Introduction

In this blog post, we will explore how to solve the LeetCode problem "1493" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a binary array nums, you should delete one element from it. Return the size of the longest non-empty subarray containing only 1's in the resulting array. Return 0 if there is no such subarray. Example 1: Input: nums = [1,1,0,1] Output: 3 Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's. Example 2: Input: nums = [0,1,1,1,0,1,1,0,1] Output: 5 Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1]. Example 3: Input: nums = [1,1,1] Output: 2 Explanation: You must delete one element. Constraints: 1 <= nums.length <= 105 nums[i] is either 0 or 1.

Explanation

Here's the breakdown of the solution:

• Sliding Window: Use a sliding window to find the longest subarray containing at most one zero. The window expands to the right, and when the number of zeros exceeds one, the window contracts from the left.
• Track Zeroes: Keep track of the number of zeroes within the current window. When the count exceeds 1, shrink the window from the left until the zero count goes down to 1.

• Handle All Ones: Account for the edge case where the array consists entirely of ones. In this scenario, the longest subarray after deleting one element is simply len(nums) - 1.

• Runtime Complexity: O(n) & Storage Complexity: O(1)

Code

    def longestSubarray(nums):
    """
    Finds the size of the longest non-empty subarray containing only 1's after deleting one element.

    Args:
        nums: A binary array.

    Returns:
        The size of the longest non-empty subarray containing only 1's.
    """

    left = 0
    zero_count = 0
    max_len = 0

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

        while zero_count > 1:
            if nums[left] == 0:
                zero_count -= 1
            left += 1

        max_len = max(max_len, right - left)

    # Handle the case where the array contains all 1s
    if max_len == len(nums):
        return max_len - 1

    return max_len