Best Time to Buy and Sell Stock with Transaction Fee

Author: anantkaushik

714-best-time-to-buy-and-sell-stock-with-transaction-fee.py

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+"""
+Problem Link: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/
+
+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.
+
+Example 2:
+Input: prices = [1,3,7,5,10,3], fee = 3
+Output: 6
+ 
+Constraints:
+1 <= prices.length <= 5 * 104
+1 <= prices[i] < 5 * 104
+0 <= fee < 5 * 104
+"""
+class Solution:
+ def maxProfit(self, prices: List[int], fee: int) -> int:
+ buy = -prices[0]
+ sell = 0
+ 
+ for i in range(1, len(prices)):
+ buy = max(buy, sell-prices[i])
+ sell = max(sell, buy + prices[i] - fee)
+ 
+ return sell
+
+# Better variable names
+class Solution:
+ def maxProfit(self, prices: List[int], fee: int) -> int:
+ 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