122. Best Time to Buy and Sell Stock II

Description

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

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can sell and buy the stock multiple times on the same day, ensuring you never hold more than one share of the stock.

Find and return the maximum profit you can achieve.

Example 1:

• Input: prices = [7,1,5,3,6,4]
• Output: 7
• Explanation:
  Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
  Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
  Total profit is 4 + 3 = 7.

Example 2:

• Input: prices = [1,2,3,4,5]
• Output: 4
• Explanation:
  Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
  Total profit is 4.

Example 3:

• Input: prices = [7,6,4,3,1]
• Output: 0
• Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

Constraints:

• 1 <= prices.length <= 3 * 10⁴
• 0 <= prices[i] <= 10⁴

Submitted Code

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

        for i in range(len(prices)-1):
            tdy, tmr = prices[i], prices[i+1]
            if tdy < tmr:
                total_profit += tmr - tdy
        
        return total_profit

Runtime: 0 ms | Beats 100.00%
Memory: 20.36 MB | Beats 52.69%

이 문제의 규칙은

• 같은 날에 주식을 팔고 나서 다시 구매할 수 있다.
• 거래 횟수의 제한이 없다.
• 미래의 가격을 모두 알고 있다.

이기 때문에 매일 바로 다음날의 가격과 비교하여 이익이 날 때만 결과에 더해주면 한 번의 순회로 최대 이익을 구할 수 있다.

121. Best Time to Buy and Sell Stock

Other Solutions

1st

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        profit = 0
        
        for i in range(1, len(prices)):
            if prices[i] > prices[i-1]:
                profit += prices[i] - prices[i-1]
        
        return profit

time complexity: 𝑂(𝑛)
space complexity: 𝑂(1)