714. Best Time to Buy and Sell Stock with Transaction Fee

1. 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).

2. Example

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

3. Constraints

• 1 <= prices.length <= $5 * 10^4$
• 1 <= prices[i] < $5 * 10^4$
• 0 <= fee < $5 * 10^4$

4. Solutions

Dynamic Programming

n = prices.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int maxProfit(const vector<int> &prices, int fee) {
        array<array<int, 2>, 2> dp{0};
        // dp[0] // have stock
        // dp[1] // don't have stock

        dp[0][0] = -prices[0], dp[0][1] = 0;
        for (int i = 1; i < prices.size(); ++i) {
            dp[1][0] = max(dp[0][1] - prices[i], dp[0][0]);
            dp[1][1] = max(dp[0][0] + prices[i] - fee, dp[0][1]);

            swap(dp[0], dp[1]);
        }

        return dp[0][1];
    }
};