#### Approach #1: Dynamic Programming [Accepted]

Intuition and Algorithm

At the end of the i-th day, we maintain cash, the maximum profit we could have if we did not have a share of stock, and hold, the maximum profit we could have if we owned a share of stock.

To transition from the i-th day to the i+1-th day, we either sell our stock cash = max(cash, hold + prices[i] - fee) or buy a stock hold = max(hold, cash - prices[i]). At the end, we want to return cash. We can transform cash first without using temporary variables because selling and buying on the same day can't be better than just continuing to hold the stock.

Python

class Solution(object):
def maxProfit(self, prices, fee):
cash, hold = 0, -prices
for i in range(1, len(prices)):
cash = max(cash, hold + prices[i] - fee)
hold = max(hold, cash - prices[i])
return cash


Java

class Solution {
public int maxProfit(int[] prices, int fee) {
int cash = 0, hold = -prices;
for (int i = 1; i < prices.length; i++) {
cash = Math.max(cash, hold + prices[i] - fee);
hold = Math.max(hold, cash - prices[i]);
}
return cash;
}
}


Complexity Analysis

• Time Complexity: , where is the number of prices.

• Space Complexity: , the space used by cash and hold.

Analysis written by: @awice.