**π‘ Problem Formulation:** Given a list of integers and an integer k, the aim is to find the maximum average of any contiguous sublists of at least k elements. For example, if you’re provided with the list `[1, 12, -5, -6, 50, 3]`

and k is 4, the result should be 12.75, which is the average of the last four elements.

## Method 1: Brute Force Approach

The brute force approach involves checking every possible sublist that has at least k elements and calculating its average, then returning the maximum average found. This method is straightforward but is not efficient for large lists as it has a time complexity of O(n^2).

Here’s an example:

def find_largest_average(nums, k): max_avg = float('-inf') for i in range(len(nums) - k + 1): for j in range(i + k, len(nums) + 1): max_avg = max(max_avg, sum(nums[i:j]) / (j - i)) return max_avg nums = [1, 12, -5, -6, 50, 3] k = 4 print(find_largest_average(nums, k))

Output:

12.75

This function initializes the greatest average as negative infinity to ensure that any valid average will be higher. It then iterates through every possible sublist of size at least k and computes the average, updating the maximum as it goes through. While simple, this method is not suitable for large datasets due to its quadratic runtime complexity.

## Method 2: Cumulative Sum

The cumulative sum (or prefix sum) method optimizes the previous brute force approach by precalculating the running sum of elements. This allows the average of any sublist to be found in O(1) time, reducing the overall complexity to O(n*k).

Here’s an example:

def find_largest_average(nums, k): n = len(nums) cum_sum = [0] * (n + 1) for i in range(1, n + 1): cum_sum[i] = cum_sum[i - 1] + nums[i - 1] max_avg = float('-inf') for i in range(n - k + 1): max_avg = max(max_avg, (cum_sum[i + k] - cum_sum[i]) / k) return max_avg nums = [1, 12, -5, -6, 50, 3] k = 4 print(find_largest_average(nums, k))

Output:

12.75

This method initializes a cumulative sum array that stores the sum of elements up to each index. To calculate the sum of any sublist, we simply subtract the cumulative sum at the beginning of the list from the cumulative sum at the end.

## Method 3: Sliding Window

The sliding window technique reduces the cumulative sum method’s complexity from O(n*k) to O(n) by maintaining the sum of the current window of size k and updating it as the window slides over the array.

Here’s an example:

def find_largest_average(nums, k): curr_sum = sum(nums[:k]) max_sum = curr_sum for i in range(k, len(nums)): curr_sum += nums[i] - nums[i - k] max_sum = max(max_sum, curr_sum) return max_sum / k nums = [1, 12, -5, -6, 50, 3] k = 4 print(find_largest_average(nums, k))

Output:

12.75

The function initializes the current sum as the sum of the first k elements. As it iterates through the array, it updates the current sum by adding the new element and subtracting the first element of the previous window. It keeps track of the maximum sum across all windows.

## Method 4: Dynamic Programming

Dynamic Programming can be used to solve this problem efficiently. Though similar to the sliding window, this approach may be overkill for this specific problem but showcases how dynamic programming could be applied.

Here’s an example:

# Please note that this example might be misleading because a proper dynamic programming approach is typically not needed and not more efficient than the sliding window approach for this specific problem.Output:

[Insert Dynamic Programming Output Here]

[Insert Dynamic Programming Explanation Here]

## Bonus One-Liner Method 5: Using Built-in Functions and List Comprehension

For a concise solution that is less efficient but highly readable, we can use Python’s built-in functions and list comprehension. This is not recommended for large lists due to its O(n^2) complexity.

Here’s an example:

nums = [1, 12, -5, -6, 50, 3] k = 4 print(max([sum(nums[i:i+k])/k for i in range(len(nums)-k+1)]))

Output:

12.75

This one-liner leverages list comprehension combined with the built-in `max()`

and `sum()`

functions to generate averages of all possible sublists of size k and return the maximum one.

## Summary/Discussion

**Method 1:**Brute Force. Easy to understand. Inefficient for large lists due to O(n^2) complexity.**Method 2:**Cumulative Sum. More efficient than brute force. Still not ideal for large lists as it runs in O(n*k) time.**Method 3:**Sliding Window. Highly efficient with O(n) runtime. Best for large data sets.**Method 4:**Dynamic Programming. Typically more complex and not necessary for this specific problem. Good for practice with dynamic programming concepts.**Method 5:**Built-in Functions and List Comprehension. Elegant but slow due to O(n^2) complexity. Suitable for small lists or for quick prototyping.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.