[TIL/05.08] Leetcode: Best Time to Buy and Sell Stock (Python)

문제

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

 

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

 

풀이

내 풀이 (시간 초과) 

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        max_profit = 0

        for l in range(len(prices)):
            r = len(prices)-1 

            while l < r:
                if prices[l] < prices[r]:
                    tmp = prices[r]-prices[l]
                    if tmp > max_profit:
                        max_profit = tmp
                r -= 1   

            if l >= len(prices)-1:
                break
                
        return max_profit

 

➤ O(n2)의 시간 복잡도

해설 풀이

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        l, r = 0, 1
        maxP = 0

        while r < len(prices):
            if prices[l] < prices[r]:
                profit = prices[r] - prices[l]
                maxP= max(maxP, profit)
            else:
                l == r 
            r += 1
        
        return maxP

 

 

배운 내용

✏️ 배움 기록

• else문에서 l += 1 이 아닌 l == r 을 함. 이후의 연산에서 해당 위치를 건너뛰는 것이 합리적이기 떄문