---
# System prepended metadata

title: 714. Best Time to Buy and Sell Stock with Transaction Fee
tags: [dp, leetcode, medium]

---


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

###### tags: `leetcode`,`dp`,`medium`


>ref: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/?envType=study-plan&id=dynamic-programming-ii
>
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).

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 <= nums.length <= 104
-104 < nums[i], target < 104
All the integers in nums are unique.
nums is sorted in ascending order.



>1. timeResolution:O(N)
>2. sc O(1)
>3. if can be iterate and subas problem use dp

```java=
class Solution {
    public int maxProfit(int[] prices, int fee) {

        // buy[i] = Math.max(sell[i-1]-prices[i],buy[i-1]);
        // sell[i] = Math.max(buy[i-1]+prices[i]-fee,sell[i-1]);

        // buy = Math.max(sell-prices[i],buy)
        // sell = Math.max(buy+prices[i]-fee,sell);

        int buy=-prices[0];
        int sell=0;

        int n=prices.length;
        for(int i=0;i<n;i++){
            buy = Math.max(sell-prices[i],buy);
            sell = Math.max(buy+prices[i]-fee,sell);
        }

        return sell;
        
    }
}
```
    
    
    
