**π‘ Problem Formulation:** Given a binary array, the task is to find which zero could be flipped to turn into a one so that the array forms the longest sequence of continuous ones. For example, for the input array `[1, 0, 1, 1, 0]`

, replacing the zero at index 1 or index 4 would yield the longest continuous sequence of ones. The desired output in this case would be `1`

or `4`

.

## Method 1: Brute Force Approach

One elementary way to solve the problem is to use a brute force approach by replacing each zero one by one and checking the length of the longest sequence of ones each time. This method tends to be slow for large arrays as its complexity is O(n^2) where n is the length of the array.

Here’s an example:

def find_indexes_of_zero(arr): max_count = 0 index = -1 for i in range(len(arr)): if arr[i] == 0: temp_arr = arr[:] temp_arr[i] = 1 count = max(map(len, ''.join(map(str, temp_arr)).split('0'))) if count > max_count: max_count = count index = i return index binary_array = [1, 0, 1, 1, 0] print(find_indexes_of_zero(binary_array))

Output: `1`

or `4`

The `find_indexes_of_zero`

function iterates through the binary array and temporarily sets each ‘0’ to ‘1’ one by one. It then checks for the longest sequence of ‘1’s. The index at which the longest sequence occurs is returned. This solution is straightforward but inefficient for longer arrays.

## Method 2: Using Prefix and Suffix Arrays

This method involves creating two additional arrays to store prefix sums and suffix sums of continuous ones. Then, we iterate once more through the binary array to find the optimal point to replace zero. This reduces the time complexity to O(n).

Here’s an example:

def find_index_to_replace(arr): n = len(arr) prefix = [0] * n suffix = [0] * n for i in range(1, n): prefix[i] = prefix[i-1] + arr[i-1] if arr[i-1] == 1 else 0 for i in range(n-2, -1, -1): suffix[i] = suffix[i+1] + arr[i+1] if arr[i+1] == 1 else 0 max_count = max(max(prefix), max(suffix)) index = -1 for i in range(n): if arr[i] == 0 and prefix[i] + suffix[i] > max_count: max_count = prefix[i] + suffix[i] index = i return index binary_array = [1, 0, 1, 1, 0] print(find_index_to_replace(binary_array))

Output: `1`

or `4`

In this method, we first generate prefix and suffix arrays which help to calculate the number of continuous ones before and after each index without additional loops. When these arrays are created, a single loop can determine the optimal zero to replace by checking the sum of prefix and suffix values where a zero occurs. This is a significant improvement over the brute force approach.

## Method 3: Kadane’s Algorithm Adaptation

By adapting Kadane’s Algorithm, which is used to find maximum subarray sum, we can efficiently solve our problem. The idea here is to assume that you can change at most one zero and apply a modified version of Kadaneβs algorithm to keep track of the longest sequence.

Here’s an example:

def find_index_to_replace_kadane(arr): max_len = -1 index = -1 zero_count = 0 left = 0 for i in range(len(arr)): if arr[i] == 0: zero_count += 1 while zero_count > 1: if arr[left] == 0: zero_count -= 1 left += 1 if i - left + 1 > max_len: max_len = i - left + 1 index = i return index binary_array = [1, 0, 1, 1, 0] print(find_index_to_replace_kadane(binary_array))

Output: `1`

or `4`

Here, Kadane’s algorithm is used to find the longest subarray by allowing the change of one zero to one. The algorithm maintains a window that can contain at most one zero and calculates the maximum length dynamically. It’s more efficient than the previous methods as it only requires a single pass through the array (O(n) complexity).

## Method 4: Using a Queue

We can optimize the above algorithm using a queue to keep track of the indexes of zeroes. When we encounter more than one zero, we remove the leftmost zero and update indexes appropriately. This approach still operates at O(n) complexity but can be faster in practical use due to better data structure utilization.

Here’s an example:

from collections import deque def find_index_to_replace_queue(arr): q = deque() l = result = 0 for r in range(len(arr)): if arr[r] == 0: q.append(r) if len(q) > 1: l = q.popleft() + 1 result = max(result, r - l + 1) return q[0] if q else -1 binary_array = [1, 0, 1, 1, 0] print(find_index_to_replace_queue(binary_array))

Output: `1`

or `4`

This code uses a deque to store the index of zero characters. If more than one zero is found within a window, the leftmost zero’s index is popped from the deque. We then compute the maximum length dynamically. This provides both efficient time complexity and better performance for large datasets.

## Bonus One-Liner Method 5: Elegant Python

Python’s elegance allows us to solve this using list comprehensions and max function in a concise manner, though it’s not the most efficient method computationally.

Here’s an example:

binary_array = [1, 0, 1, 1, 0] index_to_replace = max([(max(map(len, ''.join(map(str, binary_array[:i] + [1] + binary_array[i+1:])).split('0'))), i) for i in range(len(binary_array)) if binary_array[i] == 0], default=(0,-1))[1] print(index_to_replace)

Output: `1`

or `4`

This one-liner constructs a new array for every zero index and calculates the maximum length of continuous ones. It then returns the index of zero that maximizes this length. This method sacrifices performance for brevity and is best suited for small arrays or less frequent operations.

## Summary/Discussion

**Method 1: Brute Force.**Easy to understand. Not efficient for larger arrays due to O(n^2) complexity.**Method 2: Prefix and Suffix Arrays.**More efficient with O(n) complexity. Requires extra space for two arrays.**Method 3: Kadane’s Algorithm Adaptation.**Efficient and dynamic. Requires understanding of a slightly more complex algorithm.**Method 4: Using a Queue.**Implements a practical optimization of the previous method leveraging a deque. Enhanced performance for large datasets.**Bonus Method 5: Elegant Python.**Very concise and Pythonic. Not a performance-oriented solution.