0004EasyArrayDynamic Programming

One Flip on the Card Market

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Trains the technique from

LeetCode 121Best Time to Buy and Sell Stock

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 collectors' marketplace publishes one closing quote per trading session for a single rare card. You are given the integer array quotes, which lists those closing quotes in session order.

You may complete at most one round trip on that card: acquire it at the quote of some session, then release it at the quote of a session that lands strictly later. The gain of a round trip is its release quote minus its acquisition quote.

Return the largest gain any single round trip can reach. When no session is quoted above any session that precedes it, no round trip can finish ahead, so staying out of the market is optimal and the answer is 0.

Examples

Example 1

Input: quotes = [9, 4, 6, 3, 8, 5]
Output: 5

The lowest quote available is the 3 in session 3, and the highest quote after it is the 8 in session 4, a spread of 5. No other ordered pair of sessions spreads wider.

Example 2

Input: quotes = [12, 11, 9, 6]
Output: 0

Each session is quoted under the one before it, so any completed round trip destroys value and sitting out wins.

Example 3

Input: quotes = [3]
Output: 0

One session leaves nothing to release into, so no round trip is possible.

Constraints

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

Java

public int maxFlipGain(int[] quotes)

The intended approach

Time O(n)Space O(1)

Sweep the sessions once while remembering the cheapest quote seen so far. At each session, the best round trip that releases here is the current quote minus that running minimum, so keep the largest such value and never let it drop below zero.

Pythonentry point: max_flip_gain

def max_flip_gain(quotes):
    cheapest = quotes[0]
    best = 0
    for i in range(1, len(quotes)):
        quote = quotes[i]
        gain = quote - cheapest
        if gain > best:
            best = gain
        if quote < cheapest:
            cheapest = quote
    return best

The obvious version

Try every ordered pair of sessions, 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_profit(quotes):
    best = 0
    for i in range(len(quotes)):
        for j in range(i + 1, len(quotes)):
            if quotes[j] - quotes[i] > best:
                best = quotes[j] - quotes[i]
    return best

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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