What is the time complexity of Fibonacci Number?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Fibonacci Number?

This problem uses Math, Dynamic Programming, Recursion, Memoization. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Fibonacci Number?

Insight 1: The Fibonacci sequence is defined recursively, but a simple iterative approach is more efficient. Insight 2: We only need to keep track of the last two Fibonacci numbers to compute the next one, making space optimization possible.

Which companies ask LeetCode 509. Fibonacci Number in interviews?

This problem is frequently asked at Accenture, Capgemini, Cognizant, EY, Infosys and 5 more companies.

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Fibonacci Number - LeetCode 509 Solution

Q: How to solve Fibonacci Number?

The provided code uses an iterative approach with constant space to calculate the nth Fibonacci number. It initializes two variables, 'a' and 'b', to represent the two preceding Fibonacci numbers. It then iterates from 2 to n, updating 'a' and 'b' in each iteration to represent the next Fibonacci number in the sequence. The final value of 'b' after the loop is the nth Fibonacci number.

Fibonacci Number - Complete Solution Guide

Fibonacci Number is LeetCode problem 509, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

The Fibonacci numbers , commonly denoted F(n) form a sequence, called the Fibonacci sequence , such that each number is the sum of the two preceding ones, starting from 0 and 1 . That is, F(0) = 0, F(1) = 1 F(n) = F(n - 1) + F(n - 2), for n > 1. Given n , calculate F(n) . Example 1: Input: n = 2 Output: 1 Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1. Example 2: Input: n = 3 Output: 2 Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2. Example 3: Input: n = 4 Output: 3 Explanation: F(4) = F(3) + F(2) = 2

Detailed Explanation

The problem asks us to calculate the nth Fibonacci number. The Fibonacci sequence starts with 0 and 1. Each subsequent number is the sum of the two preceding numbers. Given an integer 'n', we need to return the nth Fibonacci number (F(n)). The constraints are 0 <= n <= 30.

Solution Approach

The provided code uses an iterative approach with constant space to calculate the nth Fibonacci number. It initializes two variables, 'a' and 'b', to represent the two preceding Fibonacci numbers. It then iterates from 2 to n, updating 'a' and 'b' in each iteration to represent the next Fibonacci number in the sequence. The final value of 'b' after the loop is the nth Fibonacci number.

Step-by-Step Algorithm

Step 1: Handle the base cases: If n is 0 or 1, return n directly.
Step 2: Initialize two variables, 'a' to 0 and 'b' to 1. These will hold F(n-2) and F(n-1) respectively.
Step 3: Iterate from i = 2 to n. In each iteration:
Step 4: Calculate the next Fibonacci number (F(i)) as the sum of 'a' and 'b'.
Step 5: Update 'a' to the value of 'b', and 'b' to the calculated sum.
Step 6: After the loop finishes, return the value of 'b', which now holds F(n).

Key Insights

Insight 1: The Fibonacci sequence is defined recursively, but a simple iterative approach is more efficient.
Insight 2: We only need to keep track of the last two Fibonacci numbers to compute the next one, making space optimization possible.
Insight 3: The base cases F(0) = 0 and F(1) = 1 are critical for the sequence to start correctly.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming, Recursion, Memoization.

Companies

Asked at: Accenture, Capgemini, Cognizant, EY, Infosys, J.P. Morgan, Nvidia, SAP, Zoox, tcs.

Fibonacci Number | LeetCode 509

Easy

Math Dynamic Programming Recursion Memoization

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int fib(int n) {
        if (n <= 1) {
            return n;
        }
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = 1;
        for (int i = 2; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
}

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Problem Info

Description

Problem 509: Fibonacci Number - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Recursion Memoization

Companies

AccentureCapgeminiCognizantEYInfosysJ.P. MorganNvidiaSAP

Stats

Difficulty:Easy

Languages:4

Companies:10

Problem #:509

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Handle the base cases: If n is 0 or 1, return n directly.
2
Step 2: Initialize two variables, 'a' to 0 and 'b' to 1. These will hold F(n-2) and F(n-1) respectively.
3
Step 3: Iterate from i = 2 to n. In each iteration:
4
Step 4: Calculate the next Fibonacci number (F(i)) as the sum of 'a' and 'b'.
5
Step 5: Update 'a' to the value of 'b', and 'b' to the calculated sum.
6
Step 6: After the loop finishes, return the value of 'b', which now holds F(n).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because the loop iterates 'n-1' times. The space complexity is O(1) because we are only using a fixed number of variables (a, b, and possibly a temporary variable) regardless of the input size 'n'.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach

A direct implementation of the Fibonacci sequence's recursive definition. F(n) = F(n-1) + F(n-2).

Time: O(2^n)Space: O(n) due to the call stack.

Trade-offs: Simple to write, but very inefficient for larger values of 'n' due to repeated calculations. Avoid for practical use beyond small values of n.

Dynamic Programming (Memoization)

A recursive approach with memoization to store previously computed Fibonacci numbers, avoiding redundant calculations.

Time: O(n)Space: O(n) to store the memoized Fibonacci numbers.

Trade-offs: More efficient than the naive recursive approach, but uses more space than the iterative approach. Suitable when the call stack depth is not a concern, and a top-down approach is preferred.

Matrix Exponentiation

Uses matrix exponentiation to compute the nth Fibonacci number in logarithmic time.

Time: O(log n)Space: O(1)

Trade-offs: Most efficient solution, but more complex to understand and implement. Suitable for very large values of 'n'.

✨

Key Insights

1Insight 1: The Fibonacci sequence is defined recursively, but a simple iterative approach is more efficient.
2Insight 2: We only need to keep track of the last two Fibonacci numbers to compute the next one, making space optimization possible.
3Insight 3: The base cases F(0) = 0 and F(1) = 1 are critical for the sequence to start correctly.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Recursion

A function that calls itself to solve smaller versions of the same problem.

Edge Cases

Edge case 1: n = 0: The code correctly returns 0 because the initial 'if' condition catches this case.
Edge case 2: n = 1: The code correctly returns 1 because the initial 'if' condition catches this case.
Edge case 3: n = 2: The code correctly calculates F(2) = F(1) + F(0) = 1 + 0 = 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using a naive recursive approach without memoization, which leads to exponential time complexity and becomes very slow for n > 20.

✗Common mistake 2: Not handling the base cases (n = 0 and n = 1) correctly, which can lead to incorrect results or infinite recursion.

✗Common mistake 3: Using an array of size 'n' instead of just two variables in the iterative approach, leading to unnecessary space usage when constant space is achievable.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Modeling population growth: The Fibonacci sequence can be used to model the growth of populations under certain idealized conditions.

🏦Real-world scenario 2: Computer algorithms: The Fibonacci sequence and its related concepts appear in various computer algorithms, such as those used for searching, sorting, and data compression.

🎮Real-world scenario 3: Financial markets: Some traders use Fibonacci retracement levels to identify potential support and resistance levels in financial markets.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Fibonacci Number"?

For this easy-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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