Length of Longest Fibonacci Subsequence

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- Leetcode: Split Array into Fibonacci Sequence
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- Tag: #dynamicprogramming, #fibonacci

A sequence X_1, X_2, …, X_n is fibonacci-like if:

n >= 3

X_i + X_{i+1} = X_{i+2} for all i + 2 <= n

Given a strictly increasing array A of positive integers forming a sequence, find the length of the longest fibonacci-like 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 fibonacci-like: [1,2,3,5,8].

Example 2:

Input: [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].

Note:

- 3 <= A.length <= 1000
- 1 <= A[0] < A[1] < … < A[A.length – 1] <= 10^9

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution: hasmap

// https://code.dennyzhang.com/length-of-longest-fibonacci-subsequence // Basic Ideas: hashmap // Complexity: Time O(n*n*log(n)), Space O(n) // Why the time complexity is not O(n^3)? // After choosing two pilots, we almost double the value each time. func lenLongestFibSubseq(A []int) int { m := map[int]bool{} for _, num := range A { m[num] = true } res := 0 for i:=0; i<len(A)-2; i++ { for j:=i+1; j<len(A)-1; j++ { prev, cur := A[i], A[j] count, next := 2, prev+cur for m[next] { prev, cur = cur, next next = prev+cur count++ } if count>res { res = count } } } if res>=3 { return res } else { return 0 } }