개요

LeetCode #121, Best Time to Buy and Sell Stock 문제를 풀어봅니다.

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.

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

최솟값으로 꽤 큰 수를 지정해놓고 가격과 비교해서 최솟값을 업데이트하고, 이익은 가격과 최솟값의 차이를 계속 구하면서 업데이트합니다.

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        profit = 0
        min_price = 100000000000
        
        # update min_price. update profit
        for price in prices:
            min_price = min(min_price, price)
            profit = max(profit, price-min_price)
            
        return profit