<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 f(int curr,int prev,vector<int>& nums,int n){
			int t=nums[curr]+nums[prev];
			int i=lower_bound(nums.begin(),nums.end(),t)-nums.begin();
			if(i<n && nums[i]==t) return 1+f(i,curr,nums,n);
			return 1;
		}

		int lenLongestFibSubseq(vector<int>& nums) {
			int n=nums.size();
			int maxi=0;
			for(int prev2=0;prev2<n;prev2++){
				for(int prev1=prev2+1;prev1<n;prev1++){
					int temp=nums[prev1]+nums[prev2];
					int ind=lower_bound(nums.begin(),nums.end(),temp)-nums.begin();
					if(ind<n && nums[ind]==temp) maxi=max(maxi,2+f(ind,prev1,nums,n));
				}
			}
			return maxi;
		}
	};
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