Best Time to Buy and Sell Stock with Transaction Fee

Description

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

Code

DP

Time Complexity: $O(n)$, Space Complexity: $O(n)$

此題為 Best Time to Buy and Sell Stock II 的變形，只需要加上

T[i][0] = max(T[i - 1][0], T[i - 1][1] + prices[i - 1] - fee); 即可，在賣掉股票之後代表交易完成，才需要收手續費。

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int days = prices.size();
        int T[days + 1][2];
 
        // On day 0, no way ending up with stock in hand
        T[0][0] = 0;
        T[0][1] = -1e7;
 
        for(int i = 1; i < days + 1; i++) {
            T[i][0] = max(T[i - 1][0], T[i - 1][1] + prices[i - 1] - fee);
            T[i][1] = max(T[i - 1][1], T[i - 1][0] - prices[i - 1]);
        }
 
        return T[days][0];
    }
};

Source

• Best Time to Buy and Sell Stock with Transaction Fee - LeetCode