<p>A sequence <code>x<sub>1</sub>, x<sub>2</sub>, ..., x<sub>n</sub></code> is <em>Fibonacci-like</em> if:</p>

<ul>
	<li><code>n &gt;= 3</code></li>
	<li><code>x<sub>i</sub> + x<sub>i+1</sub> == x<sub>i+2</sub></code> for all <code>i + 2 &lt;= n</code></li>
</ul>

<p>Given a <b>strictly increasing</b> array <code>arr</code> of positive integers forming a sequence, return <em>the <strong>length</strong> of the longest Fibonacci-like subsequence of</em> <code>arr</code>. If one does not exist, return <code>0</code>.</p>

<p>A <strong>subsequence</strong> is derived from another sequence <code>arr</code> by deleting any number of elements (including none) from <code>arr</code>, without changing the order of the remaining elements. For example, <code>[3, 5, 8]</code> is a subsequence of <code>[3, 4, 5, 6, 7, 8]</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [1,2,3,4,5,6,7,8]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The longest subsequence that is fibonacci-like: [1,2,3,5,8].</pre>

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

<pre>
<strong>Input:</strong> arr = [1,3,7,11,12,14,18]
<strong>Output:</strong> 3
<strong>Explanation</strong>:<strong> </strong>The longest subsequence that is fibonacci-like: [1,11,12], [3,11,14] or [7,11,18].</pre>

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

<ul>
	<li><code>3 &lt;= arr.length &lt;= 1000</code></li>
	<li><code>1 &lt;= arr[i] &lt; arr[i + 1] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& arr) {
        int maxlen=0;
        int n=arr.size();
        int prev1=-1;
        int prev2=-1;
        for(int i=0;i<n;i++)
        {
           
            for(int j=i+1;j<n-1;j++)
            {
                prev2=arr[i];
                prev1=arr[j];
                int currLen=2;
                for(int k=j+1;k<n;k++)
                {
                    if(arr[i])
                    if(arr[k]==prev1+prev2)
                    {
                        currLen++;
                        prev2=prev1;
                        prev1=arr[k];
                    }
                    else if(arr[k]>prev1+prev2)
                    {
                        break;
                    }
                }
                maxlen=max(maxlen,currLen);
            }
        }
        if(maxlen<3)
        {
            return 0;
        }
        return maxlen;
    }
};
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