<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 {
public:
    int maxTurbulenceSize(vector<int>& arr) {
        int n=arr.size();
        vector<pair<int, int>>v;
        int i=0;
        while(i<n-1){
            int a=0,b=0;
            if(i%2==0){
                if(arr[i]<arr[i+1]){
                    a=1;
                }
                if(arr[i]>arr[i+1]){
                    b=1;
                }
            }
        else {
            if(arr[i]>arr[i+1]){
                a=1;
            }
            if(arr[i]<arr[i+1]){
                b=1;
            }
        }
         v.push_back({a,b});
         i++;
        }
       // v.push_back({1,1});
        int maxi=0,a1=0,a2=0;
        i=0;
        while(i<n-1){
            int a= v[i].first, b= v[i].second;
            if(a==0){
                a1=0;
            }
        else a1+=1;
            if(b==0){
                a2=0;
            }
        else a2+=1;
            maxi= max(maxi, max(a1, a2));
            i++;
        }
        return maxi+1;
    }
};
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