LeetCode 3708: Longest Fibonacci Subarray

Meta Data

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Difficulty: medium First Attempt: 2025-10-11

• Submissions: Accepted (first try)

Approach

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For a contiguous run, track only the current length (Array Template 1). Any two adjacent numbers can start a new Fibonacci subarray, so a break resets the length to 2, not 1.

My Solution

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class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        n = len(nums)
        if n <= 2:
            return n
        cur = best = 2
        for i in range(2, n):
            cur = cur + 1 if nums[i] == nums[i - 1] + nums[i - 2] else 2
            best = max(best, cur)
        return best

Complexity

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Time O(n), space O(1).

Mistakes to Avoid

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• A break resets to 2, since the last two numbers start a new run.
• With fewer than 3 elements, return the length.

Interview Explanation

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From the third number on, I extend the current length when it equals the sum of the previous two, otherwise restart from the last two with length 2. Time O(n).