Best Time to Buy and Sell Stock with Transaction Fee - LeetCode
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.
- code
class Solution:
def maxProfit(self, prices, fee):
n = len(prices)
if n < 2:
return 0
ans = 0
minimum = prices[0]
for i in range(1, n):
if prices[i] < minimum:
minimum = prices[i]
elif prices[i] > minimum + fee:
ans += prices[i] - fee - minimum
minimum = prices[i] - fee #if next time a certain price even is lower than prices[i] - fee we sell it, we can get more
return ans
- code
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
dp_hold, dp_not_hold = -float('inf'), 0
for stock_price in prices:
# either keep not hold, or sell out today at stock price
dp_not_hold = max(dp_not_hold, dp_hold + stock_price)
# either keep hold, or buy in today at stock price and pay transaction fee for this trade
dp_hold = max(dp_hold, dp_not_hold - stock_price - fee)
# maximum profit must be in not-hold state
return dp_not_hold
- code fee once in either place
class Solution(object):
def maxProfit(self, prices, fee):
cash, hold = 0, -prices[0]
for i in range(1, len(prices)):
cash = max(cash, hold + prices[i] - fee)
hold = max(hold, cash - prices[i])
return cash
- code
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
dp = [[0, 0] for _ in range(len(prices) + 1)]
for pos in reversed(range(len(prices))):
for bought in [True, False]:
max_profit = 0
if not bought:
# Buy stock
max_profit = max(max_profit, dp[pos + 1][True] - prices[pos] - fee)
else:
# Sell stock
max_profit = max(max_profit, dp[pos + 1][False] + prices[pos])
# Do nothing
max_profit = max(max_profit, dp[pos + 1][bought])
dp[pos][bought] = max_profit
return dp[0][False]