Find Max Profit By Buying And Selling Stock with Fees
In the given problem, we are provided with an array named "prices," where prices[i] represents the current price of a stock on day i. The task is to determine the maximum profit that can be achieved from selling the stock. There is a price that is charged for a set of buy and sell of a stock. This problem can be found in leetcode at this link .
In the greedy approach solution presented below, we iterate through the prices and see if its selling worthy or buying worthy price or nothing. We also incorporate fees as we go along.
class Solution:
def maxProfit(self, prices: List[int], fee: int) -> int:
n = len(prices)
maxProfit = 0
cost_price = float('inf')
for i in range(0,n):
# at each iteration, we check to see if the price is viable for min price so that we can buy it
# .. if not, we see if its okay to sell it on that day
if prices[i]
The above solution works and passes 97% solution in Leetcode, but I am not confident that this works well in call the cases.
Solution: Dynamic Solution
Following is a dynamic approach that beats only 57% of submission in leetcode, but is more clear.
class Solution: def maxProfit(self, prices: List[int], fee: int) -> int: n = len(prices) if n<=1: return 0 letsbuy_profit = [0]*n letssell_profit = [0]*n # since we are buying on first day, profit is negative the cost letsbuy_profit[0] = -prices[0] # if we are selling on the first day, we do not have anything to sell, so # .. initialization is working for now for i in range(1,n): # max profit today if we decide to buy letsbuy_profit[i] = max(letsbuy_profit[i-1],letssell_profit[i-1]-prices[i]) # max profit today if we decide to sell letssell_profit[i] = max(letssell_profit[i-1],letsbuy_profit[i-1]+prices[i]-fee) return letssell_profit[n-1]