121. Best Time to Buy and Sell Stock

#Problem

You are given an array prices where prices[i] is the price of a given stock on the iᵗʰ day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/

Example 1:

Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

Constraints:

• 1 <= prices.length <= 10⁵
• 0 <= prices[i] <= 10⁴

#Test Cases

class Solution:
    def maxProfit(self, prices: List[int]) -> int:

import pytest

from solution import Solution


@pytest.mark.parametrize('prices, expected', [
    ([7,1,5,3,6,4], 5),
    ([7,6,4,3,1], 0),

    ([1,2], 1),
])
class Test:
    def test_solution(self, prices, expected):
        sol = Solution()
        assert sol.maxProfit(prices) == expected

#Thoughts

对于任意的一天，如果当天卖出，想要收益最大应该在之前最低价那天买入，二者差值记为最大收益（如果当天就是历史最低价，那就无法获得收益，相当于收益为 0）。然后看哪一天卖出的收益最大即可。

即对于 1 <= i <= n，profit(i) = prices(i) - min{prices(1..i)}，过程中记录可能的最大收益。

#Code

class Solution:
    def maxProfit(self, prices: list[int]) -> int:
        min_price = prices[0]
        max_profit = 0
        for price in prices:
            min_price = min(min_price, price)
            max_profit = max(max_profit, price - min_price)

        return max_profit