714. 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).

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 * 10^4

• 1 <= prices[i] < 5 * 10^4

• 0 <= fee < 5 * 10^4

解題

Runtime: 80 ms, faster than 100%

Memory Usage: 7.1 MdB, less than 96.92%

func maxProfit(prices []int, fee int) int {
    if len(prices) <= 1 { return 0 }
    buy := make([]int, len(prices))
    sell := make([]int, len(prices))

    buy[0] = -(prices[0] + fee)
    for i:=1; i<len(prices); i++ {
        buy[i] = max(buy[i - 1], sell[i - 1] - (prices[i] + fee)) // max(前一天的狀態, 買入股票)
        sell[i] = max(sell[i - 1], buy[i-1] + prices[i]) // max(前一天狀態, 賣出股票)
    } 

    return sell[len(sell) - 1]
}


func max(a, b int) int {
    if a > b { return a }
    return b
}