Length of Longest Fibonacci Subsequence

Description

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

Solution

Both the time complexity and the space complexity are $O(n^2)$.

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        int n = A.size();
        // dp[j][i]: length of the longest Fibonacci sequence ending with (A[j], A[i]).
        vector<vector<int>> dp(n, vector<int>(n));
        unordered_map<int, int> num2idx;
        for(int i = 0; i < n; ++i){
            num2idx[A[i]] = i;
        }
        int len = 0;
        for(int i = 1; i < A.size(); ++i){
            for(int j = 0; j < i; ++j){
                int a = A[i] - A[j];
                int b = A[j];
                int c = A[i];
                // check whether there is a k such that
                // k < j < i and (A[k], A[j], A[i]) is fibonacci-like
                if(a >= b){
                    // not possible to find A[k] because A[k] < A[j]
                    dp[j][i] = 2;
                }else{
                    auto it = num2idx.find(a);
                    if(it != num2idx.end()){
                        dp[j][i] = dp[it->second][j] + 1;
                        len = max(len, dp[j][i]);
                    }else{
                        dp[j][i] = 2;
                    }
                }
            }
        }
        return len;
    }
};