README.md

873. Length of Longest Fibonacci Subsequence

A sequence x1, x2, ..., xn is Fibonacci-like if:

• n >= 3
• xi + xi+1 == xi+2 for all i + 2 <= n

Given a strictly increasing array arr of positive integers forming a sequence, return the length of the longest Fibonacci-like subsequence of arr. If one does not exist, return 0.

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

Example 1:

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

Example 2:

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

Constraints:

• 3 <= arr.length <= 1000
• 1 <= arr[i] < arr[i + 1] <= 109

Solutions (Rust)

1. Solution

use std::collections::HashMap;

impl Solution {
    pub fn len_longest_fib_subseq(arr: Vec<i32>) -> i32 {
        let mut lengths = HashMap::new();
        let last = *arr.last().unwrap();
        let mut ret = 0;

        for i in 1..arr.len() {
            for j in 0..i {
                if arr[i] - arr[j] < arr[j] {
                    let x = *lengths.get(&(arr[i] - arr[j], arr[j])).unwrap_or(&1);

                    if x > 1 {
                        lengths.insert((arr[j], arr[i]), x + 1);
                        ret = ret.max(x + 1);
                    } else if arr[i] + arr[j] <= last {
                        lengths.insert((arr[j], arr[i]), 2);
                    }
                } else if arr[i] + arr[j] <= last {
                    lengths.insert((arr[j], arr[i]), 2);
                }
            }
        }

        ret
    }
}