<p>Given an integer array <code>arr</code>, return <em>the length of a maximum size turbulent subarray of</em> <code>arr</code>.</p>

<p>A subarray is <strong>turbulent</strong> if the comparison sign flips between each adjacent pair of elements in the subarray.</p>

<p>More formally, a subarray <code>[arr[i], arr[i + 1], ..., arr[j]]</code> of <code>arr</code> is said to be turbulent if and only if:</p>

<ul>
	<li>For <code>i &lt;= k &lt; j</code>:

	<ul>
		<li><code>arr[k] &gt; arr[k + 1]</code> when <code>k</code> is odd, and</li>
		<li><code>arr[k] &lt; arr[k + 1]</code> when <code>k</code> is even.</li>
	</ul>
	</li>
	<li>Or, for <code>i &lt;= k &lt; j</code>:
	<ul>
		<li><code>arr[k] &gt; arr[k + 1]</code> when <code>k</code> is even, and</li>
		<li><code>arr[k] &lt; arr[k + 1]</code> when <code>k</code> is odd.</li>
	</ul>
	</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [9,4,2,10,7,8,8,1,9]
<strong>Output:</strong> 5
<strong>Explanation:</strong> arr[1] &gt; arr[2] &lt; arr[3] &gt; arr[4] &lt; arr[5]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [4,8,12,16]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [100]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 4 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def maxTurbulenceSize(self, arr: List[int]) -> int:
        if len(arr) == 1:
            return 1
        
        max_len = 1  # To store the max length of turbulent subarray
        left = 0     # Left boundary of the window
        
        for right in range(1, len(arr)):
            # Check the relationship between arr[right-1] and arr[right]
            if arr[right] == arr[right-1]:
                # If the elements are equal, reset the window
                left = right
            elif right == 1 or (arr[right-2] < arr[right-1] and arr[right-1] > arr[right]) or (arr[right-2] > arr[right-1] and arr[right-1] < arr[right]):
                # Continue expanding the window if turbulent
                max_len = max(max_len, right - left + 1)
            else:
                # If the turbulent condition breaks, reset the window to start from current `right`
                left = right - 1
                max_len = max(max_len, 2)  # At least 2 elements turbulent
        
        return max_len

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