0509MediumArrayGreedy

Repeatable Trading Windows

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Trains the technique from

LeetCode 3689Maximum Total Subarray Value I

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.

prices holds one settlement price per trading day, in day order. A window is any non-empty run of consecutive days, and the spread of a window is its highest settlement price minus its lowest.

A desk must file exactly k windows for the quarter. The same window may be filed more than once, and there is no rule against overlapping windows. The quarter's score is the sum of the spreads of the k filed windows.

Return the largest score the desk can file.

Examples

Example 1

Input: prices = [3, 1, 4], k = 3
Output: 9

The window covering all three days has highest price 4 and lowest price 1, a spread of 3. Filing that window three times scores 9.

Example 2

Input: prices = [1, 5, 2, 6], k = 2
Output: 10

The window covering all four days has highest price 6 and lowest price 1, a spread of 5, and filing it twice scores 10.

Example 3

Input: prices = [0, 1000000000], k = 100000
Output: 100000000000000

The record has highest price 1000000000 and lowest price 0, so the two-day window spreads 1000000000, and filing it 100000 times scores 100000000000000.

Constraints

1 <= prices.length <= 5 * 10^4
0 <= prices[i] <= 10^9
1 <= k <= 10^5

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_total_spread(prices: list[int], k: int) -> int:

Java

public long maxTotalSpread(int[] prices, int k)

The intended approach

Time O(n): a single scan for the highest and lowest price.Space O(1): two running extremes.

No window can hold a price above the record's highest or below its lowest, so no window's spread can beat the difference between those two extremes, and the window covering every day attains it. Because a window may be filed repeatedly, all `k` filings can be that same window, so one linear scan for the two extremes and a single multiplication finish the job.

Pythonentry point: max_total_spread

def max_total_spread(prices, k):
    # The spread of a stretch can never beat the spread of the whole record, and
    # the whole record is itself a legal stretch, so one stretch already reaches
    # the largest spread available. Repeats are allowed, so charging that same
    # stretch k times is the whole answer.
    # O(n) time, O(1) space.
    lo = hi = prices[0]
    for v in prices:
        if v < lo:
            lo = v
        if v > hi:
            hi = v
    return k * (hi - lo)

The obvious version

Grow every window from every starting day, tracking its running highest and lowest price to get its spread, keep the largest spread seen, and multiply by `k`. 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_max_total_spread(prices, k):
    n = len(prices)
    best = 0
    for i in range(n):
        lo = hi = prices[i]
        for j in range(i, n):
            if prices[j] < lo:
                lo = prices[j]
            if prices[j] > hi:
                hi = prices[j]
            if hi - lo > best:
                best = hi - lo
    return best * k

How this problem was checked

Reference solution executed against 12 test cases — 12 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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