<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) {
    unordered_map<int, int> index;  
    unordered_map<int, int> dp;     
    int max_len = 0;
    int n = arr.size();

    for (int i = 0; i < n; ++i) {
        index[arr[i]] = i;
    }

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < i; ++j) {
            int k = arr[i] - arr[j];  // Find the potential third element in the Fibonacci sequence
            if (index.count(k) && index[k] < j) {
                int key = j * n + i;  
                int prevKey = index[k] * n + j;
                dp[key] = dp.count(prevKey) ? dp[prevKey] + 1 : 3; 
                max_len = max(max_len, dp[key]);
            }
        }
    }

    return max_len >= 3 ? max_len : 0;  
}
};
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