0270HardArrayDynamic Programming

Two Resale Rounds on the Scrap Board

Tracked in this browser only

Trains the technique from

LeetCode 123Best Time to Buy and Sell Stock III

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 scrap dealer posts one price per day for a pallet of recycled aluminium, and trades in either direction at that price. quotes[i] is the posted price on day i.

You may own at most one pallet at any moment, and a pallet you buy has to be sold on a strictly later day. A round is one purchase followed by its sale, and you may complete at most two rounds; the day a pallet is sold may also be the day the next one is bought.

Return the largest profit these rounds can add up to. Sitting the whole stretch out is allowed, so the answer is never below 0.

Examples

Example 1

Input: quotes = [12, 5, 9, 3, 11, 4, 10]
Output: 14

Buy on day 3 at 3 and sell on day 4 at 11 for 8, then buy on day 5 at 4 and sell on day 6 at 10 for 6. The two rounds add up to 14.

Example 2

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

Every later day posts a lower price than every earlier one, so no purchase can be sold for more than it cost and no round is opened.

Example 3

Input: quotes = [4, 6]
Output: 2

Buy on day 0 at 4 and sell on day 1 at 6 for 2. Only one round fits in two days, and using fewer than two rounds is allowed.

Constraints

1 <= quotes.length <= 10^5
0 <= quotes[i] <= 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 best_two_rounds(quotes: list[int]) -> int:

Java

public int bestTwoRounds(int[] quotes)

The intended approach

Time O(n)Space O(1)

Track four running balances as the days go by: the best balance while holding a first pallet, after selling it, while holding a second pallet, and after selling that one. Each day every balance is refreshed from the one before it in that order, so a purchase can be funded by the profit already banked. The answer is the last balance after the final day.

Pythonentry point: best_two_rounds

def best_two_rounds(quotes):
    # Four running bests, each refreshed once per day in this order:
    #   hold1 - best balance while holding the first pallet
    #   done1 - best balance after that pallet has been sold
    #   hold2 - best balance while holding the second pallet
    #   done2 - best balance after the second pallet has been sold
    # Time: O(n), one pass.  Space: O(1), four scalars.
    NEG = float('-inf')
    hold1 = NEG
    done1 = 0
    hold2 = NEG
    done2 = 0
    for p in quotes:
        if -p > hold1:
            hold1 = -p
        if hold1 + p > done1:
            done1 = hold1 + p
        if done1 - p > hold2:
            hold2 = done1 - p
        if hold2 + p > done2:
            done2 = hold2 + p
    return done2

The obvious version

For every cut day, recompute by nested loops the best single round entirely on or before that day and the best single round entirely on or after it, and take the largest sum over all cuts.

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_two_rounds(quotes):
    n = len(quotes)
    if n < 2:
        return 0

    def best_single(lo, hi):
        best = 0
        for i in range(lo, hi + 1):
            for j in range(i + 1, hi + 1):
                if quotes[j] - quotes[i] > best:
                    best = quotes[j] - quotes[i]
        return best

    ans = 0
    for k in range(n):
        total = best_single(0, k) + best_single(k, n - 1)
        if total > ans:
            ans = total
    return ans

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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in