What is the brute force solution for Best Time to Buy and Sell Stock with Transaction Fee?

The brute force approach for Best Time to Buy and Sell Stock with Transaction Fee is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible buy and sell combinations to find the maximum profit. This method is straightforward but inefficient, as it examines every possible transaction. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Best Time to Buy and Sell Stock with Transaction Fee?

Best Time to Buy and Sell Stock with Transaction Fee uses the following concepts: Array, Dynamic Programming, Greedy. The recommended patterns to study are: Dynamic Programming, Greedy.

What are the interview tips for Best Time to Buy and Sell Stock with Transaction Fee?

Clearly explain your thought process as you work through the problem. Consider edge cases, such as when prices are always decreasing. Practice explaining your solution to someone else to reinforce your understanding.

What are common mistakes when solving Best Time to Buy and Sell Stock with Transaction Fee?

Students often forget to account for the transaction fee when calculating profits. Not considering the case where no transactions yield a profit.

#714

Best Time to Buy and Sell Stock with Transaction Fee

Medium

ArrayDynamic ProgrammingGreedyDynamic ProgrammingGreedy

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses dynamic programming to keep track of two states: the maximum profit when holding a stock and when not holding a stock. This allows us to efficiently calculate the maximum profit without redundant calculations.

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Algorithm

5 steps

1Step 1: Initialize two variables: cash (profit when not holding a stock) and hold (profit when holding a stock). Set cash to 0 and hold to negative infinity.
2Step 2: Iterate through each price in the prices array.
3Step 3: Update the cash to be the maximum of its current value and hold plus the current price minus the fee (selling the stock).
4Step 4: Update the hold to be the maximum of its current value and cash minus the current price (buying the stock).
5Step 5: After iterating through all prices, cash will contain the maximum profit.

solution.py12 lines

1# Full working Python code
2prices = [1, 3, 2, 8, 4, 9]
3fee = 2
4
5def maxProfit(prices, fee):
6    cash, hold = 0, float('-inf')
7    for price in prices:
8        cash = max(cash, hold + price - fee)
9        hold = max(hold, cash - price)
10    return cash
11
12print(maxProfit(prices, fee))

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Complexity note: The time complexity is O(n) because we only iterate through the prices array once. The space complexity is O(1) since we are using a constant amount of space for the cash and hold variables.

1Understanding the difference between holding and not holding stock is crucial.
2Dynamic programming allows us to avoid redundant calculations.

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