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Best Time to Buy and Sell Stock IV

Hard

Apple

9 views

Topics:

ArraysDynamic Programming

You are given an integer array prices where prices[i] is the price of a given stock on the i<sup>th</sup> day, and an integer k.

Find the maximum profit you can achieve. You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

For example:

If k = 2 and prices = [2,4,1], the maximum profit is 2 because you can buy on day 1 (price = 2) and sell on day 2 (price = 4), resulting in a profit of 4-2 = 2.
If k = 2 and prices = [3,2,6,5,0,3], the maximum profit is 7 because you can buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Explain how to efficiently find the maximum profit you can achieve with at most k transactions.

Best Time to Buy and Sell Stock IV

Hard

Apple

9 views

Topics:

ArraysDynamic Programming

You are given an integer array prices where prices[i] is the price of a given stock on the i<sup>th</sup> day, and an integer k.

Find the maximum profit you can achieve. You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

For example:

If k = 2 and prices = [2,4,1], the maximum profit is 2 because you can buy on day 1 (price = 2) and sell on day 2 (price = 4), resulting in a profit of 4-2 = 2.
If k = 2 and prices = [3,2,6,5,0,3], the maximum profit is 7 because you can buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Explain how to efficiently find the maximum profit you can achieve with at most k transactions.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum value of `k` (the number of transactions allowed), and what happens if `k` is larger than the number of days in the `prices` array?
Can the `prices` array be empty, and if so, what should I return?
Can the prices in the `prices` array be negative, zero, or only positive?
If no transaction is profitable, should I return 0?
Are we aiming for the absolute maximum profit, or is there a specific criteria for choosing which transactions to perform if multiple transaction combinations yield the same maximum profit?

Brute Force Solution

Approach

The brute force approach to maximizing stock profits with multiple transactions means we'll explore every single possible combination of buying and selling days. We'll consider every way to make up to the allowed number of transactions and then pick the best one.

Here's how the algorithm would work step-by-step:

First, imagine we do nothing. No buying or selling at all. This is one possibility.
Next, consider all the single possible buy and sell days. Buy on the first day, sell on the second. Buy on the first, sell on the third, and so on. Buy on the second, sell on the third, and so on. Calculate the profit for each of these single transactions.
Now, consider all possible combinations of two transactions. This means trying every combination of two separate buy and sell periods. Make sure these periods don't overlap (you can't buy and sell the same stock twice at the same time).
Continue this process, considering all possible combinations of three transactions, four transactions, and so on, up to the maximum number of transactions allowed.
For each of these combinations, calculate the total profit from all the transactions.
Finally, compare the total profit of all the possible combinations (including the no-transaction case) and choose the one with the highest profit. That's the best possible outcome.

Code Implementation

def max_profit_brute_force(max_transactions, prices):
    max_profit = 0
    number_of_days = len(prices)

    def calculate_profit(transactions):
        current_profit = 0
        for buy_day, sell_day in transactions:
            current_profit += prices[sell_day] - prices[buy_day]
        return current_profit

    def find_all_combinations(current_transaction_list, start_day, transaction_count):
        nonlocal max_profit

        # Base case: No more transactions allowed.
        if transaction_count == max_transactions:
            max_profit = max(max_profit, calculate_profit(current_transaction_list))
            return

        # This allows for skipping transaction combos
        max_profit = max(max_profit, calculate_profit(current_transaction_list))

        for buy_day in range(start_day, number_of_days):
            for sell_day in range(buy_day + 1, number_of_days):
                new_transaction_list = current_transaction_list + [(buy_day, sell_day)]
                # Only explore transactions that start after the current one ends.
                find_all_combinations(new_transaction_list, sell_day + 1, transaction_count + 1)

    # Start the recursion with an empty transaction list
    find_all_combinations([], 0, 0)
    return max_profit

Big(O) Analysis

Time Complexity

O(n^(2k)) – The brute force approach involves exploring all possible combinations of up to k transactions, where k is the maximum number of allowed transactions and n is the number of days (price points). For each possible transaction, we iterate through all possible buy and sell days, leading to approximately n^2 possibilities for a single transaction. Since we are considering up to k non-overlapping transactions, the total number of combinations grows to O(n^(2k)). This is because for each of the k transactions we have to find all possible buy and sell days which amounts to iterating all the n days for k times, and for each day, we may have to iterate all the n days which amounts to iterating n days for k times, hence n^(2k).

Space Complexity

O(1) – The described brute force approach does not use any significant auxiliary space. It iterates through combinations of buy and sell days, calculating profits on the fly without storing intermediate combinations or profits in data structures that scale with the input size. The number of variables needed to keep track of the current combination and the maximum profit remains constant regardless of the number of prices (N). Therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

This problem asks for the most profit achievable with a limited number of stock trades. The clever approach involves making decisions at each price point, tracking our maximum profit after each possible buy or sell. By remembering the best state after each transaction, we avoid exploring every trade combination.

Here's how the algorithm would work step-by-step:

Consider two key states: holding a stock and not holding a stock, for each possible transaction.
If we are allowed zero transactions, our maximum profit will always be zero.
If the number of allowed transactions exceeds half the number of days (prices), then we can make a transaction on every possible price dip and increase, making it a simpler problem.
For each day and for each possible number of transactions allowed, we examine if we should buy or sell stock to increase our profit.
If we don't currently hold stock, determine if buying stock today would yield a better outcome than our current best profit.
If we currently hold stock, determine if selling stock today would yield a better outcome than our current best profit.
Keep track of the maximum possible profit after each transaction, for each allowed transaction, and on each day.
The maximum profit at the end after all allowed transactions represents the best outcome.

Code Implementation

def max_profit_with_k_transactions(max_transactions, stock_prices):
    number_of_days = len(stock_prices)

    if number_of_days <= 1:
        return 0

    # For a large number of transactions, use a simpler approach.
    if max_transactions >= number_of_days // 2:
        max_profit = 0
        for day in range(1, number_of_days):
            if stock_prices[day] > stock_prices[day - 1]:
                max_profit += stock_prices[day] - stock_prices[day - 1]
        return max_profit

    # Initialize tables to store max profits when holding and not holding stock
    dp_holding = [[float('-inf')] * (max_transactions + 1) for _ in range(number_of_days)]
    dp_not_holding = [[0] * (max_transactions + 1) for _ in range(number_of_days)]

    for transaction in range(1, max_transactions + 1):
        dp_holding[0][transaction] = -stock_prices[0]

    for day in range(1, number_of_days):
        for transaction in range(1, max_transactions + 1):
            # Decide whether to buy stock or not
            dp_holding[day][transaction] = max(
                dp_holding[day - 1][transaction],
                dp_not_holding[day - 1][transaction - 1] - stock_prices[day]
            )

            # Decide whether to sell stock or not
            dp_not_holding[day][transaction] = max(
                dp_not_holding[day - 1][transaction],
                dp_holding[day - 1][transaction] + stock_prices[day]
            )

    # Return the maximum profit with k transactions
    return dp_not_holding[number_of_days - 1][max_transactions]

Big(O) Analysis

Time Complexity

O(k*n) – The primary driver of the time complexity is the nested loop structure. The outer loop iterates 'k' times, where 'k' is the maximum number of transactions allowed. The inner loop iterates 'n' times, where 'n' is the number of prices (days). Inside the inner loop, constant-time operations are performed to update the maximum profit. Therefore, the overall time complexity is proportional to k multiplied by n, or O(k*n).

Space Complexity

O(k*n) – The algorithm uses two 2D arrays to store maximum profits, one for holding stock (hold) and one for not holding stock (not_hold). Both arrays have dimensions (k+1) x n, where k is the maximum number of transactions allowed and n is the number of prices (days). Therefore, the auxiliary space is proportional to k*n, dominating any other constant space variables. Thus, the overall space complexity is O(k*n).

Edge Cases

Case	How to Handle
Empty prices array or k=0	Return 0 profit immediately as no transactions are possible.
k is greater than prices.length / 2	Treat as unlimited transactions and use a simplified greedy approach for efficiency to prevent Time Limit Exceeded (TLE).
Prices array contains only one element.	Return 0 profit since buying and selling on the same day yields no profit.
Prices are monotonically increasing (always going up)	The algorithm should maximize profit by buying at the beginning and selling at the end for the allowed number of transactions or perform k single-day transactions.
Prices are monotonically decreasing (always going down)	Return 0 profit since no profitable transaction can be made.
Large k value combined with large prices array causing Integer Overflow	Use long data types to prevent integer overflow in calculations of profit.
Prices array contains duplicate values repeatedly.	The algorithm should handle this and choose the optimal buy/sell points to maximize profit even if prices are the same for multiple days.
Negative prices or prices containing very large values	The problem statement constraints should explicitly disallow this, but a check should be put in place to validate and return an error or adjust to zero if encountered

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum value of `k` (the number of transactions allowed), and what happens if `k` is larger than the number of days in the `prices` array?
Can the `prices` array be empty, and if so, what should I return?
Can the prices in the `prices` array be negative, zero, or only positive?
If no transaction is profitable, should I return 0?
Are we aiming for the absolute maximum profit, or is there a specific criteria for choosing which transactions to perform if multiple transaction combinations yield the same maximum profit?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

First, imagine we do nothing. No buying or selling at all. This is one possibility.
Next, consider all the single possible buy and sell days. Buy on the first day, sell on the second. Buy on the first, sell on the third, and so on. Buy on the second, sell on the third, and so on. Calculate the profit for each of these single transactions.
Now, consider all possible combinations of two transactions. This means trying every combination of two separate buy and sell periods. Make sure these periods don't overlap (you can't buy and sell the same stock twice at the same time).
Continue this process, considering all possible combinations of three transactions, four transactions, and so on, up to the maximum number of transactions allowed.
For each of these combinations, calculate the total profit from all the transactions.
Finally, compare the total profit of all the possible combinations (including the no-transaction case) and choose the one with the highest profit. That's the best possible outcome.

Code Implementation

def max_profit_brute_force(max_transactions, prices):
    max_profit = 0
    number_of_days = len(prices)

    def calculate_profit(transactions):
        current_profit = 0
        for buy_day, sell_day in transactions:
            current_profit += prices[sell_day] - prices[buy_day]
        return current_profit

    def find_all_combinations(current_transaction_list, start_day, transaction_count):
        nonlocal max_profit

        # Base case: No more transactions allowed.
        if transaction_count == max_transactions:
            max_profit = max(max_profit, calculate_profit(current_transaction_list))
            return

        # This allows for skipping transaction combos
        max_profit = max(max_profit, calculate_profit(current_transaction_list))

        for buy_day in range(start_day, number_of_days):
            for sell_day in range(buy_day + 1, number_of_days):
                new_transaction_list = current_transaction_list + [(buy_day, sell_day)]
                # Only explore transactions that start after the current one ends.
                find_all_combinations(new_transaction_list, sell_day + 1, transaction_count + 1)

    # Start the recursion with an empty transaction list
    find_all_combinations([], 0, 0)
    return max_profit

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Consider two key states: holding a stock and not holding a stock, for each possible transaction.
If we are allowed zero transactions, our maximum profit will always be zero.
If the number of allowed transactions exceeds half the number of days (prices), then we can make a transaction on every possible price dip and increase, making it a simpler problem.
For each day and for each possible number of transactions allowed, we examine if we should buy or sell stock to increase our profit.
If we don't currently hold stock, determine if buying stock today would yield a better outcome than our current best profit.
If we currently hold stock, determine if selling stock today would yield a better outcome than our current best profit.
Keep track of the maximum possible profit after each transaction, for each allowed transaction, and on each day.
The maximum profit at the end after all allowed transactions represents the best outcome.

Code Implementation

def max_profit_with_k_transactions(max_transactions, stock_prices):
    number_of_days = len(stock_prices)

    if number_of_days <= 1:
        return 0

    # For a large number of transactions, use a simpler approach.
    if max_transactions >= number_of_days // 2:
        max_profit = 0
        for day in range(1, number_of_days):
            if stock_prices[day] > stock_prices[day - 1]:
                max_profit += stock_prices[day] - stock_prices[day - 1]
        return max_profit

    # Initialize tables to store max profits when holding and not holding stock
    dp_holding = [[float('-inf')] * (max_transactions + 1) for _ in range(number_of_days)]
    dp_not_holding = [[0] * (max_transactions + 1) for _ in range(number_of_days)]

    for transaction in range(1, max_transactions + 1):
        dp_holding[0][transaction] = -stock_prices[0]

    for day in range(1, number_of_days):
        for transaction in range(1, max_transactions + 1):
            # Decide whether to buy stock or not
            dp_holding[day][transaction] = max(
                dp_holding[day - 1][transaction],
                dp_not_holding[day - 1][transaction - 1] - stock_prices[day]
            )

            # Decide whether to sell stock or not
            dp_not_holding[day][transaction] = max(
                dp_not_holding[day - 1][transaction],
                dp_holding[day - 1][transaction] + stock_prices[day]
            )

    # Return the maximum profit with k transactions
    return dp_not_holding[number_of_days - 1][max_transactions]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty prices array or k=0	Return 0 profit immediately as no transactions are possible.
k is greater than prices.length / 2	Treat as unlimited transactions and use a simplified greedy approach for efficiency to prevent Time Limit Exceeded (TLE).
Prices array contains only one element.	Return 0 profit since buying and selling on the same day yields no profit.
Prices are monotonically increasing (always going up)	The algorithm should maximize profit by buying at the beginning and selling at the end for the allowed number of transactions or perform k single-day transactions.
Prices are monotonically decreasing (always going down)	Return 0 profit since no profitable transaction can be made.
Large k value combined with large prices array causing Integer Overflow	Use long data types to prevent integer overflow in calculations of profit.
Prices array contains duplicate values repeatedly.	The algorithm should handle this and choose the optimal buy/sell points to maximize profit even if prices are the same for multiple days.
Negative prices or prices containing very large values	The problem statement constraints should explicitly disallow this, but a check should be put in place to validate and return an error or adjust to zero if encountered