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++.)

Bruteforce Algorithm to Find the Longest Fibonacci Sequence

Given the A[i] constrains that the maximum number of A[i] is no more than 10^9 and the fact that the fibonacci grows exponentially, we know roughly that there are at most 43 elements in the Fibonacci subsequences.

We remember the numbers using a set. Then we can bruteforce the pairs in O(N^2), and iteratively extending the sequence using set.find in O(1) time. The overall complexity is O(N^2 LogM) where M is the maximum value of the A[i].

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class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        int n = A.size();
        if (n <= 2) return 0;
        unordered_set<int> S(A.begin(), A.end());
        int ans = 0;
        for (int i = 0; i < n; ++ i) {
            for (int j = i + 1; j < n; ++ j) {
                int x = A[j], y = A[i] + A[j];
                int len = 2;
                while (S.count(y)) {
                    int z = x + y;
                    x = y;
                    y = z;
                    ans = max(ans, ++len);
                }                
            }
        }
        return ans >= 3 ? ans : 0;
    }
};

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        int n = A.size();
        if (n <= 2) return 0;
        unordered_set<int> S(A.begin(), A.end());
        int ans = 0;
        for (int i = 0; i < n; ++ i) {
            for (int j = i + 1; j < n; ++ j) {
                int x = A[j], y = A[i] + A[j];
                int len = 2;
                while (S.count(y)) {
                    int z = x + y;
                    x = y;
                    y = z;
                    ans = max(ans, ++len);
                }                
            }
        }
        return ans >= 3 ? ans : 0;
    }
};

Finding the Longest Fibonacci Sequence using Dynamic Programming Algorithm

Let’s consider this problem in DP manner where we define dp[i][j] is the length of the Fibonacci subsequence that ends at A[i] and A[j]. Then we can deduce the previous number in the Fibonacci subsequence is A[j] – A[i].

Using a hash map to remember the index of the numbers, we can find out if the A[j]-A[i] is in the array. If it this in the array, and the index is smaller than i, then we know the dp[j][k] will be dp[i][j] plus 1.

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class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        int n = A.size();
        if (n <= 2) return 0;
        unordered_map<int, int> index;        
        for (int i = 0; i < n; ++ i) {
            index[A[i]] = i;
        }
        int ans = 0;
        unordered_map<int, unordered_map<int, int>> dp;
        for (int k = 0; k < n; ++ k) {
            for (int j = 0; j < k; ++ j) {
                int prev = A[k] - A[j];
                if (index.find(prev) != index.end()) {
                    int i = index[prev];
                    if (i < j) {
                        dp[j][k] = max(2, dp[i][j]) + 1;
                        ans = max(ans, dp[j][k]);
                    }
                }
            }
        }
        return ans >= 3 ? ans : 0;
    }
};

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& A) {
        int n = A.size();
        if (n <= 2) return 0;
        unordered_map<int, int> index;        
        for (int i = 0; i < n; ++ i) {
            index[A[i]] = i;
        }
        int ans = 0;
        unordered_map<int, unordered_map<int, int>> dp;
        for (int k = 0; k < n; ++ k) {
            for (int j = 0; j < k; ++ j) {
                int prev = A[k] - A[j];
                if (index.find(prev) != index.end()) {
                    int i = index[prev];
                    if (i < j) {
                        dp[j][k] = max(2, dp[i][j]) + 1;
                        ans = max(ans, dp[j][k]);
                    }
                }
            }
        }
        return ans >= 3 ? ans : 0;
    }
};

Special case has to be dealt with when the answer is less than 3 – which would not form a valid Fibonacci sequence. The Dynamic Programming Algorithm runs at O(N^2) time and using O(N) space.

–EOF (The Ultimate Computing & Technology Blog) —