<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>

#include <vector>
using namespace std;

class Solution {
public:
    int longestMountain(vector<int>& arr) {
        int n = arr.size();
        if (n < 3) return 0;  // A valid mountain needs at least 3 elements
        
        int longest = 0;
        
        for (int i = 1; i < n - 1; i++) {
            // Check if arr[i] is a peak
            if (arr[i - 1] < arr[i] && arr[i] > arr[i + 1]) {
                // Expand left to find how far the mountain goes on the left
                int left = i - 1;
                while (left > 0 && arr[left - 1] < arr[left]) {
                    left--;
                }
                
                // Expand right to find how far the mountain goes on the right
                int right = i + 1;
                while (right < n - 1 && arr[right] > arr[right + 1]) {
                    right++;
                }
                
                // Calculate the length of the mountain
                int mountainLength = right - left + 1;
                longest = max(longest, mountainLength);
                
                // Move i to right because we don't need to recheck this part
                i = right;
            }
        }
        
        return longest;
    }
};

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