Leetcode: Length of Longest Fibonacci Subsequence

Length of Longest Fibonacci Subsequence



Similar Problems:


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
// Blog link: 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
    }
}
linkedin
github
slack

Share It, If You Like It.

Leave a Reply

Your email address will not be published.