<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 {
public:
    int longestMountain(vector<int>& arr) {
        int n=arr.size();
        int ans=0;
        int i=1;
        int upper,lower;
        
        while(i<n)
       {    upper=0;lower=0;                  //Each time after traversing one mountain or not,we have to set upper & lower to 0
            while(i<n && arr[i-1]==arr[i])    //   for flat array,where elements are equal
              i++;                            //This also handles when we get same adjacent elements
            while(i<n && arr[i-1]<arr[i])     //Increasing part of mountain
            {  upper++;i++;
                
            }
                    
            
           while(i<n && arr[i-1]>arr[i])    //Decreasing part of mountain 
           {
               lower++;
               i++;    
           } 
          if(upper && lower)       //Mountain only exists if we have both incresing part and decreasing part adjacently
           { ans=max(ans,lower+upper+1);
           }
        
        }
        
        return ans;
     
    }
};
21300