How do you solve Length of Longest Fibonacci Subsequence on LeetCode?

Length of Longest Fibonacci Subsequence (LeetCode #873) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The Fibonacci-like condition requires the last two numbers to sum up to the next number, which can be efficiently checked using a hash map.

What is the brute force solution for Length of Longest Fibonacci Subsequence?

The brute force approach for Length of Longest Fibonacci Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the array and checking which ones are Fibonacci-like. This is simple but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Length of Longest Fibonacci Subsequence?

Length of Longest Fibonacci Subsequence uses the following concepts: Array, Hash Table, Dynamic Programming. The recommended patterns to study are: Hash Map, Dynamic Programming, Two Pointers.

What are the interview tips for Length of Longest Fibonacci Subsequence?

Always clarify the problem constraints and requirements before diving into coding. Think about edge cases, such as arrays that do not contain any Fibonacci-like subsequences. Explain your thought process while coding to demonstrate your understanding of the problem.

What are common mistakes when solving Length of Longest Fibonacci Subsequence?

Failing to check if the indices are valid when using the hash map. Not considering that the sequence must be strictly increasing.

#873

Length of Longest Fibonacci Subsequence

Medium

ArrayHash TableDynamic ProgrammingHash MapDynamic ProgrammingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(n)

The optimal solution uses dynamic programming and a hash map to efficiently find pairs of numbers that can form a Fibonacci-like sequence. This approach significantly reduces the number of checks needed.

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Algorithm

4 steps

1Step 1: Create a hash map to store the indices of each number in the array for quick access.
2Step 2: Initialize a 2D array dp where dp[i][j] represents the length of the Fibonacci-like subsequence ending with arr[i] and arr[j].
3Step 3: Iterate through pairs of indices (i, j) and check if arr[i] + arr[j] exists in the hash map. If it does, update dp[j][k] where k is the index of arr[i] + arr[j].
4Step 4: Keep track of the maximum length found during the updates.

solution.py17 lines

1# Full working Python code
2from collections import defaultdict
3
4def lenLongestFibSubseq(arr):
5    index_map = {x: i for i, x in enumerate(arr)}
6    n = len(arr)
7    dp = defaultdict(int)
8    max_length = 0
9
10    for j in range(n):
11        for i in range(j):
12            if arr[j] - arr[i] in index_map:
13                k = index_map[arr[j] - arr[i]]
14                if k < i:
15                    dp[i, j] = dp[k, i] + 1
16                    max_length = max(max_length, dp[i, j] + 2)
17    return max_length if max_length >= 3 else 0

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops. However, we use O(n) space for the hash map and the dp array, which allows for quick lookups and efficient updates.

1The Fibonacci-like condition requires the last two numbers to sum up to the next number, which can be efficiently checked using a hash map.
2Dynamic programming allows us to build solutions incrementally, storing results of subproblems to avoid redundant calculations.

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