**π‘ Problem Formulation:** The task is to find the length of the smallest contiguous sublist within a list whose sum is greater than or equal to a given target value. For instance, given the list [2, 3, 1, 2, 4, 3] and a target value of 7, the desired output is 2, since the sublist [4, 3] meets the criteria and is the smallest.

## Method 1: Brute Force Approach

The brute force method involves generating all possible sublists and their sums, checking against the target value, and identifying the smallest sublist that meets the criteria. This is the most straightforward approach, albeit with a high time complexity of O(n^2).

Here’s an example:

def min_sublist_length(nums, target): n = len(nums) min_length = float('inf') for start in range(n): sum = 0 for end in range(start, n): sum += nums[end] if sum >= target: min_length = min(min_length, end - start + 1) break return min_length if min_length != float('inf') else 0 # Example usage print(min_sublist_length([2, 3, 1, 2, 4, 3], 7))

The output of this code snippet:

2

This function iterates through all possible starting and ending points of sublists, checks if their sum is at least the target, and records the length of the smallest such sublist. This method works but is inefficient for large lists due to its quadratic time complexity.

## Method 2: Sliding Window Technique

The sliding window technique efficiently handles this problem by expanding and contracting a window on the list, thus keeping a running sum. This approach reduces the time complexity significantly to O(n).

Here’s an example:

def min_sublist_length(nums, target): n = len(nums) start = 0 sum = 0 min_length = float('inf') for end in range(n): sum += nums[end] while sum >= target: min_length = min(min_length, end - start + 1) sum -= nums[start] start += 1 return min_length if min_length != float('inf') else 0 # Example usage print(min_sublist_length([2, 3, 1, 2, 4, 3], 7))

The output of this code snippet:

2

This code creates a window that moves over the list, adding elements until the target sum is reached, then contracting from the start to find the smallest window that still meets the sum requirement. It’s a more efficient solution compared to the brute force approach and works well with large lists.

## Method 3: Prefix Sum and Binary Search

This method utilizes a prefix sum array to keep a cumulative sum of elements and then applies binary search on this array to find the minuscule length sublist that exceeds the target sum, providing a balance between time and space efficiency with O(n log n) complexity.

Here’s an example:

import bisect def min_sublist_length(nums, target): n = len(nums) min_length = float('inf') prefix_sum = [0] for num in nums: prefix_sum.append(prefix_sum[-1] + num) for i in range(1, len(prefix_sum)): required = target + prefix_sum[i-1] bound = bisect.bisect_left(prefix_sum, required) if bound != len(prefix_sum): min_length = min(min_length, bound - (i - 1)) return min_length if min_length != float('inf') else 0 # Example usage print(min_sublist_length([2, 3, 1, 2, 4, 3], 7))

The output of this code snippet:

2

By using the prefix sum array and binary search, the function efficiently determines the minimum length of the sublist necessary to meet or exceed the target sum. It’s more time-efficient than the brute force approach for longer lists, though not as efficient as the sliding window technique.

## Method 4: Optimized Prefix Sum with Hashing

An optimized version of the prefix sum approach that uses a hash table to store previous sums and their corresponding indices, allowing for faster lookup times and thereby achieving an O(n) time complexity for the problem.

Here’s an example:

def min_sublist_length(nums, target): n = len(nums) min_length = float('inf') sum = 0 sum_indices = {0: -1} for i, num in enumerate(nums): sum += num if sum - target in sum_indices: min_length = min(min_length, i - sum_indices[sum - target]) sum_indices[sum] = i return min_length if min_length != float('inf') else 0 # Example usage print(min_sublist_length([2, 3, 1, 2, 4, 3], 7))

The output of this code snippet:

2

By using a hash table to keep track of the indices of the prefix sums, the function quickly checks if there’s a previous sum that, if subtracted from the current sum, would equal or exceed the target. It’s a space-optimized method that offers good performance on large lists.

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

In this bonus one-liner, Python’s itertools library is used to generate all possible contiguous sublists, which are then filtered and processed using list comprehension to find the smallest sublist meeting the target sum, although this method is more convenient than efficient.

Here’s an example:

from itertools import accumulate def min_sublist_length(nums, target): cumsums = list(accumulate(nums, initial=0)) return min((j - i for i, c in enumerate(cumsums) for j in range(i + 1, len(cumsums)) if cumsums[j] - c >= target), default=0) # Example usage print(min_sublist_length([2, 3, 1, 2, 4, 3], 7))

The output of this code snippet:

2

This clever one-liner makes use of Python’s powerful itertools library to create a list of cumulative sums, then uses list comprehension with a generator expression to find the smallest sublist that meets or exceeds the target sum. However, due to its less efficient nature, it’s best used for smaller lists or quick scripts.

## Summary/Discussion

**Method 1:**Brute Force Approach. Simple to implement, but has high time complexity and is not suitable for long lists.**Method 2:**Sliding Window Technique. Highly efficient with a linear time complexity, ideal for large lists, and easy to understand.**Method 3:**Prefix Sum and Binary Search. Balances between time and space efficiency. Suitable for medium-sized lists with better performance than brute force.**Method 4:**Optimized Prefix Sum with Hashing. Fast and space-efficient, it offers a significant improvement for large datasets and time-sensitive applications.**Method 5:**Bonus One-Liner. Aesthetically pleasing and compact, but not recommended for performance-critical solutions.