714. Best Time to Buy and Sell Stock with Transaction Fee (M)

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

Solution:

每次交易要支付手续费，只要把手续费从利润中减去即可。改写方程：

dp[i][0] = max(dp[i-1][0], dp[i-1][1] + prices[i])
dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i] - fee)
解释：相当于买入股票的价格升高了。
在第一个式子里减也是一样的，相当于卖出股票的价格减小了。

如果直接把 fee 放在第一个式子里减，会有测试用例无法通过，错误原因是整型溢出而不是思路问题。一种解决方案是把代码中的 int 类型都改成 long 类型，避免 int 的整型溢出。

直接翻译成代码，注意状态转移方程改变后 base case 也要做出对应改变：

// 原始版本
int maxProfit_with_fee(int[] prices, int fee) {
    int n = prices.length;
    int[][] dp = new int[n][2];
    for (int i = 0; i < n; i++) {
        if (i - 1 == -1) {
            // base case
            dp[i][0] = 0;
            dp[i][1] = -prices[i] - fee;
            //   dp[i][1]
            // = max(dp[i - 1][1], dp[i - 1][0] - prices[i] - fee)
            // = max(dp[-1][1], dp[-1][0] - prices[i] - fee)
            // = max(-inf, 0 - prices[i] - fee)
            // = -prices[i] - fee
            continue;
        }
        dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] + prices[i]);
        dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] - prices[i] - fee);
    }
    return dp[n - 1][0];
}

// 空间复杂度优化版本
int maxProfit_with_fee(int[] prices, int fee) {
    int n = prices.length;
    int dp_i_0 = 0, dp_i_1 = Integer.MIN_VALUE;
    for (int i = 0; i < n; i++) {
        int temp = dp_i_0;
        dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i]);
        dp_i_1 = Math.max(dp_i_1, temp - prices[i] - fee);
    }
    return dp_i_0;
}