How do you solve Longest Fibonacci Subarray on LeetCode?

Longest Fibonacci Subarray (LeetCode #3708) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Any two elements can start a Fibonacci sequence.

What is the time complexity of Longest Fibonacci Subarray?

The optimal solution for Longest Fibonacci Subarray runs in O(n) time and uses O(1) space. Single pass through the array results in linear time complexity.

What is the brute force solution for Longest Fibonacci Subarray?

The brute force approach for Longest Fibonacci Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible subarray and verify if it follows the Fibonacci condition. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Fibonacci Subarray?

Longest Fibonacci Subarray uses the following concepts: Array. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Fibonacci Subarray?

Clarify the definition of a Fibonacci subarray. Discuss edge cases, like arrays with all identical elements. Explain your thought process while coding.

What are common mistakes when solving Longest Fibonacci Subarray?

Forgetting to reset the length when the condition fails. Assuming subarrays of length 1 or 2 need checking.

#3708

Longest Fibonacci Subarray

Medium

ArrayHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Use a single pass to check the Fibonacci condition, leveraging the fact that any two elements can start a Fibonacci sequence.

⚙️

Algorithm

3 steps

1Step 1: Initialize length to 2 (since any two elements form a Fibonacci sequence).
2Step 2: Iterate through the array starting from the third element.
3Step 3: If the current element equals the sum of the two preceding elements, increase the length; otherwise, reset it to 2.

solution.py10 lines

1def lenLongestFibSubarray(nums):
2    max_len = 2
3    current_len = 2
4    for i in range(2, len(nums)):
5        if nums[i] == nums[i - 1] + nums[i - 2]:
6            current_len += 1
7            max_len = max(max_len, current_len)
8        else:
9            current_len = 2
10    return max_len if max_len >= 3 else 0

ℹ

Complexity note: Single pass through the array results in linear time complexity.

1Any two elements can start a Fibonacci sequence.
2Resetting the length to 2 helps track valid sequences efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.