0714. Best Time to Buy and Sell Stock with Transaction Fee

Medium | Array + DP | 688 ms (75.06%), 21.1 MB (99.14%)

❓ Problem Statement

Source: LeetCode - Best Time to Buy and Sell Stock with Transaction Fee GitHub: Solution / Performance

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).

✍🏻 Constraints & Example

Constraints:

• 1 <= prices.length <= 5 * 104

• 1 <= prices[i] < 5 * 104

• 0 <= fee < 5 * 104

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.

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

💡 Ideas

We can use Final State Machine to solve the series of these problems.

Note that this problem requires transaction fee after selling the stock.

• Hold state means you hold the stock.

  • Since you don't have any stocks at the beginning, the profit at this state is initialized with negative infinity.

  • The profit in this state is calculated by: previous no-hold state's profit - current stock price (= buy)

• No-Hold state means you do not hold any stocks.

  • Since you don't have any stocks at the beginning (you're initially at this state ), the profit at this state is initialized with 0.

  • The profit in this state is calculated by: previous hold state's profit + current stock price - fee (= sell)

  • Return the profit at the no-hold state as the maximum profit since the hold state still has stock on hold.

🤖 Python3

class Solution:
    def maxProfit(self, prices: List[int], fee: int) -> int:
        # (base case)
        if len(prices) == 0 or len(prices) == 1: return 0

        # ==================================================
        #  Array + Dynamic Programming              (FSM)  =
        # ==================================================
        # time  : O(n)
        # space : O(1)
        
        hold, noHold = float('-inf'), 0
        
        for price in prices:
            preHold, preNoHold = hold, noHold
            
            hold   = max(preHold,   preNoHold - price)
            noHold = max(preNoHold, preHold   + price - fee)
            
        return noHold

🤖 Java

class Solution {
    /**
     * @time  : O(n)
     * @space : O(1)

    public int maxProfit(int[] prices, int fee) {
        /* base case */
        if(prices.length == 1) return 0;
        
        int hold = -prices[0], noHold = 0;
        
        for(int i=1 ; i<prices.length ; i++) {
            int preHold = hold, preNoHold = noHold;
            
            hold   = Math.max(preHold,   preNoHold - prices[i]);
            noHold = Math.max(preNoHold, preHold   + prices[i] - fee);
        }
        
        return noHold;
    }
}