**π‘ Problem Formulation:** There’s a common algorithmic task where we want to find the number of contiguous sublists within a list of binary values (0s and 1s) that add up to a specific sum ‘k’. For example, given the list [1,0,1,0,1] and sum k=2, the desired output is 4, representing sublists [1,0,1], [0,1,0], and starting at the first and third indices [1,0,1]. This article discusses five methods to solve this problem using Python.

## Method 1: Brute Force

Using brute force, we iterate over all possible sublists and count those whose sum equals ‘k’. The strength of this method lies in its simplicity and straightforwardness. However, its weakness is the inefficiency, resulting from its O(n^2) time complexity, making it unsuitable for long lists.

Here’s an example:

def count_sublists_brute_force(binary_list, k): count = 0 for i in range(len(binary_list)): for j in range(i+1, len(binary_list)+1): if sum(binary_list[i:j]) == k: count += 1 return count print(count_sublists_brute_force([1,0,1,0,1], 2))

The output is `4`

.

This brute force method steps through each possible sublist, uses `sum`

to evaluate its total, and increments a counter when a sublist’s sum equals ‘k’. It’s simple but not efficient for large inputs.

## Method 2: Using Prefix Sums

Prefix sums reduce the need to repeatedly calculate the sum of overlapping sublists. The prefix sum array is first computed, after which the difference of two prefix sums is used to obtain the sum of any sublist. This improves efficiency, especially for large lists, reducing the time complexity to O(n).

Here’s an example:

def count_sublists_prefix_sum(binary_list, k): prefix_sums = {0: 1} total, count = 0, 0 for num in binary_list: total += num if total - k in prefix_sums: count += prefix_sums[total - k] prefix_sums[total] = prefix_sums.get(total, 0) + 1 return count print(count_sublists_prefix_sum([1,0,1,0,1], 2))

The output is `4`

.

This method leverages a dictionary to track prefix sums and their frequency, enabling it to calculate the number of sublists that sum to ‘k’ quickly. It is an efficient approach for finding contiguous sublists with a given sum.

## Method 3: Two Pointers Technique

Two pointers technique is applicable here since the list contains only binary values. By maintaining two pointers, we can keep track of a running sum and adjust the pointers to include or exclude elements to match the sum ‘k’. This method operates in linear time and is very efficient for binary lists.

Here’s an example:

def count_sublists_two_pointers(binary_list, k): start, end, count, running_sum = 0, 0, 0, 0 while end k and start <= end: running_sum -= binary_list[start] start += 1 if running_sum == k: count += 1 end += 1 return count print(count_sublists_two_pointers([1,0,1,0,1], 2))

The output is `4`

.

This method moves two pointers across the binary list, adjusting the running sum and count accordingly. It’s efficient due to its linear time performance and constant space usage.

## Method 4: Using Hash Table with Early Stopping

Similar to the prefix sum method, we use a hash table to store the sum frequencies but with an early stopping condition when the sum exceeds ‘k’. This is particularly useful for binary lists where the sum can only increase.

Here’s an example:

def count_sublists_hash_table(binary_list, k): sum_freq = {0: 1} total, count = 0, 0 for i in binary_list: total += i count += sum_freq.get(total - k, 0) if total > k: break sum_freq[total] = sum_freq.get(total, 0) + 1 return count print(count_sublists_hash_table([1,0,1,0,1], 2))

The output is `4`

.

Keeping track of sum frequencies using a hash table allows for fast calculations, while the early stopping condition helps prevent unnecessary iterations for binary lists.

## Bonus One-Liner Method 5: List Comprehension with Cumulative Sum

A Pythonic one-liner using list comprehension and the `itertools.accumulate`

function can solve the problem concisely. It’s not the most efficient but appeals for its simplicity and clever usage of built-in functions.

Here’s an example:

from itertools import accumulate def count_sublists_oneliner(binary_list, k): return sum(1 for i, total in enumerate(accumulate(binary_list)) if total - k in accumulate(binary_list[:i+1])) print(count_sublists_oneliner([1,0,1,0,1], 2))

The output is `4`

.

This concise solution generates cumulative sums on the fly and counts sublists where the running sum minus ‘k’ exists in previous sums. It’s an example of Python’s powerful one-liner capabilities.

## Summary/Discussion

**Method 1:**Brute Force. Simple but slow for large input lists due to its O(n^2) complexity.

**Method 2:**Prefix Sums. Efficient and fast with time complexity O(n), excellent for large lists.

**Method 3:**Two Pointers Technique. Highly efficient for binary lists, operates in linear time, and uses constant space.

**Method 4:**Hash Table with Early Stopping. Fast computation with beneficial early stopping condition, best for binary lists where the sum cannot decrease.

**Method 5:**List Comprehension with Cumulative Sum. Pythonic and concise; however, not the most efficient for performance-critical applications.