0026MediumArrayDynamic ProgrammingDivide and Conquer

Best Trading Day Run

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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 market stall balances its books every evening. daily[i] is the money the stall cleared on day i: a positive figure on a day it came out ahead, a negative figure on a day it went backwards, and 0 on a day that broke even.

The owner wants to quote her strongest run of trading. A run is the days from some day i through some day j with i <= j, taken back to back with no days left out in between, and its worth is the sum of the figures on those days. Return the worth of the strongest run in daily.

A run must contain at least one day; skipping every day is not allowed. So when the stall went backwards on all of its days, the strongest run is the single day it lost the least on, and the answer you return is negative. Do not report 0 for that case.

Examples

Example 1

Input: daily = [3, -2, 7, -8, 4]
Output: 8

Days 0 through 2 clear 3 - 2 + 7 = 8. Taking day 2 by itself yields only 7, and carrying on into day 3 gives back 8 of it, so no other run beats 8.

Example 2

Input: daily = [-6, -2, -9]
Output: -2

Every run of two or more days is worse than its own worst day, so the strongest choice is day 1 alone at -2. Since a run cannot be empty, a negative answer is the correct one.

Example 3

Input: daily = [5]
Output: 5

One day of trading means exactly one run is available and its worth is 5.

Constraints

1 <= daily.length <= 10^5
-10^4 <= daily[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 best_run_total(daily: list[int]) -> int:

Java

public int bestRunTotal(int[] daily)

The intended approach

Time O(n)Space O(1)

Sweep the days once while tracking the worth of the strongest run that ends on the current day. Extending yesterday's best run only helps while that worth is above zero; the moment it has dipped below zero, starting fresh on today's figure is at least as good. Keep the largest of those per-day values seen so far and report it at the end. Because the first day seeds both trackers rather than zero, an all-negative book still yields its least costly day.

Pythonentry point: best_run_total

def best_run_total(daily):
    best = daily[0]
    ending_here = daily[0]
    for figure in daily[1:]:
        if ending_here < 0:
            ending_here = figure
        else:
            ending_here += figure
        if ending_here > best:
            best = ending_here
    return best

The obvious version

Total up every run explicitly by fixing a starting day and extending the end day one at a time, keeping the largest total. O(n^2).

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_best_run(daily):
    best = daily[0]
    for start in range(len(daily)):
        running = 0
        for end in range(start, len(daily)):
            running += daily[end]
            if running > best:
                best = running
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 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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