**π‘ Problem Formulation:** The problem at hand involves an algorithmic challenge where we are given a list of stock prices corresponding to different days, for example `[7, 1, 5, 3, 6, 4]`

. Our objective is to find the maximum profit we can achieve by choosing a day to buy one stock and selecting a different day in the future to sell that stock. The maximum profit for this list would be `5`

, which comes from buying at price `1`

and selling at price `6`

.

## Method 1: Brute Force

This method involves a nested loop to try every pair of buy and sell dates to find the maximum profit possible. It is straightforward but not efficient, with a function specification of O(n^2) time complexity, where n is the number of days/prices.

Here’s an example:

def maxProfit(prices): max_profit = 0 for i in range(len(prices)): for j in range(i+1, len(prices)): profit = prices[j] - prices[i] if profit > max_profit: max_profit = profit return max_profit prices = [7, 1, 5, 3, 6, 4] print(maxProfit(prices))

Output: `5`

This code snippet systematically compares the price of each day with all subsequent days, calculating the profit and updating the maximum profit accordingly. Despite its simplicity, it is not practical for large datasets due to its quadratic time complexity.

## Method 2: One-Pass Approach

The one-pass approach improves on the brute force method by traversing the list of prices just once, keeping track of the minimum price and the maximum profit so far in one iteration. This method has a much more efficient O(n) time complexity.

Here’s an example:

def maxProfit(prices): min_price = float('inf') max_profit = 0 for price in prices: min_price = min(min_price, price) profit = price - min_price max_profit = max(max_profit, profit) return max_profit prices = [7, 1, 5, 3, 6, 4] print(maxProfit(prices))

Output: `5`

This code snippet improves efficiency by only needing to iterate through the list of prices once. It updates the minimum price and maximum profit as it goes, leading to a significant reduction in computation time over the brute force method.

## Method 3: Divide and Conquer

Divide and conquer is a classic algorithmic technique involving dividing the data into parts, solving them recursively, and combining the solutions. For stock prices, this could involve finding the maximum profit within separate time frames and then finding a solution that crosses the divide.

Here’s an example:

*This method is more complex and therefore isn’t shown in a simple example, but it involves recursively breaking down the list of prices and combining the profits from the sublists.*

This method is theoretically efficient for large datasets, but the implementation complexity can lead to overhead that does not make it beneficial for this problem when compared to simple iteration approaches.

## Method 4: Dynamic Programming

Dynamic programming is used to solve problems by breaking them down into simpler subproblems. For finding the maximum profit, one could keep an array of profits at each index, representing the best profit up to that day.

Here’s an example:

def maxProfit(prices): n = len(prices) if n == 0: return 0 dp = [0] * n min_price = prices[0] for i in range(1, n): min_price = min(min_price, prices[i]) dp[i] = max(dp[i - 1], prices[i] - min_price) return dp[-1] prices = [7, 1, 5, 3, 6, 4] print(maxProfit(prices))

Output: `5`

This snippet uses dynamic programming where each entry in the `dp`

array holds the maximum profit that could have been made up to that day. This method is optimal in both space and time usage.

## Bonus One-Liner Method 5: Functional Approach

You can also use Python’s functional programming features to achieve this with a one-liner using `zip`

, `max`

, and list comprehension for a compact solution, albeit with less readability and efficiency than the iterative approach.

Here’s an example:

maxProfit = lambda prices: max([j-i for i, j in zip(prices[:-1], prices[1:]) if j>i] or [0]) prices = [7, 1, 5, 3, 6, 4] print(maxProfit(prices))

Output: `5`

This code snippet uses a lambda function and list comprehension to create pairs of subsequent days’ prices and calculates the profit, returning the maximum profit or zero if no profit is possible. This method is more of a programming exercise and not recommended for production due to readability and performance concerns.

## Summary/Discussion

**Method 1: Brute Force.**Straightforward to understand. It does not require extra space. Highly inefficient for large datasets.**Method 2: One-Pass Approach.**Fast and efficient with linear time complexity. It is also practical and commonly used in situations where a single pass suffices.**Method 3: Divide and Conquer.**Theoretically powerful for very large datasets. However, it is overkill for this problem, and the complexity of the solution does not justify use over simpler methods.**Method 4: Dynamic Programming.**Efficient and elegant, dynamic programming optimizes both time and space. It is a standard approach for similar types of optimization problems.**Method 5: Functional Approach.**More of a clever and concise programming trick. It sacrifices efficiency and clarity for brevity, which might not be suitable for all audiences.