-
Notifications
You must be signed in to change notification settings - Fork 0
/
LC_Best_Time_to_Buy_and_Sell_Stock_with_Fee.py
44 lines (38 loc) · 1.5 KB
/
LC_Best_Time_to_Buy_and_Sell_Stock_with_Fee.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
"""
714: Best Time to Buy and Sell Stock with Transaction Fee
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.
Note:
0 < prices.length <= 50000.
0 < prices[i] < 50000.
0 <= fee < 50000.
"""
class Solution(object):
def maxProfit(self, prices, fee):
"""
:type prices: List[int]
:type fee: int
:rtype: int
"""
# At every i+1 th day, we either sell or dont sell.
# At the end we return cash
cash, hold = 0, -prices[0]
for i in range(1,len(prices)):
# hold - holds the -buyprice so hold+prices[i] will give us the profit
# subtract the fees.
# sell stock
cash = max(cash, hold + prices[i] - fee)
# day one we dont have any cash. so it will be -prices
# buy stock
hold = max(hold, cash - prices[i])
return cash