Last updated on October 5th, 2024 at 05:30 pm

Here, We see Best Time to Buy and Sell Stock with Transaction Fee LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Best Time to Buy and Sell Stock with Transaction Fee LeetCode Solution

Problem Statement

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

1. Best Time to Buy and Sell Stock with Transaction Fee LeetCode Solution C++

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int buy = INT_MIN;
        int sell = 0;
        for (int price : prices) {
            buy = max(buy, sell - price);
            sell = max(sell, buy + price - fee);
        }
        return sell;
    }
};

2. Best Time to Buy and Sell Stock with Transaction Fee LeetCode Solution Java

class Solution {
    public int maxProfit(int[] prices, int fee) {
        int buy = Integer.MIN_VALUE;
        int sell = 0;
        for (int price : prices) {
            buy = Math.max(buy, sell - price);
            sell = Math.max(sell, buy + price - fee);
        }
        return sell;        
    }
}

3. Best Time to Buy and Sell Stock with Transaction Fee Solution JavaScript

var maxProfit = function(prices, fee) {
    let len = prices.length, buying = 0, selling = -prices[0]
    for (let i = 1; i < len; i++) {
        buying = Math.max(buying, selling + prices[i] - fee)
        selling = Math.max(selling, buying - prices[i])
    }
    return buying    
};

4. Best Time to Buy and Sell Stock with Transaction Fee Solution Python

class Solution(object):
    def maxProfit(self, prices, fee):
        buy = float('-inf')
        sell = 0
        for price in prices:
            buy = max(buy, sell - price)
            sell = max(sell, buy + price - fee)
        return sell