---
# System prepended metadata

title: 714. Best Time to Buy and Sell Stock with Transaction Fee
tags: [Leetcode, greedy, medium, python, array, c++, dynamic programming]

---

###### tags: `Leetcode` `medium` `dynamic programming` `array` `greedy` `python` `c++`

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

## [題目連結:] https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/

## 題目:
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
```

## 解題想法:
* 此題為股票題，給一系列股票交易價格，每次交易都會有手續費(fee)，求最大獲益
* 使用DP:
    * 定義兩組
        * unHold=[0]* len(prices) **手中沒持有股票的獲利**
        * Hold=[0]* len(prices)  **手中持有股票的獲利**
        * 初始Hold[0]=-prices[0]
    * 轉移方程:
        * unHold[i]: 表示今天手裡沒有股票了
            * max(昨天手中就沒有股票, 昨天有股票+今天賣-手續費)
        * Hold[i]: 表示今天手裡有股票
            * max(昨天持有的, 昨天手中沒股票-今天買)

## Python:
``` python=
class Solution(object):
    def maxProfit(self, prices, fee):
        """
        :type prices: List[int]
        :type fee: int
        :rtype: int
        """
        unHold=[0]*len(prices) #手中沒持有股票的獲利
        Hold=[0]*len(prices)  #手中持有股票的獲利
        Hold[0]=-prices[0]
        for i in range(1,len(prices)):
            #昨天的vs昨天有買+今天賣-手續費
            unHold[i]=max(unHold[i-1],Hold[i-1]+prices[i]-fee)
            #昨天持有的vs昨天積蓄-今天買
            Hold[i]=max(Hold[i-1],unHold[i-1]-prices[i])
            print(unHold,Hold)
        return unHold[-1]

result=Solution()
ans=result.maxProfit(prices = [1,3,2,8,4,9], fee = 2)
print(ans)
```

## C++:
``` cpp=
#include<bits/stdc++.h>
using namespace std;

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int n=prices.size();
        vector<int> Hold(n, 0), unHold(n, 0);
        Hold[0]=-prices[0];
        for (int i=1; i<n; i++){
            unHold[i]=max(unHold[i-1], Hold[i-1]+prices[i]-fee);
            Hold[i]=max(Hold[i-1], unHold[i-1]-prices[i]);
        }
        return unHold[n-1];
    }
};
```