873. Length of Longest Fibonacci Subsequence

Problem

A sequence x1, x2, ..., xn is Fibonacci-like if:

• n >= 3

• xi + xi+1 == xi+2 for all i + 2 <= n

Given a strictly increasing array arr of positive integers forming a sequence, return **the *length* of the longest Fibonacci-like subsequence of** arr. If one does not exist, return 0.

A subsequence is derived from another sequence arr by deleting any number of elements (including none) from arr, 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: arr = [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: arr = [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].

  Constraints:

• 3 <= arr.length <= 1000

• 1 <= arr[i] < arr[i + 1] <= 10^9

Solution (Java)

class Solution {
    public int lenLongestFibSubseq(int[] arr) {
        if (arr == null || arr.length < 3) {
            return 0;
        }
        int len = arr.length;
        int[][] dp = new int[len][len];
        int ans = 0;
        for (int i = 2; i < len; i++) {
            int left = 0;
            int right = i - 1;
            while (left < right) {
                if (arr[left] + arr[right] < arr[i]) {
                    left++;
                } else if (arr[left] + arr[right] > arr[i]) {
                    right--;
                } else {
                    // dp[right][i] = Math.max(dp[right][i], dp[left][right] + 1);
                    dp[right][i] = dp[left][right] + 1;
                    ans = Math.max(ans, dp[right][i]);
                    left++;
                    right--;
                }
            }
        }
        return ans > 0 ? ans + 2 : 0;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).