Length of Longest Fibonacci Subsequence
<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>
Array; Hash Table; Dynamic Programming
use std::collections::HashMap;
impl Solution {
    pub fn len_longest_fib_subseq(arr: Vec<i32>) -> i32 {
        let n = arr.len();
        let mut f = vec![vec![0; n]; n];
        let mut d = HashMap::new();
        for i in 0..n {
            d.insert(arr[i], i);
            for j in 0..i {
                f[i][j] = 2;
            }
        }
        let mut ans = 0;
        for i in 2..n {
            for j in 1..i {
                let t = arr[i] - arr[j];
                if let Some(&k) = d.get(&t) {
                    if k < j {
                        f[i][j] = f[i][j].max(f[j][k] + 1);
                        ans = ans.max(f[i][j]);
                    }
                }
            }
        }
        ans
    }
}
