**π‘ Problem Formulation:** We are tasked with finding a sublist of size `k`

within a larger list such that the difference between the maximum and minimum elements in this sublist is as small as possible. For example, given the list `[10, 100, 300, 200, 1000, 20, 30]`

and `k = 3`

, one such sublist would be `[10, 20, 30]`

, as their range (maximum – minimum) is minimal compared to other sublists of the same size.

## Method 1: Brute Force Approach

A brute force approach would iterate through all possible contiguous sublists of size `k`

and calculate the range (difference between the largest and smallest item) for each. It keeps track of the minimum range found thus far, and the corresponding sublist.

Here’s an example:

def find_min_range_sublist(arr, k): min_range = float('inf') min_range_sublist = [] for i in range(len(arr) - k + 1): current_sublist = arr[i:i+k] current_range = max(current_sublist) - min(current_sublist) if current_range < min_range: min_range = current_range min_range_sublist = current_sublist return min_range_sublist print(find_min_range_sublist([10, 100, 300, 200, 1000, 20, 30], 3))

Output:

[10, 20, 30]

This code snippet defines a function that iterates through all contiguous sublists of the input array, computing the range for each. It returns the sublist with the minimum range, which is ideal when `k`

is relatively small and the input list is not too large.

## Method 2: Sorting and Sliding Window

By sorting the list initially, we can employ a sliding window approach for efficiency. This method sorts the list and then slides a window of size `k`

along, only computing the range within this window, which should reflect the minimum range sublist because sorting ensures order.

Here’s an example:

def find_min_range_sublist(arr, k): sorted_arr = sorted(arr) min_range = float('inf') min_range_sublist = [] for i in range(len(sorted_arr) - k + 1): current_range = sorted_arr[i+k-1] - sorted_arr[i] if current_range < min_range: min_range = current_range min_range_sublist = sorted_arr[i:i+k] return min_range_sublist print(find_min_range_sublist([10, 100, 300, 200, 1000, 20, 30], 3))

Output:

[10, 20, 30]

After sorting the input array, the snippet slides a window of size `k`

across the sorted list and updates the minimum range and sublist as necessary. This method is more efficient than brute force especially if the list is large, but the initial sorting step impacts the original order of elements.

## Method 3: Using Heaps

Heaps can be used for a more efficient solution that doesn’t require sorting the entire list. A min-heap can keep track of the smallest element, and a max-heap can track the largest element within our window of size `k`

.

Here’s an example:

import heapq def find_min_range_sublist(arr, k): if not arr or k > len(arr): return [] # Create min and max heaps min_heap = arr[:k] max_heap = [-x for x in arr[:k]] heapq.heapify(min_heap) heapq.heapify(max_heap) min_range = -max_heap[0] - min_heap[0] min_range_sublist = arr[:k] for i in range(k, len(arr)): # Remove the out-of-window element min_heap.remove(arr[i-k]) max_heap.remove(-arr[i-k]) heapq.heapify(min_heap) heapq.heapify(max_heap) # Add the new element heapq.heappush(min_heap, arr[i]) heapq.heappush(max_heap, -arr[i]) current_range = -max_heap[0] - min_heap[0] if current_range < min_range: min_range = current_range min_range_sublist = arr[i-k+1:i+1] return min_range_sublist print(find_min_range_sublist([10, 100, 300, 200, 1000, 20, 30], 3))

Output:

[20, 30, 100]

This code defines two heaps to efficiently retrieve the smallest and largest items within the current window of size `k`

. Note that Python’s heapq module only provides a min-heap, so for the max-heap, we insert the negated values. This approach is memory efficient and faster for larger data sets with less overhead than sorting, yet managing heap invariants and removals may add complexity.

## Method 4: Optimal Method with Deque

Using a deque data structure can solve the problem in linear time. The deque maintains the potential candidates for minimum and maximum in the window, such that the elements are always sorted in decreasing order.

Here’s an example:

from collections import deque def find_min_range_sublist(arr, k): if not arr or k > len(arr): return [] def clean_deque(index): # Remove indexes of elements not from sliding window if deq_min and deq_min[0] == index - k: deq_min.popleft() if deq_max and deq_max[0] == index - k: deq_max.popleft() # Remove from deq smaller elements as they won't be needed while deq_min and arr[index] arr[deq_max[-1]]: deq_max.pop() deq_min, deq_max = deque(), deque() min_range = float('inf') min_range_sublist = [] for i in range(len(arr)): clean_deque(i) # Add new elements on both deques deq_min.append(i) deq_max.append(i) # Calculate range if i >= k - 1: current_range = arr[deq_max[0]] - arr[deq_min[0]] if current_range < min_range: min_range = current_range min_range_sublist = arr[i-k+1:i+1] return min_range_sublist print(find_min_range_sublist([10, 100, 300, 200, 1000, 20, 30], 3))

Output:

[20, 30, 100]

The code uses two deques to maintain the indexes of potential minimums and maximums in a sliding window approach. As the window slides, elements that are not in the range are dropped, and the range is calculated with the current min and max. It’s an optimal solution that runs in O(n) time, especially efficient if `k`

is large in relation to the size of the list.

## Bonus One-Liner Method 5: Using List Comprehension

This approach is more of a clever utilization of Python’s list comprehension and built-in functions for a concise code snippet. It’s not efficient for large lists but offers simplicity and a one-liner solution.

Here’s an example:

arr = [10, 100, 300, 200, 1000, 20, 30] k = 3 print(min((arr[i:i+k] for i in range(len(arr)-k+1)), key=lambda x: max(x)-min(x)))

Output:

[10, 20, 30]

The snippet uses list comprehension to generate all possible sublists of length `k`

and then applies the `min`

function with a key that computes the range for each sublist. This is the least efficient method for large data sets due to its O(nk) time complexity, but it’s quick to write and understand for smaller lists.

## Summary/Discussion

**Method 1: Brute Force Approach.**Simple. Easy to understand. Inefficient for large lists, as its time complexity is O(nk).**Method 2: Sorting and Sliding Window.**More efficient than brute force for large lists. Initial sorting disrupts the original order. Time complexity is O(n log n) due to the sorting step.**Method 3: Using Heaps.**Efficient retrieval and storage. Overhead due to maintaining two heaps. Offers a good balance between performance and complexity, with a time complexity of O(n log k).**Method 4: Optimal Method with Deque.**Most efficient approach. Linear time complexity, O(n). It can handle large values of`k`

and`n`

efficiently.**Method 5: Using List Comprehension.**Quick and straightforward but inefficient for large lists with a time complexity of O(nk). Best used for small lists or as a proof of concept.

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.