## Algorithm

Problem Name: 714. Best Time to Buy and Sell Stock with Transaction Fee

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 <= prices.length <= 5 * 104`
• `1 <= prices[i] < 5 * 104`
• `0 <= fee < 5 * 104`

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
int _max(a, b) {
if (a > b) return a;
return b;
}
int maxProfit(int* prices, int pricesSize, int fee) {

int i, p;

#if 1
sell = 0;

for (i = 0; i  <  pricesSize; i ++) {
p = prices[i];
sell = _max(sell, buy  + p - fee);
}
#else
sell = 0;

for (i = 1; i  <  pricesSize; i ++) {
p = prices[i];
sell = _max(sell, buy  + p - fee);
}
#endif

return sell;
}
``````
Copy The Code &

Input

cmd
prices = [1,3,2,8,4,9], fee = 2

Output

cmd
8

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const maxProfit = function(prices, fee) {
let cash = 0,
hold = -prices[0];
for (let i = 1; i  <  prices.length; i++) {
cash = Math.max(cash, hold + prices[i] - fee);
hold = Math.max(hold, cash - prices[i]);
}
return cash;
};

console.log(maxProfit([1, 3, 2, 8, 4, 9], 2));
``````
Copy The Code &

Input

cmd
prices = [1,3,2,8,4,9], fee = 2

Output

cmd
8

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def maxProfit(self, prices, fee):
pre = [0, -float("inf")]
for p in prices:
p0, p1 = pre[1] + p - fee, pre[0] - p
if p0 > pre[0]: pre[0] = p0
if p1 > pre[1]: pre[1] = p1
return pre[0]
``````
Copy The Code &

Input

cmd
prices = [1,3,7,5,10,3], fee = 3

Output

cmd
6