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714. Best Time To Buy And Sell Stock With Transaction Fee

Array Dynamic Programming Greedy

Problem - Best Time To Buy And Sell Stock With Transaction Fee

Medium

You are given an array prices where prices[i] is the price of a given stock on the ith day, and an integer fee representing a transaction fee.

Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.

Note:

  • You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
  • The transaction fee is only charged once for each stock purchase and sale.

 

Example 1:

Input: prices = [1,3,2,8,4,9], fee = 2
Output: 8
Explanation: The maximum profit can be achieved by:
- Buying at prices[0] = 1
- Selling at prices[3] = 8
- Buying at prices[4] = 4
- Selling at prices[5] = 9
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Example 2:

Input: prices = [1,3,7,5,10,3], fee = 3
Output: 6

 

Constraints:

  • 1 <= prices.length <= 5 * 104
  • 1 <= prices[i] < 5 * 104
  • 0 <= fee < 5 * 104

Solutions

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class Solution:
    def maxProfit(self, prices: List[int], fee: int) -> int:
        # n = len(prices)
        # dp = [[0] * 2 for _ in range(n)]
        # dp[0][1] = -prices[0]

        # for i in range(1, n):
        #     dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + prices[i] - fee)
        #     dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i])

        # return dp[-1][0]

        hold = -prices[0]
        cash = 0
        for price in prices[1:]:
            hold = max(hold, cash - price)
            cash = max(cash, hold + price - fee)

        return cash

Submission Stats:

  • Runtime: 94 ms (83.73%)
  • Memory: 23.6 MB (58.89%)