<p>You may recall that an array <code>arr</code> is a <strong>mountain array</strong> if and only if:</p>

<ul>
	<li><code>arr.length &gt;= 3</code></li>
	<li>There exists some index <code>i</code> (<strong>0-indexed</strong>) with <code>0 &lt; i &lt; arr.length - 1</code> such that:
	<ul>
		<li><code>arr[0] &lt; arr[1] &lt; ... &lt; arr[i - 1] &lt; arr[i]</code></li>
		<li><code>arr[i] &gt; arr[i + 1] &gt; ... &gt; arr[arr.length - 1]</code></li>
	</ul>
	</li>
</ul>

<p>Given an integer array <code>arr</code>, return <em>the length of the longest subarray, which is a mountain</em>. Return <code>0</code> if there is no mountain subarray.</p>

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

<pre>
<strong>Input:</strong> arr = [2,1,4,7,3,2,5]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The largest mountain is [1,4,7,3,2] which has length 5.
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,2,2]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no mountain.
</pre>

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

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

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<ul>
	<li>Can you solve it using only one pass?</li>
	<li>Can you solve it in <code>O(1)</code> space?</li>
</ul>

class Solution:
    def longestMountain(self, arr: List[int]) -> int:
        if len(arr) < 3:
            return 0
        
        result = 0

        start = 0
        while start < len(arr) - 2:
            end = start
            if arr[start + 1] > arr[start]:
                while (end + 1) < len(arr) and arr[end + 1] > arr[end]:
                    end += 1
                if (end + 1) < len(arr) and arr[end + 1] < arr[end]:
                    while (end + 1) < len(arr) and arr[end + 1] < arr[end]:
                        end += 1
                    
                    result = max(result, end - start + 1)
            start = max(start + 1, end)
        
        return result

17800