<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) {
        vector<int> temp;

        if(arr.size()==1)return 1;
        for(int i=0;i<arr.size()-1;i++){
            if(arr[i]>arr[i+1]){
                temp.push_back(1);
            }
            else if(arr[i]<arr[i+1]){
                temp.push_back(-1);
            }
            else temp.push_back(0);
        }
        int ans=1;
        int i=0;
        int j=0;
        int n=temp.size();
        int prev=-temp[0];
        while(i<n){
            if(temp[i]==0){
                ans=max(ans,i-j+1);
                if(i+1<n)prev=-temp[i+1];
                j=i+1;
                i++;
                continue;
            }
            if(temp[i]==prev){
                ans=max(ans,i-j+1);
                j=i;
                prev=-temp[i];
                continue;
            }
            ans=max(ans,i-j+2);
            prev=temp[i];
            i++;
        }
        return ans;
    }
};
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