Introduction

In this blog post, we will explore how to solve the LeetCode problem "2765" 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

You are given a 0-indexed integer array nums. A subarray s of length m is called alternating if: m is greater than 1. s1 = s0 + 1. The 0-indexed subarray s looks like [s0, s1, s0, s1,...,s(m-1) % 2]. In other words, s1 - s0 = 1, s2 - s1 = -1, s3 - s2 = 1, s4 - s3 = -1, and so on up to s[m - 1] - s[m - 2] = (-1)m. Return the maximum length of all alternating subarrays present in nums or -1 if no such subarray exists. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [2,3,4,3,4] Output: 4 Explanation: The alternating subarrays are [2, 3], [3,4], [3,4,3], and [3,4,3,4]. The longest of these is [3,4,3,4], which is of length 4. Example 2: Input: nums = [4,5,6] Output: 2 Explanation: [4,5] and [5,6] are the only two alternating subarrays. They are both of length 2. Constraints: 2 <= nums.length <= 100 1 <= nums[i] <= 104

Explanation

Here's a breakdown of the solution:

• Sliding Window: Use a sliding window approach to iterate through the nums array. The window expands as long as the alternating pattern continues.
• Pattern Check: For each subarray (window), verify if it follows the required alternating pattern (s1 = s0 + 1, s2 = s1 - 1, etc.).

• Maximum Length: Keep track of the maximum length of the alternating subarray found so far.

• Time Complexity: O(n), where n is the length of the input array.

• Space Complexity: O(1)

Code

    def alternatingSubarray(nums):
    """
    Finds the maximum length of an alternating subarray in nums.

    Args:
        nums: A 0-indexed integer array.

    Returns:
        The maximum length of all alternating subarrays present in nums or -1 if no such subarray exists.
    """

    max_length = -1
    n = len(nums)

    for i in range(n - 1):
        for j in range(i + 1, n):
            length = j - i + 1
            if length > 1:
                is_alternating = True
                for k in range(1, length):
                    if nums[i + k] - nums[i + k - 1] != (1 if k % 2 == 1 else -1):
                        is_alternating = False
                        break

                if is_alternating:
                    max_length = max(max_length, length)

    return max_length