0334EasyArrayPrefix Sum

Peak Balance On A Ledger

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Trains the technique from

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 petty-cash ledger is opened with a balance of 0 and then n entries are posted in order. Entry i changes the balance by gain[i], which is negative for money paid out and positive for money taken in.

The balance is only ever inspected between postings, so the figures that matter are the opening 0 and the balance right after each of the n entries.

Return the largest balance the ledger ever shows. Because the opening figure counts, the answer is never below 0.

Examples

Example 1

Input: gain = [3, -2, 5]
Output: 6

The balances on the ledger read 0, 3, 1 and 6, and the largest of those four figures is 6.

Example 2

Input: gain = [5, -10, 2]
Output: 5

The balances read 0, 5, -5 and -3, so the ledger tops out at 5.

Example 3

Input: gain = [-4, -2, -3]
Output: 0

Every posting pays money out, so the balances read 0, -4, -6 and -9 and the opening figure is the largest.

Example 4

Input: gain = [100, -100, 100, -100]
Output: 100

The balances read 0, 100, 0, 100 and 0; the ledger reaches 100 twice and never goes higher.

Constraints

n == gain.length
1 <= n <= 100
-100 <= gain[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 peak_balance(deltas: list[int]) -> int:

Java

public int peakBalance(int[] deltas)

The intended approach

Time O(n)Space O(1)

Carry a running balance that starts at zero and add each entry to it in turn, keeping the largest value the balance has taken. Seeding the best figure with zero is what makes the opening balance one of the candidates.

Pythonentry point: peak_balance

def peak_balance(deltas):
    # Running balance, opened at zero, with the best figure seen so far.
    # Time O(n). Space O(1).
    balance = 0
    peak = 0
    for d in deltas:
        balance += d
        if balance > peak:
            peak = balance
    return peak

The obvious version

For every posting count, add that prefix of the entries up from scratch and keep the largest total against a floor of zero.

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_peak_balance(deltas):
    best = 0
    for k in range(1, len(deltas) + 1):
        total = 0
        for i in range(k):
            total += deltas[i]
        if total > best:
            best = total
    return best

How this problem was checked

Reference solution executed against 9 test cases — 9 passed
4 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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