Formatted question description: https://leetcode.ca/all/873.html
873. Length of Longest Fibonacci Subsequence
Level
Medium
Description
A sequence X_1, X_2, ..., X_n
is fibonaccilike if:
n >= 3
X_i + X_{i+1} = X_{i+2}
for alli + 2 <= n
Given a strictly increasing array A
of positive integers forming a sequence, find the length of the longest fibonaccilike subsequence of A
. If one does not exist, return 0.
(Recall that a subsequence is derived from another sequence A by deleting any number of elements (including none) from A
, 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: [1,2,3,4,5,6,7,8]
Output: 5
Explanation:
The longest subsequence that is fibonaccilike: [1,2,3,5,8].
Example 2:
Input: [1,3,7,11,12,14,18]
Output: 3
Explanation:
The longest subsequence that is fibonaccilike: [1,11,12], [3,11,14] or [7,11,18].
Note:
3 <= A.length <= 1000
1 <= A[0] < A[1] < ... < A[A.length  1] <= 10^9
 (The time limit has been reduced by 50% for submissions in Java, C, and C++.)
Solution
Use a set to store all the numbers in array A
. For each indices pair (i, j)
where i < j
, try to generate a fibonaccilike subsequence of A
, where all the numbers from the third number to the end are determined by the sum of the two previous numbers. A fibonaccilike subsequence ends until a number is not in the set. If the subsequence has length greater than or equal to 3, update the longest length. Finally, return the longest length.

class Solution { public int lenLongestFibSubseq(int[] A) { int length = A.length; Set<Integer> set = new HashSet<Integer>(); for (int i = 0; i < length; i++) set.add(A[i]); int longestLength = 0; int end1 = length  2, end2 = length  1; for (int i = 0; i < end1; i++) { for (int j = i + 1; j < end2; j++) { int num1 = A[i]; int num2 = A[j]; int curLength = 2; while (set.contains(num1 + num2)) { int num3 = num1 + num2; curLength++; num1 = num2; num2 = num3; } if (curLength >= 3) longestLength = Math.max(longestLength, curLength); } } return longestLength; } }

Todo

class Solution(object): def lenLongestFibSubseq(self, A): """ :type A: List[int] :rtype: int """ s = set(A) n = len(A) res = 0 for i in range(n  1): for j in range(i + 1, n): a, b = A[i], A[j] count = 2 while a + b in s: a, b = b, a + b count += 1 res = max(res, count) return res if res > 2 else 0