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# Best Time to Buy and Sell Stock with Transaction Fee

## Description

Your are given an array of integers `prices`, for which the `i`-th element is the price of a given stock on day `i`; and a non-negative integer `fee` representing a transaction fee.

You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)

Return the maximum profit you can make.

**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] = 1Selling at prices[3] = 8Buying at prices[4] = 4Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
```

**Note:**

* `0 < prices.length <= 50000`.
* `0 < prices[i] < 50000`.
* `0 <= fee < 50000`.

## Solutions

### My solution

```cpp
class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        if(n < 2) return 0;
        vector<int> stock(n); // stock[i]: max profit at day i with stock
        vector<int> cash(n); // cash[i]: max profit at day i without holding stock
        stock[0] = -prices[0];
        cash[0] = 0;
        for(int i = 1; i < n; ++i){
            stock[i] = max(stock[i - 1], cash[i - 1] - prices[i]);
            cash[i] = max(cash[i - 1], stock[i - 1] + prices[i] - fee);
        }
        return cash.back();
    }
};
```

### Optimal solution

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