<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>& a) 
    {
         int n = a.size();

         map<int,int> idx;
         for( int i = 0; i < n; i++ ) idx[a[i]] = i;

         int dp[1001][1001][3];
         memset(dp , -1 , sizeof(dp));

         function<int(int,int,int)> f =[&](int i , int j , int k)
         {
             if( i == n ) return 0;
             if( dp[i][j][k] != -1 ) return dp[i][j][k];

             int ans = 0;

             if( k == 0 ) ans = max( 1 + f(i + 1 , i , 1 ) , f(i + 1 , j , 0) );
             else if( k == 1 )
             {
                if( idx.find(a[i] + a[j]) != idx.end() )
                   ans = max(1 + f(idx[a[i] + a[j]] , i , 2) , f(i + 1 , j , 1) );
                else 
                   ans = f(i + 1 , j , k);   
             }
             else
             {
                if( idx.find(a[i] + a[j]) != idx.end() )
                   ans = 1 + f(idx[a[i] + a[j]] , i , 2);
             }
             return dp[i][j][k] = ans;
         };

         int res = f(0 , 0 , 0);
         
         if( res < 2 ) return 0;
         return res + 1;
    }
};
30900