Best Time to Buy and Sell Stock

너무 쉽게 생각했는데 이중 for 문으로 memory error가 생겼음

https://leetcode.com/problems/best-time-to-buy-and-sell-stock/submissions/1519067390

문제 링크

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

내 답안

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        profit = []

        for i in range(0, len(prices)):
            for j in range(i, len(prices)):
                profit.append(prices[j] - prices[i])
        if max(profit) <= 0:
            return 0
        else:
            return max(profit)

Leet Code

class Solution:
    def maxProfit(self,prices):
        left = 0 #Buy
        right = 1 #Sell
        max_profit = 0
        while right < len(prices):
            currentProfit = prices[right] - prices[left] #our current Profit
            if prices[left] < prices[right]:
                max_profit =max(currentProfit,max_profit)
            else:
                left = right
            right += 1
        return max_profit

1. buy 와 sell을 설정 (sell은 buy의 다음날인 1로 initialize)
2. current profit을 계산
3. 만약 buy가격 < sell가격: max_profit 업데이트
4. 아니라면, buy타이밍을 sell타이밍 allocate
5. 항상 right += 1
6. 최종적으로는 max_profit 리턴

핵심은 가격기준으로 Update를 생각