = 0
        for value in order:
            total += value
            if total > 0:
                count += 1
        if count > best:
            best = count
    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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