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.

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

My code

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if len(prices) < 2:
            return 0
        lo = prices[0]
        hi = prices[0]
        maxprofit = 0
        for p in prices:
            if p > hi:
                hi = p
                maxprofit = max(hi-lo, maxprofit)
            if p < lo:
                lo = p
                hi = p
        return maxprofit

Review

[실행 결과]
Runtime: 936 ms / Memory: 25.2MB

[접근법]
낮은 값과 높은 값을 저장하는 공간을 만들어서 새 숫자가 나올때마다 두 값과 비교하여 교체해줌
높은값 - 낮은값의 최대값을 저장하는 공간을 만듦
최대값 출력

[느낀점]
한 칸씩 옮겨 중복을 막기 위해서는 for loop가 필수