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873. Length of Longest Fibonacci Subsequence

难度: Meidum

刷题内容

原题连接

内容描述

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
(The time limit has been reduced by 50% for submissions in Java, C, and C++.)

解题方案

思路 1 - 时间复杂度: O(N^2LgM)- 空间复杂度: O(N)*****

思路见lee215

Save array A to a hash set s.
Start from base (A[i], A[j]) as the first two element in the sequence,
we try to find the Fibonacci like subsequence as long as possible,

Initial (a, b) = (A[i], A[j])
While the set s contains a + b, we update (a, b) = (b, a + b).
In the end we update the longest length we find.

Time Complexity:
O(N^2logM), where M is the max(A).

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

思路 2 - 时间复杂度: O(N^2)- 空间复杂度: O(NlgM)******

dp[a, b] represents the length of fibo sequence ends up with (a, b)
Then we have dp[a, b] = dp[b - a, a] or 2
The complexity reduce to O(N^2).

beats 83.98%

class Solution:
    def lenLongestFibSubseq(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        dp = {}
        s, res = set(A), 0
        for i in range(len(A)):
            for j in range(i+1, len(A)):
                if A[j] - A[i] < A[i] and A[j] - A[i] in s:
                    dp[A[i], A[j]] = dp.get((A[j] - A[i], A[i]), 2) + 1
                    res = max(res, dp[(A[i], A[j])])
        return res if res > 2 else 0