Best Time to Buy and Sell Stock IV

We can solve this problem using dynamic programming. We’ll create a 2D table dp where dp[i][j] represents the maximum profit with at most i transactions up to day j. The states can be defined as follows:

dp[i][j] = max(dp[i][j-1], prices[j] + max(-prices[m] + dp[i-1][m])) for m = 0 to j-1

The above state tries to sell the stock on day j and calculates the profit with all possible buying days, and also considers the case of not making any transaction on that day.

Since we are iterating through previous days to calculate the inner maximum, the time complexity becomes quadratic. We can optimize it by keeping track of the maximum value in a variable.

Here’s the code:

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class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        n = len(prices)

        # If there are no prices or k is 0, the profit is 0
        if n == 0 or k == 0:
            return 0

        # If k is greater than n//2, then it is the same as having unlimited transactions
        if k > n // 2:
            return sum(max(prices[i + 1] - prices[i], 0) for i in range(n - 1))

        # Initialize dp table
        dp = [[0] * n for _ in range(k + 1)]

        # Iterate through transactions
        for i in range(1, k + 1):
            maxVal = -prices[0]
            for j in range(1, n):
                # Update dp value for current transaction and day
                dp[i][j] = max(dp[i][j - 1], prices[j] + maxVal)
                maxVal = max(maxVal, -prices[j] + dp[i - 1][j])

        return dp[k][n - 1]

This code will return the maximum profit with at most k transactions. For the given examples, it would return 2 and 7.

Identifying Problem Isomorphism

“Best Time to Buy and Sell Stock IV” can be mapped to “Maximum Subarray III”.

“Best Time to Buy and Sell Stock IV” is about finding the maximum profit from at most ‘k’ transactions, given a list of stock prices for each day. In this case, a transaction is defined as a complete buy and sell cycle.

“Maximum Subarray III” involves finding ‘k’ non-overlapping subarrays which have the largest sum.

The reasoning for this mapping is that both problems require you to split the input into ‘k’ parts (transactions or subarrays) to maximize a certain value (profit or sum).

“Maximum Subarray III” is simpler because it only requires finding maximum sums without the constraints of a buying and selling cycle found in “Best Time to Buy and Sell Stock IV”.

“Best Time to Buy and Sell Stock IV” is a variation of the classic dynamic programming problem about stock trading. Here are 10 problems to prepare for the concepts needed to solve this problem:

These cover how to handle different dynamic programming situations, especially when the states of the dynamic programming depend on each other in a complex way.

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

Can you help me with finding the isomorphism for this problem?

Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?

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class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        if k == 0: return 0

        dp = [[1000, 0] for _ in range(k + 1)]
        for price in prices:
            for i in range(1, k + 1):
                dp[i][0] = min(dp[i][0], price - dp[i - 1][1])
                dp[i][1] = max(dp[i][1], price - dp[i][0])

        return dp[k][1]

Best Time to Buy and Sell Stock III - This problem requires identifying the maximum profit that can be obtained by performing at most two non-overlapping transactions. This problem requires understanding and analyzing price trends, hence Financial Analysis / Optimization.

Language Agnostic Coding Drills

This problem involves dynamic programming, a strategy used to solve problems by breaking them down into smaller subproblems, and reusing solutions to the subproblems to construct solutions to the overall problem.

Understanding Problem Statement: This problem requires finding the maximum profit with at most ‘k’ transactions from the given price array.
Understanding Dynamic Programming (DP): DP is a method for solving complex problems by breaking them down into simpler subproblems and then combining the solutions of the subproblems to reach the final solution.
Understanding 2D Arrays: This solution uses a 2D array ‘dp’ where the first dimension represents the transaction count, and the second dimension represents the state of the transaction (buying or selling).
Array Initialization: Learn how to initialize an array with a given value. Here, the ‘dp’ array is initialized with 1000 for buying and 0 for selling.
Iterative Logic: Understand how to iterate over an array (the given prices).
Nested Iteration: Nested iteration is used here to loop over each transaction for each price.
Conditional Statements and Comparison: This problem uses conditional statements and comparison operations to update the buying and selling prices for each transaction.
Array Manipulation: Practice accessing and modifying elements of an array.
Returning Results: The final step of the problem is to return the maximum profit, which is the selling price after the last transaction.

The approach to solving this problem involves initializing the DP array and then iterating over the prices and transactions to find the minimum buying price and maximum selling price for each transaction. After completing all transactions, the solution (maximum profit) is returned.

Targeted Drills in Python

Let’s breakdown the problem into individual coding drills. Each drill is directly related to the final solution and can be coded separately in Python:

Understanding Problem Statement:
The problem statement is directly provided and doesn’t require a coding drill.

Understanding Dynamic Programming (DP):

Let’s practice dynamic programming on a simple problem like Fibonacci series. This will help us understand how we use recursion and memoization in dynamic programming.

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def fibonacci(n, memo = {}):
    if n in memo:
        return memo[n]
    if n <= 2:
        return 1
    memo[n] = fibonacci(n-1, memo) + fibonacci(n-2, memo)
    return memo[n]

print(fibonacci(10))  # 55

Understanding 2D Arrays:

To understand 2D arrays, let’s create a simple 2D array and perform some operations on it like reading and modifying elements.

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arr = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
print(arr[0][1])  # 2
arr[1][1] = 10
print(arr)  # [[1, 2, 3], [4, 10, 6], [7, 8, 9]]

Array Initialization:

Create a 2D array initialized with certain values.

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dp = [[1000, 0] for _ in range(5)]
print(dp)  # [[1000, 0], [1000, 0], [1000, 0], [1000, 0], [1000, 0]]

Iterative Logic:
Let’s practice iterating over a simple list of numbers.
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for num in [1, 2, 3, 4, 5]: print(num)

Nested Iteration:

Let’s practice nested iteration with a simple example.

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for i in range(1, 4):
    for j in range(1, 4):
        print(i, j)

Conditional Statements and Comparison:

Let’s practice conditional statements and comparisons with a simple example of finding the minimum and maximum of an array.

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arr = [5, 3, 2, 6, 7, 1]
min_val, max_val = float('inf'), float('-inf')
for num in arr:
    min_val = min(min_val, num)
    max_val = max(max_val, num)
print(min_val, max_val)  # 1 7

Array Manipulation:

Accessing and modifying elements of an array can be practiced with a simple list.

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arr = [1, 2, 3, 4, 5]
print(arr[2])  # 3
arr[2] = 10
print(arr)  # [1, 2, 10, 4, 5]

Returning Results:
In the context of a function, we should be able to return results. Let’s write a simple function that returns the sum of two numbers.
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def add(a, b): return a + b print(add(3, 4)) # 7


After mastering these individual drills, combining them into the final solution would be the next step. The problem is solved using dynamic programming where we keep track of the minimum buying price and maximum selling price for each transaction to eventually return the maximum profit.