# Solving Leetcode Interviews in Seconds with AI: Longest Turbulent Subarray


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "978" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 an integer array arr, return the length of a maximum size turbulent subarray of arr. A subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray. More formally, a subarray [arr[i], arr[i + 1], ..., arr[j]] of arr is said to be turbulent if and only if:  For i <= k < j:  	 arr[k] > arr[k + 1] when k is odd, and arr[k] < arr[k + 1] when k is even.   Or, for i <= k < j: 	 arr[k] > arr[k + 1] when k is even, and arr[k] < arr[k + 1] when k is odd.      Example 1:  Input: arr = [9,4,2,10,7,8,8,1,9] Output: 5 Explanation: arr[1] > arr[2] < arr[3] > arr[4] < arr[5]  Example 2:  Input: arr = [4,8,12,16] Output: 2  Example 3:  Input: arr = [100] Output: 1    Constraints:  1 <= arr.length <= 4 * 104 0 <= arr[i] <= 109  

	# Explanation
	*   **Sliding Window:** Use a sliding window approach to iterate through the array. Expand the window as long as the turbulent condition is met.
*   **Track Expected Sign:** Maintain variables to track the expected sign (greater or lesser) at each position within the potential turbulent subarray.
*   **Reset Window:** When the turbulent condition is violated, reset the window to start from the next element or the element causing the violation.

*   **Time Complexity:** O(n), where n is the length of the array. **Space Complexity:** O(1).

	
	# Code
	```python
	def maxTurbulenceSize(arr: list[int]) -> int:
    """
    Finds the length of a maximum size turbulent subarray of arr.
    """
    n = len(arr)
    if n < 2:
        return n

    max_len = 1
    start = 0
    end = 0

    for i in range(1, n):
        if arr[i] == arr[i - 1]:
            start = i
        else:
            if (i - start) < 2:
                end = i
            else:
                if (arr[i - 2] < arr[i - 1] and arr[i - 1] > arr[i]) or \
                   (arr[i - 2] > arr[i - 1] and arr[i - 1] < arr[i]):
                    end = i
                else:
                    start = i - 1
                    end = i

            max_len = max(max_len, end - start + 1)

    return max_len
	```