How do you solve Fibonacci Number on LeetCode?

Fibonacci Number (LeetCode #509) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Recursive definitions can lead to exponential time complexity if not optimized.

What is the brute force solution for Fibonacci Number?

The brute force approach for Fibonacci Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach directly implements the recursive definition of Fibonacci numbers. It calculates F(n) by recursively calling F(n-1) and F(n-2) until reaching the base cases. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Fibonacci Number?

Fibonacci Number uses the following concepts: Math, Dynamic Programming, Recursion, Memoization. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Fibonacci Number?

Clearly explain your thought process and approach before coding. Consider edge cases and test your solution with them. If using recursion, mention the potential inefficiencies and how you would optimize it.

What are common mistakes when solving Fibonacci Number?

Forgetting to handle base cases in recursion. Not using memoization or dynamic programming when applicable.

#509

Fibonacci Number

Easy

MathDynamic ProgrammingRecursionMemoizationDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to store previously computed Fibonacci numbers, avoiding redundant calculations. This significantly reduces the number of recursive calls.

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Algorithm

4 steps

1Step 1: Create an array to store Fibonacci numbers up to n.
2Step 2: Initialize the first two values: F(0) = 0 and F(1) = 1.
3Step 3: Use a loop to fill in the array from F(2) to F(n) using the relation F(n) = F(n-1) + F(n-2).
4Step 4: Return the value at index n.

solution.py8 lines

1def fib(n):
2    if n == 0:
3        return 0
4    fibs = [0] * (n + 1)
5    fibs[1] = 1
6    for i in range(2, n + 1):
7        fibs[i] = fibs[i - 1] + fibs[i - 2]
8    return fibs[n]

ℹ

Complexity note: The time complexity is O(n) because we compute each Fibonacci number once. The space complexity is O(n) due to the storage of the Fibonacci numbers in an array.

1Recursive definitions can lead to exponential time complexity if not optimized.
2Dynamic programming can significantly reduce redundant calculations.

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