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/

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] = 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.

Thoughts

max问题，且每天的状态是从前面一天跳转过来的，用DP。action分别为buy和sell。难点在于能想到要针对action分别维持两个不同类型的状态。buy[i] = max(sell[i - 2] - price, buy(i - 1)), 表示前i天的任意序列中以buy为结尾的最大利润，可以分别是前面i-1天中买的，也可以是在第i天买的。sell[i] = max(buy(i - 1) + price - fee, sell(i - 1)), 表示前i天的任意序列中以sell为结尾的最大利润。同理可以是当天卖出，也可以是前面卖出。全局最优肯定不能以持有结束，只会是sell[N-1]。

Code

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        const int N = prices.size(); 
        vector<int> buy(N), sell(N);
        buy[0] = -prices[0];
        for (int i = 1; i < N; ++i) {
            buy[i] = max(buy[i - 1], sell[i - 1] - prices[i]);
            sell[i] = max(sell[i - 1], buy[i - 1] + prices[i] - fee);
        }
        return sell[N - 1];
    }
};

Analysis

Errors:

1. 不是buy和sell都有fee, 两者选一个加入fee即可.

时间复杂度O(N), 空间可以优化成O(1).