**π‘ Problem Formulation:** You have an array composed of 0s and 1s. Your task is to determine the longest subarray of 1s that can be achieved by removing at most one element from the array. For example, given input `[1, 1, 0, 1, 1]`

, the desired output after deletion is `4`

because by deleting the single 0, the longest subarray of 1s would be of length 4.

## Method 1: Brute Force Approach

This approach iterates through the array, simulating the removal of every element and then counting the longest continuous subarray of 1s that can be formed. It is simple but not the most efficient, with a function specification having a time complexity of O(n^2).

Here’s an example:

def longest_subarray_of_1s(arr): max_length = 0 for i in range(len(arr)): tmp_arr = arr[:i] + arr[i+1:] max_length = max(max_length, max(map(len, ''.join(map(str, tmp_arr)).split('0')))) return max_length print(longest_subarray_of_1s([1, 1, 0, 1, 1]))

Output: `4`

This code snippet defines a function that iteratively removes each element from the array, evaluates the longest stretch of 1s, and then tracks the maximum length found. The usage of map functions and string operations makes the code compact but might be less readable for beginners.

## Method 2: Sliding Window Technique

The sliding window technique involves creating a window that can expand and contract based on specific conditions. This method is highly efficient for this problem and runs in O(n) time complexity, making it suitable for large arrays.

Here’s an example:

def find_max_consecutive_ones(arr): max_count = count = zero_index = 0 for i in range(len(arr)): if arr[i] == 1: count += 1 else: count = i - zero_index zero_index = i max_count = max(max_count, count) return max_count print(find_max_consecutive_ones([1, 1, 0, 1, 1]))

Output: `4`

This snippet slides a window through the array, counting consecutive ones and keeping track of the position of the last zero encountered. Upon finding a zero, count is reset. The function maintains the highest count of 1s between zeros as the maximum length. This is more efficient and concise than the brute force approach.

## Method 3: Prefix Sums

Prefix sums are used to calculate the cumulative total up to a certain index, which simplifies the process of finding subarray sums efficiently. By utilizing prefix sums, we can improve the performance of the task only slightly worse than the sliding window by having an O(n) time complexity with additional space complexity for storing sum arrays.

Here’s an example:

def longest_subarray_after_deletion(arr): prefix_sum = [0] for num in arr: prefix_sum.append(prefix_sum[-1] + (1 if num == 1 else 0)) max_length = 0 for i in range(1, len(prefix_sum)): if arr[i-1] == 0: length = prefix_sum[i-1] + (prefix_sum[-1] - prefix_sum[i]) max_length = max(max_length, length) return max_length print(longest_subarray_after_deletion([1, 1, 0, 1, 1]))

Output: `4`

This code calculates the prefix sum of 1s then iterates through the array once again, checking each potential deletion point to find the longest subarray. The maximum sum of the prefix up to a zero and the suffix from that point gives the longest subarray after a deletion.

## Method 4: Improved Sliding Window with Deletion Check

Enhancing the sliding window technique from Method 2, this method explicitly accounts for a single deletion, thus providing better insight into the problem dynamics. Still efficient, its O(n) time complexity stands, with a more direct application to the problem.

Here’s an example:

def longest_subarray(arr): max_length = left = zero_count = 0 for right in range(len(arr)): if arr[right] == 0: zero_count += 1 while zero_count > 1: if arr[left] == 0: zero_count -= 1 left += 1 max_length = max(max_length, right - left) return max_length + 1 print(longest_subarray([1, 1, 0, 1, 1]))

Output: `4`

The code monitors the number of zeros within the current window and shrinks the window from the left whenever more than one zero is encountered. The length of each valid window is compared to find the maximum, where the additional one accounts for the actual deletion of a zero, if present.

## Bonus One-Liner Method 5: Functional Programming Twist

For the functional programming aficionados, this one-liner uses list comprehensions and max function prowess to solve the problem succinctly. However, being a one-liner, it lacks readability and could be inefficient for large arrays due to the use of O(n^2) operations inside the list comprehension.

Here’s an example:

print(max(len(max((''.join(map(str, arr[:i] + arr[i+1:]))).split('0'), key=len)) for i in range(len(arr))))

Output: `4`

This one-liner combines map, str and join operations to simulate the deletion of each element, then finds the max length of 1s using split and max functions. It is an elegant yet less practical solution for those who value brevity over clarity.

## Summary/Discussion

**Method 1: Brute Force Approach.**Easy to implement and understand. Best suited for small arrays due to its inefficiency with large datasets.**Method 2: Sliding Window Technique.**Efficient and concise. Ideal for large datasets, but requires understanding of the sliding window concept.**Method 3: Prefix Sums.**Efficient use of cumulative sums which makes it slightly slower than the sliding window but still practical for medium-sized arrays.**Method 4: Improved Sliding Window with Deletion Check.**Directly solves the problem with an efficient approach. Slightly more complex logic but excellent for large datasets.**Method 5: Functional Programming Twist.**Elegant one-liner that leverages Pythonβs functional programming features. Not recommended for large arrays or those new to Python due to its reduced readability and potential performance issues.