LeetCode link: 714. Best Time to Buy and Sell Stock with Transaction Fee
You are given an array prices where prices[i] is the price of a given stock on the i-th 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.
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[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.
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[Example 2]
Input: prices = [1,3,7,5,10,3], fee = 3
Output: 6
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[Constraints]
1 <= prices.length <= 5 * 10000
1 <= prices[i] < 5 * 10000
0 <= fee < 5 * 10000
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This problem can be solved using Dynamic programming.
Detailed solutions will be given later, and now only the best practices in 3 to 7 languages are given.
- Time:
O(n). - Space:
O(n).
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
# states:
# 0: hold stock
# 1) keep holding
# 2) today just bought
# 1: no stock
# 1) keep no stock
# 2) today just sold
dp = [-prices[0], 0]
for price in prices[1:]:
dc = dp.copy()
dp[0] = max(dc[0], dc[1] - price)
dp[1] = max(dc[1], dc[0] + price - fee)
return dp[1]// Welcome to create a PR to complete the code of this language, thanks!// Welcome to create a PR to complete the code of this language, thanks!// Welcome to create a PR to complete the code of this language, thanks!var maxProfit = function (prices, fee) {
const dp = [-prices[0], 0]
for (let i = 1; i < prices.length; i++) {
const dc = [...dp]
dp[0] = Math.max(dc[0], dc[1] - prices[i])
dp[1] = Math.max(dc[1], dc[0] + prices[i] - fee)
}
return dp[1]
};func maxProfit(prices []int, fee int) int {
dp := []int{-prices[0], 0}
for i := 1; i < len(prices); i++ {
dc := slices.Clone(dp)
dp[0] = max(dc[0], dc[1] - prices[i])
dp[1] = max(dc[1], dc[0] + prices[i] - fee)
}
return dp[1]
}# Welcome to create a PR to complete the code of this language, thanks!// Welcome to create a PR to complete the code of this language, thanks!// Welcome to create a PR to complete the code of this language, thanks!