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

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

Topics

Array, Dynamic Programming, Greedy

Companies

Bloomberg, Facebook

Level of Question

Medium

Best Time to Buy and Sell Stock with Transaction Fee 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
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