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Best Time to Buy and Sell Stock with Cooldown

Medium

Google

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Topics:

ArraysDynamic Programming

You are given an array prices where prices[i] is the price of a given stock on the i<sup>th</sup> day. Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times) with the following restrictions:

After you sell your stock, you cannot buy stock on the next day (i.e., cooldown one day).

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

Example 1:

Input: prices = [1,2,3,0,2]
Output: 3
Explanation: transactions = [buy, sell, cooldown, buy, sell]

Example 2:

Input: prices = [1]
Output: 0

Constraints:

1 <= prices.length <= 5000
0 <= prices[i] <= 1000

Can you implement a function to solve this problem efficiently? Explain your approach and analyze its time and space complexity. Also, consider edge cases and how your solution handles them.

Best Time to Buy and Sell Stock with Cooldown

Medium

Google

4 views

Topics:

ArraysDynamic Programming

After you sell your stock, you cannot buy stock on the next day (i.e., cooldown one day).

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

Example 1:

Input: prices = [1,2,3,0,2]
Output: 3
Explanation: transactions = [buy, sell, cooldown, buy, sell]

Example 2:

Input: prices = [1]
Output: 0

Constraints:

1 <= prices.length <= 5000
0 <= prices[i] <= 1000

Can you implement a function to solve this problem efficiently? Explain your approach and analyze its time and space complexity. Also, consider edge cases and how your solution handles them.

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 range of values for the stock prices in the input array?
Can the input array of prices be empty or null?
If no transaction is possible (i.e., no profit can be made), what should I return?
Is there a transaction fee associated with each buy or sell?
Is the cooldown period exactly one day after selling?

Brute Force Solution

Approach

The brute force method to maximize stock profit involves trying every possible combination of buying and selling stocks. This means checking every single day as a potential buy day, and then every subsequent day as a potential sell day. We must also respect the cooldown period after a sale by skipping the next day.

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

Consider each day as a potential day to buy a stock.
For each buy day, consider every following day as a potential day to sell the stock, but only if it's after the buy day.
Calculate the profit made from each of these buy-sell combinations.
Since there is a cooldown day after selling, skip the day immediately after the sell day when considering the next buy.
Continue exploring all valid buy and sell opportunities after each cooldown.
Keep track of the maximum profit found across all these buy-sell combinations, taking into account the cooldown.
Ultimately, the largest profit found by trying all combinations will be the answer.

Code Implementation

def max_profit_brute_force(prices):
    max_profit_overall = 0
    number_of_days = len(prices)

    for buy_day in range(number_of_days):
        for sell_day in range(buy_day + 1, number_of_days):
            current_profit = prices[sell_day] - prices[buy_day]

            # Ensures we skip the cooldown day
            next_buy_day = sell_day + 2

            if next_buy_day < number_of_days:
                max_profit_after_cooldown = 0
                for subsequent_buy_day in range(next_buy_day, number_of_days):
                    for subsequent_sell_day in range(subsequent_buy_day + 1, number_of_days):

                        profit_after_cooldown = prices[subsequent_sell_day] - prices[subsequent_buy_day]
                        max_profit_after_cooldown = max(max_profit_after_cooldown, profit_after_cooldown)

                current_profit += max_profit_after_cooldown

            max_profit_overall = max(max_profit_overall, current_profit)

    # Consider the scenario with no transactions
    return max(0, max_profit_overall)

Big(O) Analysis

Time Complexity

O(n³) – The algorithm considers each day as a potential buy day. For each buy day, it iterates through the remaining days as potential sell days. After each sale, it skips one cooldown day and then continues searching for more buy/sell opportunities. Because of the skipping and nested iterations through potential buy/sell days and recursing for future profits after the cooldown, the time complexity grows cubically. This process results in approximately n * n * n operations in the worst case, which simplifies to O(n³).

Space Complexity

O(1) – The provided brute force method explores all buy-sell combinations directly without using auxiliary data structures to store intermediate states or visited combinations. While it iterates through the input array, it doesn't create any extra data structures that scale with the input size. Thus, the auxiliary space used is constant, regardless of the number of prices N.

Optimal Solution

Approach

This problem is about maximizing profit from stock trading with a twist: after selling, you need to wait a day before buying again. The best approach involves tracking the maximum profit you can have at each day in different states: holding stock, not holding stock, or being in a cooldown period after selling.

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

Imagine you can be in one of three states each day: holding a stock, not holding a stock, or in a cooldown period after a sale.
For each day, figure out the maximum profit you can have in each of these three states.
If you're holding a stock today, your profit is either what it was yesterday (if you didn't buy or sell), or it's the profit from yesterday when you weren't holding anything, minus the price of the stock today (if you bought it).
If you're not holding a stock today, your profit is either what it was yesterday (if you didn't buy or sell), or it's the profit from yesterday when you were in a cooldown period (meaning you sold the day before).
If you're in a cooldown period today, your profit is the profit from yesterday when you were holding a stock, plus the price of the stock today (because you sold it).
Continue this process for each day, updating the maximum profit for each state based on the possible transitions from the previous day.
At the end, the maximum profit you can have is the larger of the profit when you are not holding any stock and the profit you have when you are in a cooldown period, since you can't be holding stock when the process ends.

Code Implementation

def best_time_to_buy_and_sell_stock_with_cooldown(prices):
    if not prices:
        return 0

    number_of_days = len(prices)

    # Initialize arrays to store maximum profits for each state.
    holding_stock = [0] * number_of_days
    not_holding_stock = [0] * number_of_days
    cooldown = [0] * number_of_days

    # Base case: For the first day,
    # either hold stock (buy) or don't.
    holding_stock[0] = -prices[0]
    not_holding_stock[0] = 0
    cooldown[0] = 0

    for day in range(1, number_of_days):
        # If holding stock today, either held yesterday,
        # or bought today.
        holding_stock[day] = max(holding_stock[day - 1],
                               not_holding_stock[day - 1] - prices[day],
                               cooldown[day-1] - prices[day])

        # If not holding stock today, either didn't hold yesterday
        # or sold yesterday and are now cooling down.
        not_holding_stock[day] = max(not_holding_stock[day - 1],
                                   cooldown[day - 1])

        # If in cooldown today, must have sold yesterday.
        cooldown[day] = holding_stock[day - 1] + prices[day]

    # The max profit is the larger of not holding stock or being in cooldown.
    return max(not_holding_stock[number_of_days - 1], cooldown[number_of_days - 1])

Big(O) Analysis

Time Complexity

O(n) – The solution iterates through the input array of prices once. For each day (price), it updates the maximum profit achievable in each of the three states: holding, not holding, and cooldown. The updates involve constant-time operations based on the previous day's state values. Therefore, the time complexity is directly proportional to the number of days (prices), which is n. The overall time complexity is O(n).

Space Complexity

O(1) – The algorithm maintains three variables for each day to store the maximum profit for the 'holding', 'not holding', and 'cooldown' states. These variables are updated iteratively and do not depend on the size of the input (N, the number of days). Therefore, the auxiliary space required is constant, regardless of the input size, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Empty prices array	Return 0 as no transaction is possible with no prices.
Prices array with only one price	Return 0 as a transaction requires buying and selling which is not possible with one price.
Prices array with all prices being the same	The DP transition should still correctly result in 0 profit, as no buying and selling will yield a profit.
Prices array with monotonically decreasing prices	The DP should correctly deduce that no profitable transaction is possible, and return 0.
Large prices array exceeding memory constraints for DP table	Optimize space complexity by using a rolling array or state compression technique to reduce memory footprint.
Integer overflow when calculating profit with extremely large prices	Use long data types to store profit to prevent overflow during calculations if prices can be very large.
Prices fluctuate wildly up and down	The dynamic programming approach correctly considers all possible buy/sell/cooldown sequences to find the optimal profit.
Input contains very large numbers, which can lead to calculation errors	Validate numerical inputs to prevent calculation errors by raising errors if numbers are very large.

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 range of values for the stock prices in the input array?
Can the input array of prices be empty or null?
If no transaction is possible (i.e., no profit can be made), what should I return?
Is there a transaction fee associated with each buy or sell?
Is the cooldown period exactly one day after selling?

Brute Force Solution

Approach

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

Consider each day as a potential day to buy a stock.
For each buy day, consider every following day as a potential day to sell the stock, but only if it's after the buy day.
Calculate the profit made from each of these buy-sell combinations.
Since there is a cooldown day after selling, skip the day immediately after the sell day when considering the next buy.
Continue exploring all valid buy and sell opportunities after each cooldown.
Keep track of the maximum profit found across all these buy-sell combinations, taking into account the cooldown.
Ultimately, the largest profit found by trying all combinations will be the answer.

Code Implementation

def max_profit_brute_force(prices):
    max_profit_overall = 0
    number_of_days = len(prices)

    for buy_day in range(number_of_days):
        for sell_day in range(buy_day + 1, number_of_days):
            current_profit = prices[sell_day] - prices[buy_day]

            # Ensures we skip the cooldown day
            next_buy_day = sell_day + 2

            if next_buy_day < number_of_days:
                max_profit_after_cooldown = 0
                for subsequent_buy_day in range(next_buy_day, number_of_days):
                    for subsequent_sell_day in range(subsequent_buy_day + 1, number_of_days):

                        profit_after_cooldown = prices[subsequent_sell_day] - prices[subsequent_buy_day]
                        max_profit_after_cooldown = max(max_profit_after_cooldown, profit_after_cooldown)

                current_profit += max_profit_after_cooldown

            max_profit_overall = max(max_profit_overall, current_profit)

    # Consider the scenario with no transactions
    return max(0, max_profit_overall)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine you can be in one of three states each day: holding a stock, not holding a stock, or in a cooldown period after a sale.
For each day, figure out the maximum profit you can have in each of these three states.
If you're holding a stock today, your profit is either what it was yesterday (if you didn't buy or sell), or it's the profit from yesterday when you weren't holding anything, minus the price of the stock today (if you bought it).
If you're not holding a stock today, your profit is either what it was yesterday (if you didn't buy or sell), or it's the profit from yesterday when you were in a cooldown period (meaning you sold the day before).
If you're in a cooldown period today, your profit is the profit from yesterday when you were holding a stock, plus the price of the stock today (because you sold it).
Continue this process for each day, updating the maximum profit for each state based on the possible transitions from the previous day.
At the end, the maximum profit you can have is the larger of the profit when you are not holding any stock and the profit you have when you are in a cooldown period, since you can't be holding stock when the process ends.

Code Implementation

def best_time_to_buy_and_sell_stock_with_cooldown(prices):
    if not prices:
        return 0

    number_of_days = len(prices)

    # Initialize arrays to store maximum profits for each state.
    holding_stock = [0] * number_of_days
    not_holding_stock = [0] * number_of_days
    cooldown = [0] * number_of_days

    # Base case: For the first day,
    # either hold stock (buy) or don't.
    holding_stock[0] = -prices[0]
    not_holding_stock[0] = 0
    cooldown[0] = 0

    for day in range(1, number_of_days):
        # If holding stock today, either held yesterday,
        # or bought today.
        holding_stock[day] = max(holding_stock[day - 1],
                               not_holding_stock[day - 1] - prices[day],
                               cooldown[day-1] - prices[day])

        # If not holding stock today, either didn't hold yesterday
        # or sold yesterday and are now cooling down.
        not_holding_stock[day] = max(not_holding_stock[day - 1],
                                   cooldown[day - 1])

        # If in cooldown today, must have sold yesterday.
        cooldown[day] = holding_stock[day - 1] + prices[day]

    # The max profit is the larger of not holding stock or being in cooldown.
    return max(not_holding_stock[number_of_days - 1], cooldown[number_of_days - 1])

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty prices array	Return 0 as no transaction is possible with no prices.
Prices array with only one price	Return 0 as a transaction requires buying and selling which is not possible with one price.
Prices array with all prices being the same	The DP transition should still correctly result in 0 profit, as no buying and selling will yield a profit.
Prices array with monotonically decreasing prices	The DP should correctly deduce that no profitable transaction is possible, and return 0.
Large prices array exceeding memory constraints for DP table	Optimize space complexity by using a rolling array or state compression technique to reduce memory footprint.
Integer overflow when calculating profit with extremely large prices	Use long data types to store profit to prevent overflow during calculations if prices can be very large.
Prices fluctuate wildly up and down	The dynamic programming approach correctly considers all possible buy/sell/cooldown sequences to find the optimal profit.
Input contains very large numbers, which can lead to calculation errors	Validate numerical inputs to prevent calculation errors by raising errors if numbers are very large.