Formatted question description: https://leetcode.ca/all/873.html

873. Length of Longest Fibonacci Subsequence

Level

Medium

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

Solution

Use a set to store all the numbers in array A. For each indices pair (i, j) where i < j, try to generate a fibonacci-like subsequence of A, where all the numbers from the third number to the end are determined by the sum of the two previous numbers. A fibonacci-like subsequence ends until a number is not in the set. If the subsequence has length greater than or equal to 3, update the longest length. Finally, return the longest length.

• Java
• C++
• Python

• class Solution {
      public int lenLongestFibSubseq(int[] A) {
          int length = A.length;
          Set<Integer> set = new HashSet<Integer>();
          for (int i = 0; i < length; i++)
              set.add(A[i]);
          int longestLength = 0;
          int end1 = length - 2, end2 = length - 1;
          for (int i = 0; i < end1; i++) {
              for (int j = i + 1; j < end2; j++) {
                  int num1 = A[i];
                  int num2 = A[j];
                  int curLength = 2;
                  while (set.contains(num1 + num2)) {
                      int num3 = num1 + num2;
                      curLength++;
                      num1 = num2;
                      num2 = num3;
                  }
                  if (curLength >= 3)
                      longestLength = Math.max(longestLength, curLength);
              }
          }
          return longestLength;
      }
  }
  

• Todo
  

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

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