**π‘ Problem Formulation:** This article aims to explore different methods to find the number of sublists (continuous sequences within a list) that sum up to a specified target value in Python. For example, given a list `[1, 2, 3, 4]`

and a target sum `5`

, the output would be `2`

, since there are two sublists that sum up to 5: `[2, 3]`

and `[1, 4]`

.

## Method 1: Brute Force Approach

The brute force approach involves generating all possible sublists and checking their sums. This method is straightforward but can become prohibitively slow for large lists as it runs in O(n^2) time, where n is the list size.

Here’s an example:

def count_sublists_with_target_sum(nums, target): count = 0 for i in range(len(nums)): for j in range(i, len(nums)): if sum(nums[i:j+1]) == target: count += 1 return count # Example use case print(count_sublists_with_target_sum([1, 2, 3, 4], 5))

Output: `2`

This snippet creates a function `count_sublists_with_target_sum`

that iterates through all sublists using two nested loops and checks if the sublist sum equals the target. It increments a counter each time it finds a matching sublist.

## Method 2: Cumulative Sum

Using cumulative sums, we first create an auxiliary array that holds the sum of elements up to each index. This optimization allows for checking the sum of any sublist in constant time. This method is more efficient than brute force with a time complexity of O(n^2).

Here’s an example:

def count_sublists_with_target_sum(nums, target): count = 0 cum_sum = [0] * (len(nums) + 1) for i in range(1, len(cum_sum)): cum_sum[i] = cum_sum[i-1] + nums[i-1] for start in range(len(nums)): for end in range(start, len(nums)): if cum_sum[end + 1] - cum_sum[start] == target: count += 1 return count # Example use case print(count_sublists_with_target_sum([1, 2, 3, 4], 5))

Output: `2`

The function establishes a list of cumulative sums which significantly reduces the number of operations required to calculate the sum of any given sublist, then counts the number of sublists that meet the target sum.

## Method 3: Hash Map for Cumulative Sums

A more advanced technique involves using a hash map (dictionary in Python) to store the cumulative sums. This yields an efficient algorithm that can potentially reach O(n) time complexity, since we only need a single traversal of the list to calculate sums and check for the target.

Here’s an example:

def count_sublists_with_target_sum(nums, target): count, cum_sum = 0, 0 sum_dict = {0: 1} for num in nums: cum_sum += num count += sum_dict.get(cum_sum - target, 0) sum_dict[cum_sum] = sum_dict.get(cum_sum, 0) + 1 return count # Example use case print(count_sublists_with_target_sum([1, 2, 3, 4], 5))

Output: `2`

This function keeps track of cumulative sums and the frequency of each sum using a hash map. It updates the count each time it finds that current cumulative sum minus target sum has already been encountered.

## Method 4: Sliding Window Technique

For a list of positive numbers, the sliding window technique can be employed. It uses two pointers to create a window which can be adjusted based on the current sum. This method offers O(n) complexity as it only requires a single pass through the list.

Here’s an example:

def count_sublists_with_target_sum(nums, target): count, current_sum = 0, 0 start = 0 for end in range(len(nums)): current_sum += nums[end] while current_sum > target and start <= end: current_sum -= nums[start] start += 1 if current_sum == target: count += 1 return count # Example use case print(count_sublists_with_target_sum([1, 2, 3, 4], 5))

Output: `2`

Suitable for positive integers, the sliding window technique maintains a window sum and slides the window along the list, adjusting its size to find sublists that sum up to the target value.

## Bonus One-Liner Method 5: Using itertools

This one-liner approach utilizes Python’s itertools library to generate all possible sublists using combinations. It is an elegant solution but not recommended for large lists due to its O(2^n) complexity.

Here’s an example:

from itertools import combinations def count_sublists_with_target_sum(nums, target): return sum(1 for i in range(len(nums) + 1) for j in combinations(nums, i) if sum(j) == target) # Example use case print(count_sublists_with_target_sum([1, 2, 3, 4], 5))

Output: `2`

This creative use of itertools.combinations iterates through all the possible sublists of each length and counts those whose sums match the target.

## Summary/Discussion

**Method 1:**Brute Force. Simple to understand and implement. Not suitable for large lists due to O(n^2) complexity.**Method 2:**Cumulative Sum. Saves time compared to brute force by using an auxiliary array. Still O(n^2) complexity but often faster in practice.**Method 3:**Hash Map for Cumulative Sums. Highly efficient with best-case O(n) complexity. May require additional memory for the hash map.**Method 4:**Sliding Window Technique. Offers O(n) complexity, but only works for lists of positive numbers.**Method 5:**Using itertools. Elegant one-liner but has exponential time complexity and is thus impractical for large lists.