873. Length of Longest Fibonacci Subsequence
Description
A sequence x1, x2, ..., xn is Fibonacci-like if:
n >= 3xi + xi+1 == xi+2for alli + 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 <= 10001 <= arr[i] < arr[i + 1] <= 109
Solutions
Solution 1: Dynamic Programming
We define $f[i][j]$ as the length of the longest Fibonacci-like subsequence, with $\textit{arr}[i]$ as the last element and $\textit{arr}[j]$ as the second to last element. Initially, for any $i \in [0, n)$ and $j \in [0, i)$, we have $f[i][j] = 2$. All other elements are $0$.
We use a hash table $d$ to record the indices of each element in the array $\textit{arr}$.
Then, we can enumerate $\textit{arr}[i]$ and $\textit{arr}[j]$, where $i \in [2, n)$ and $j \in [1, i)$. Suppose the currently enumerated elements are $\textit{arr}[i]$ and $\textit{arr}[j]$, we can obtain $\textit{arr}[i] - \textit{arr}[j]$, denoted as $t$. If $t$ is in the array $\textit{arr}$, and the index $k$ of $t$ satisfies $k < j$, then we can get a Fibonacci-like subsequence with $\textit{arr}[j]$ and $\textit{arr}[i]$ as the last two elements, and its length is $f[i][j] = \max(f[i][j], f[j][k] + 1)$. We can continuously update the value of $f[i][j]$ in this way, and then update the answer.
After the enumeration ends, return the answer.
The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$, where $n$ is the length of the array $\textit{arr}$.
Python3
class Solution:
def lenLongestFibSubseq(self, arr: List[int]) -> int:
n = len(arr)
f = [[0] * n for _ in range(n)]
d = {x: i for i, x in enumerate(arr)}
for i in range(n):
for j in range(i):
f[i][j] = 2
ans = 0
for i in range(2, n):
for j in range(1, i):
t = arr[i] - arr[j]
if t in d and (k := d[t]) < j:
f[i][j] = max(f[i][j], f[j][k] + 1)
ans = max(ans, f[i][j])
return ans
Java
class Solution {
public int lenLongestFibSubseq(int[] arr) {
int n = arr.length;
int[][] f = new int[n][n];
Map<Integer, Integer> d = new HashMap<>();
for (int i = 0; i < n; ++i) {
d.put(arr[i], i);
for (int j = 0; j < i; ++j) {
f[i][j] = 2;
}
}
int ans = 0;
for (int i = 2; i < n; ++i) {
for (int j = 1; j < i; ++j) {
int t = arr[i] - arr[j];
Integer k = d.get(t);
if (k != null && k < j) {
f[i][j] = Math.max(f[i][j], f[j][k] + 1);
ans = Math.max(ans, f[i][j]);
}
}
}
return ans;
}
}
C++
class Solution {
public:
int lenLongestFibSubseq(vector<int>& arr) {
int n = arr.size();
int f[n][n];
memset(f, 0, sizeof(f));
unordered_map<int, int> d;
for (int i = 0; i < n; ++i) {
d[arr[i]] = i;
for (int j = 0; j < i; ++j) {
f[i][j] = 2;
}
}
int ans = 0;
for (int i = 2; i < n; ++i) {
for (int j = 1; j < i; ++j) {
int t = arr[i] - arr[j];
auto it = d.find(t);
if (it != d.end() && it->second < j) {
int k = it->second;
f[i][j] = max(f[i][j], f[j][k] + 1);
ans = max(ans, f[i][j]);
}
}
}
return ans;
}
};
Go
func lenLongestFibSubseq(arr []int) (ans int) {
n := len(arr)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, n)
}
d := make(map[int]int)
for i, x := range arr {
d[x] = i
for j := 0; j < i; j++ {
f[i][j] = 2
}
}
for i := 2; i < n; i++ {
for j := 1; j < i; j++ {
t := arr[i] - arr[j]
if k, ok := d[t]; ok && k < j {
f[i][j] = max(f[i][j], f[j][k]+1)
ans = max(ans, f[i][j])
}
}
}
return
}
TypeScript
function lenLongestFibSubseq(arr: number[]): number {
const n = arr.length;
const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
const d: Map<number, number> = new Map();
for (let i = 0; i < n; ++i) {
d.set(arr[i], i);
for (let j = 0; j < i; ++j) {
f[i][j] = 2;
}
}
let ans = 0;
for (let i = 2; i < n; ++i) {
for (let j = 1; j < i; ++j) {
const t = arr[i] - arr[j];
const k = d.get(t);
if (k !== undefined && k < j) {
f[i][j] = Math.max(f[i][j], f[j][k] + 1);
ans = Math.max(ans, f[i][j]);
}
}
}
return ans;
}
Rust
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
}
}
JavaScript
/**
* @param {number[]} arr
* @return {number}
*/
var lenLongestFibSubseq = function (arr) {
const n = arr.length;
const f = Array.from({ length: n }, () => Array(n).fill(0));
const d = new Map();
for (let i = 0; i < n; ++i) {
d.set(arr[i], i);
for (let j = 0; j < i; ++j) {
f[i][j] = 2;
}
}
let ans = 0;
for (let i = 2; i < n; ++i) {
for (let j = 1; j < i; ++j) {
const t = arr[i] - arr[j];
const k = d.get(t);
if (k !== undefined && k < j) {
f[i][j] = Math.max(f[i][j], f[j][k] + 1);
ans = Math.max(ans, f[i][j]);
}
}
}
return ans;
};