To solve this problem, we need to find the maximum number of consecutive 1s in a binary string after flipping exactly one bit (either 0 to 1 or vice versa). The optimal approach to this problem is using a sliding window technique, which allows us to efficiently compute the solution in linear time.

Approach

1. Sliding Window Technique: We use two pointers (left and right) to represent the current window of the string we are considering. The goal is to find the longest window where the number of 0s is at most 1 (since we can flip exactly one 0 to 1 to get consecutive 1s).
2. Count Zeros: We need to count the total number of 0s in the string. This helps us handle the edge case where the string consists entirely of 1s (we have to flip one 1 to 0, so the maximum consecutive 1s will be the length of the string minus 1).
3. Adjust Window: As we expand the window using the right pointer, we count the number of 0s. If the number of 0s exceeds 1, we move the left pointer to the right until the number of 0s in the window is at most 1.
4. Compute Result: The maximum length of the valid window (with at most one 0) gives the answer unless the string is all 1s, in which case the answer is the length of the string minus 1.

Solution Code

def max_consecutive_ones_after_flip(s):
    if not s:
        return 0

    left = 0
    zero_count = 0
    max_len = 0
    total_zeros = s.count('0')  # Count total zeros in the string

    for right in range(len(s)):
        if s[right] == '0':
            zero_count += 1

        # Shrink the window if there are more than one zeros
        while zero_count > 1:
            if s[left] == '0':
                zero_count -= 1
            left += 1

        # Update the maximum length of the valid window
        current_len = right - left + 1
        if current_len > max_len:
            max_len = current_len

    # If there are no zeros, we have to flip one 1 to zero, so subtract 1
    if total_zeros == 0:
        return len(s) - 1
    else:
        return max_len

Explanation

• Sliding Window: The sliding window technique efficiently finds the longest subarray with at most one 0. This is done by expanding the window with the right pointer and adjusting the left pointer when the number of 0s exceeds 1.
• Edge Case Handling: When the string is all 1s, the total number of zeros is zero. In this case, flipping any 1 to 0 reduces the maximum consecutive 1s to the length of the string minus 1.
• Efficiency: The algorithm runs in O(n) time complexity, where n is the length of the string, since each character is processed at most twice (once by the right pointer and once by the left pointer). The space complexity is O(1) as we only use a few variables.

This approach ensures that we find the optimal solution efficiently, making it suitable for large input sizes.

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