5 Best Ways to Find the Minimum Number of Days to Wait to Make Profit in Python

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π‘ Problem Formulation: In the realm of stock trading, investors often need to determine the optimal time to sell their stocks for maximum profit. This article illustrates various Python programs that can compute the minimum number of days one must wait before selling stocks to make a profit. Given an array of daily stock prices, the output is the least number of days to wait to sell the stock for more than the purchase price.

Method 1: Brute Force Approach

Using the brute force approach, we iterate through each day’s stock price, looking forward to find a future day with a higher price to sell. This can be inefficient for larger datasets, but it’s conceptually simple and easy to implement.

Here’s an example:

```prices = [7, 1, 5, 3, 6, 4]

def minimum_days_to_profit(prices):
min_days = float('inf')
for i in range(len(prices)-1):
for j in range(i+1, len(prices)):
if prices[j] > prices[i] and (j-i) < min_days:
min_days = j-i
return min_days if min_days != float('inf') else -1

print(minimum_days_to_profit(prices))```

Output of this code snippet:

`1`

The function `minimum_days_to_profit` takes a list of stock prices and calculates the minimum number of days needed to make a profit through nested loops. If no profit can be made, it returns -1. This method is straightforward but may be too slow for long lists of prices.

Method 2: Optimized Search with Peak and Valley Approach

This method involves finding the next valley (local minimum) and the following peak (local maximum), calculating the time between them. It’s more efficient than the brute force method since it avoids unnecessary comparisons.

Here’s an example:

```prices = [7, 1, 5, 3, 6, 4]

def minimum_days_to_profit(prices):
for i in range(1, len(prices)):
if prices[i] > prices[i-1]:
return i
return -1

print(minimum_days_to_profit(prices))```

Output of this code snippet:

`1`

In this snippet, the `minimum_days_to_profit` function scans through the list just once, comparing adjacent days to find the first opportunity to make a profit, thus improving efficiency. The time complexity is better, however it’s still not optimal for finding the global maximum profit.

Method 3: Dynamic Programming Approach

Dynamic programming can be used to solve this problem by keeping track of the minimum price so far and the day on which it occurred, then determining the earliest day after the minimum when a profit can be made.

Here’s an example:

```prices = [7, 1, 5, 3, 6, 4]

def minimum_days_to_profit(prices):
min_price = prices[0]
min_day = 0
for i in range(1, len(prices)):
if prices[i]  0:
return i - min_day
return -1

print(minimum_days_to_profit(prices))```

Output of this code snippet:

`1`

The `minimum_days_to_profit` function implements a dynamic programming approach to keep track of the lowest price and its day index. It finds the earliest day to sell and make a profit after that low point. This is efficient for both time and space complexities.

Method 4: Divide and Conquer Approach

This approach divides the stock price data into two halves recursively to find the minimum waiting period to make a profit. It combines the results from each half to find the global solution. More complex to understand and implement but works well on large datasets.

Here’s an example:

`# Method 4 is typically too complex to illustrate in a short snippet but would involve recursively breaking down the array and combining the solutions.`

This method is theoretical for the context of this problem and generally not implemented due to its complexity over other more straightforward approaches.

Bonus One-Liner Method 5: Using List Comprehensions and min Function

The one-liner solution leverages Python’s list comprehensions and built-in functions like `min` to find the minimum waiting days to make profit in an elegant and compact manner. However, it’s not always the most readable or efficient.

Here’s an example:

```prices = [7, 1, 5, 3, 6, 4]

print(min([j-i for i in range(len(prices)) for j in range(i+1, len(prices)) if prices[j] > prices[i]], default=-1))```

Output of this code snippet:

`1`

This one-liner uses a list comprehension to generate a list of distances (days) between buying and selling days that yield a profit, then finds the minimum of those distances. The `default=-1` handles the case where no profit is possible.

Summary/Discussion

• Method 1: Brute Force. Easy to understand but inefficient for large datasets. Time complexity is O(n^2).
• Method 2: Peak and Valley. Faster than brute force and simple, but does not always find the global maximum profit. Time complexity is O(n).
• Method 3: Dynamic Programming. Optimal for this problem, it’s efficient and provides the correct minimum day profit in O(n) time complexity.
• Method 4: Divide and Conquer. Best for distributed systems and large datasets, but overkill for this problem. Complexity varies.
• Bonus Method 5: One-Liner. Elegant and compact, but has poor readability and performance for large datasets with time complexity of O(n^2).