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 >= 3X_i + X_{i+1} = X_{i+2}for alli + 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 <= 10001 <= 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.
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;
}
}