0044MediumArrayDynamic ProgrammingGreedy

Collectible Resale Run

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Trains the technique from

LeetCode 122Best Time to Buy and Sell Stock II

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 dealer trades one collectible and already knows what it will be quoted at on every trading day: quotes[i] is its price in whole credits on day i.

On each day the dealer may buy one copy, sell the copy currently held, or stand pat. Never more than one copy may be held at a time, so a copy has to be sold before another is bought, and a sale followed by a purchase on the same day is allowed. The dealer holds nothing before day 0 and must hold nothing once the last day closes.

Return the largest total profit available over the whole run.

Examples

Example 1

Input: quotes = [6, 1, 4, 2, 8, 5]
Output: 9

Buy on day 1 at 1 and sell on day 2 at 4 for 3 credits, then buy on day 3 at 2 and sell on day 4 at 8 for 6 credits, totalling 9.

Example 2

Input: quotes = [8, 6, 4, 1]
Output: 0

The quote never rises from one day to the next, so the dealer buys nothing and keeps a profit of 0.

Example 3

Input: quotes = [2, 2, 5, 5, 1, 1, 6]
Output: 8

Hold from day 1 through day 3 for 3 credits, then from day 5 through day 6 for 5 credits; flat days add nothing either way.

Constraints

1 <= quotes.length <= 3 * 10^4
0 <= quotes[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 resale_profit(quotes: list[int]) -> int:

Java

public int resaleProfit(int[] quotes)

The intended approach

Time O(n)Space O(1)

Any holding period pays exactly the sum of the day-to-day changes inside it, so a period that spans a fall can be split at that fall without losing anything, and a period that spans a rise is never worth skipping. That leaves one rule: bank every rise between consecutive days and ignore every fall. One pass, one running total.

Pythonentry point: resale_profit

def resale_profit(quotes):
    total = 0
    for i in range(1, len(quotes)):
        rise = quotes[i] - quotes[i - 1]
        if rise > 0:
            total += rise
    return total

The obvious version

Unmemoised search over the calendar: on each day, branch on standing pat and on the one trade the current position allows, spending credits on a purchase and collecting them on a sale, then take the best leaf. Selling and rebuying within one day cancels out, so at most one action per day is explored. Exponential in the number of days, and it makes no greedy assumption at all.

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_resale_profit(quotes):
    n = len(quotes)

    def best(day, holding):
        if day == n:
            return 0
        outcome = best(day + 1, holding)
        if holding:
            outcome = max(outcome, quotes[day] + best(day + 1, False))
        else:
            outcome = max(outcome, -quotes[day] + best(day + 1, True))
        return outcome

    return best(0, False)

How this problem was checked

Reference solution executed against 9 test cases — 9 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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