0587MediumArrayDynamic Programming

Reselling One Lot With a Rest Day

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Trains the technique from

LeetCode 309Best Time to Buy and Sell Stock with Cooldown

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 reseller watches one component whose quoted price on day i is prices[i].

The rules of the desk are:

At most one lot may be held at any moment, so a lot has to be sold before another is bought.
A lot bought on some day may be sold on that day or on any later day.
The day straight after a sale is a rest day: no buying is allowed on it. If a sale happens on day i, the earliest possible next purchase is day i + 2.
Buying and selling may be repeated as often as the rules allow, and it is always allowed to trade nothing at all.

Return the largest profit the reseller can finish with.

Examples

Example 1

Input: prices = [5, 10, 1, 10]
Output: 9

Buying on day 2 at 1 and selling on day 3 at 10 gains 9. Buying on day 0 at 5 and selling on day 1 at 10 gains 5, and day 2 is then a rest day, so day 3 is the only day left to buy on and there is no later day to sell on.

Example 2

Input: prices = [9, 8, 7, 6, 5]
Output: 0

Every later quote is below every earlier one, so any completed trade loses money and trading nothing is allowed.

Example 3

Input: prices = [6, 1, 3, 2, 4, 7]
Output: 6

Buying on day 1 at 1 and selling on day 2 at 3 gains 2, day 3 is the rest day, and buying on day 4 at 4 then selling on day 5 at 7 gains 3, for 5 in total.

Constraints

1 <= prices.length <= 5000
0 <= prices[i] <= 1000

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_profit(prices: list[int]) -> int:

Java

public int maxProfit(int[] prices)

The intended approach

Time O(n) — one pass over the days.Space O(1) — three running balances.

At the close of any day the desk is in one of three situations: holding a lot, having sold on that very day, or free to buy. Keep the best balance reachable in each situation and roll them forward one day at a time. Holding either continues from yesterday's holding or starts by paying today's price out of yesterday's free balance. Having just sold comes from yesterday's holding plus today's price. Being free either continues from yesterday's free balance or inherits yesterday's just-sold balance, and that one-day lag is exactly the rest day. The answer is the better of the two situations that hold no lot.

Pythonentry point: max_profit

def max_profit(prices):
    # O(n) time, O(1) space: carry three running totals across the days - the
    # best balance while holding a lot, the best balance on a day a sale just
    # happened, and the best balance while free to buy. The rest day is encoded
    # by letting the free state pick up the sale state only on the day after.
    holding = -prices[0]
    just_sold = 0
    free = 0
    for i in range(1, len(prices)):
        price = prices[i]
        new_holding = holding if holding > free - price else free - price
        new_just_sold = holding + price
        new_free = free if free > just_sold else just_sold
        holding = new_holding
        just_sold = new_just_sold
        free = new_free
    return just_sold if just_sold > free else free

The obvious version

Pick the trades directly. From the first day the desk is free, try every possible buy day and every later sell day, keep only trades that gain something, and recurse from two days after the sale so the rest day is skipped. The best total over all those choices is the answer.

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_profit(prices):
    n = len(prices)

    def best_from(day):
        top = 0
        for buy in range(day, n):
            for sell in range(buy + 1, n):
                gain = prices[sell] - prices[buy]
                if gain > 0:
                    total = gain + best_from(sell + 2)
                    if total > top:
                        top = total
        return top

    return best_from(0)

How this problem was checked

Reference solution executed against 10 test cases — 10 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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