#509

Easy

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Fibonacci Number

Calculate the nth Fibonacci number using state transition dynamic programming and recursive techniques efficiently in interviews.

math dynamic programming recursion memoization

Problem Statement

Given an integer n, compute the nth number in the Fibonacci sequence, where each number is the sum of the previous two numbers, starting from F(0) = 0 and F(1) = 1.

Implement a function that returns F(n). You may use recursion, memoization, or iterative dynamic programming, but must handle the sequence efficiently to avoid excessive redundant computations.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

F(0) = 0, F(1) = 1 F(n) = F(n - 1) + F(n - 2), for n > 1.

Example 2

Input: n = 2

Output: 1

F(2) = F(1) + F(0) = 1 + 0 = 1.

Example 3

Input: n = 3

Output: 2

F(3) = F(2) + F(1) = 1 + 1 = 2.

Constraints

0 <= n <= 30

Solution Approach

Recursive Solution with Memoization

Implement a top-down recursive function storing intermediate results in a hash map or array. Base cases are F(0) = 0 and F(1) = 1. Each recursive call returns F(n-1) + F(n-2), caching results to avoid repeated work.

Iterative Dynamic Programming

Use a bottom-up approach by initializing an array of size n+1. Set dp[0] = 0, dp[1] = 1, then iterate from 2 to n, updating dp[i] = dp[i-1] + dp[i-2]. This avoids recursion overhead and uses O(n) time and space.

Space-Optimized Iterative Approach

Keep only two variables representing the last two Fibonacci numbers instead of a full array. Update them iteratively: prev = 0, curr = 1, then for i = 2 to n, compute next = prev + curr, update prev = curr, curr = next. This reduces space to O(1).

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the approach: naive recursion is O(2^n), memoized recursion and iterative DP are O(n). Space complexity ranges from O(n) for DP array or memoization to O(1) for space-optimized iteration.

What Interviewers Usually Probe

Expect correct handling of base cases F(0) and F(1).
Check whether the candidate avoids redundant recursive calls using memoization.
Watch for space optimization in iterative solutions for follow-up questions.

Common Pitfalls or Variants

Common pitfalls

Naive recursion without memoization causes exponential time complexity and stack overflow for larger n.
Incorrectly handling base cases leads to off-by-one errors in sequence values.
Using unnecessary extra space when a simple two-variable iteration suffices.

Follow-up variants

Compute Fibonacci modulo 10^9+7 to test large n handling and modular arithmetic.
Return the first n Fibonacci numbers as an array to test iterative DP comprehension.
Implement a recursive Fibonacci with limited stack depth using tail recursion to explore optimization.

FAQ

What is the Fibonacci Number problem pattern?

It follows a state transition dynamic programming pattern where F(n) = F(n-1) + F(n-2).

Can I solve Fibonacci Number with plain recursion?

Yes, but naive recursion has O(2^n) time complexity and will fail for larger n without memoization.

How do I optimize Fibonacci Number computation?

Use iterative DP or memoization to reduce time complexity to O(n) and consider two-variable iteration to reduce space to O(1).

What are common mistakes in Fibonacci Number interviews?

Common mistakes include missing base cases, redundant recursion, and overusing space for simple iterative solutions.

Is memoization always better than iterative for Fibonacci Number?

Memoization helps clarify recursion logic, but iterative DP is often faster and uses less space, making it more efficient for large n.

terminal

Solution

Solution 1: Recurrence

We define two variables $a$ and $b$, initially $a = 0$ and $b = 1$.

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class Solution:
    def fib(self, n: int) -> int:
        a, b = 0, 1
        for _ in range(n):
            a, b = b, a + b
        return a

Solution 2: Matrix Exponentiation

We define $\textit{Fib}(n)$ as a $1 \times 2$ matrix $\begin{bmatrix} F_n & F_{n - 1} \end{bmatrix}$, where $F_n$ and $F_{n - 1}$ are the $n$-th and $(n - 1)$-th Fibonacci numbers, respectively.

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class Solution:
    def fib(self, n: int) -> int:
        a, b = 0, 1
        for _ in range(n):
            a, b = b, a + b
        return a

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