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

## 714. [Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/)

## 1. Question

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:

The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
```

**Note:**

`0 < prices.length <= 50000`.

`0 < prices[i] < 50000`.

`0 <= fee < 50000`.

## 2. Implementation

**(1) DP**

```java
class Solution {
    public int maxProfit(int[] prices, int fee) {
        if (prices == null || prices.length == 0) {
            return 0;
        }

        int n = prices.length;
        int[] buy = new int[n];
        int[] sell = new int[n];

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

        for (int i = 1; i < n; i++) {
            buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i]);
            sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i] - fee);
        }
        return Math.max(buy[n - 1], sell[n - 1]);
    }
}
```

## 3. Time & Space Complexity

**DP:** 时间复杂度O(n), 空间复杂度O(n)


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