The best time to buy and sell stocks includes handling charges (leetcode-714)

The best time to buy and sell stocks includes handling fees （LeetCode-714）

subject

Given an array of integers prices, among prices[i] It means the first one i Day's share price ; Integers fee Represents the handling fee for trading shares .

You can close the deal indefinitely , But you have to pay a handling fee for every transaction . If you have bought a stock , You can't buy any more stock until you sell it .

Return the maximum profit .

** Be careful ：** A deal here refers to the whole process of buying, holding and selling stocks , You only need to pay a handling fee for each transaction .

Example 1：

 Input ：prices = [1, 3, 2, 8, 4, 9], fee = 2
 Output ：8
 explain ： The maximum profit that can be achieved :  
 Buy... Here  prices[0] = 1
 Sell... Here  prices[3] = 8
 Buy... Here  prices[4] = 4
 Sell... Here  prices[5] = 9
 Total profit : ((8 - 1) - 2) + ((9 - 4) - 2) = 8

Example 2：

 Input ：prices = [1,3,7,5,10,3], fee = 3
 Output ：6

Tips ：

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

Ideas

and LeetCode-122 About the same type , As long as the handling fee is deducted when selling

Five steps

1. dp[i][ take 0 or 1] The meaning of
   • $dp[i][0]$ It means the first one $i$ Cash from holding the stock for days
   • $dp[i][1]$ It means the first one $i$ Cash from not holding the stock for days
2. The recursive formula
   • $dp[i][0]$ Can be derived from two states
     • The first $i-1$ Days holding shares , It is equal to $dp[i-1][0]$
     • The first $i$ Days to buy shares , Because you can buy many times , There may be a situation where there has been a round of trading to generate profits before , The cash will be after buying today $dp[i-1][1]-price[i]$
     • Choose the one with the most cash , That is, the greater of the two
   • $dp[i][1]$ Can be derived from two states
     • The first $i-1$ Days do not hold shares , It is equal to $dp[i-1][1]$
     • The first $i$ Days to sell shares , It is equal to $price[i]+dp[i-1][0]-fee$
     • Choose the one with the most cash , That is, the greater of the two
3. Array initialization
   • dp[0][0] It means the first one 0 Days holding shares , So it's equal to $-price[0]$
   • dp[0][1] It means the first one 0 Days do not hold shares , be equal to 0
4. traversal order
   • Before and after

Code display

class Solution
{
    
public:
    int maxProfit(vector<int> &prices, int fee)
    {
    
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(2));
        dp[0][0] = -prices[0];
        dp[0][1] = 0;
        for (int i = 1; i < n; i++)
        {
    
            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);
        }
        return dp[n - 1][1];
    }
};