What is the time complexity of Fizz Buzz?

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

What data structures are used to solve Fizz Buzz?

This problem uses Math, String, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Fizz Buzz?

Insight 1: The core operation is checking divisibility using the modulo operator (%). Insight 2: Implementing the conditions in the correct order is important. Checking for divisibility by both 3 and 5 before checking individual divisibility by 3 or 5 avoids incorrect results.

Which companies ask LeetCode 412. Fizz Buzz in interviews?

This problem is frequently asked at Cisco, Citadel, Cloudflare, Cognizant, IBM and 6 more companies.

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Fizz Buzz - LeetCode 412 Solution

Fizz Buzz - Complete Solution Guide

Fizz Buzz is LeetCode problem 412, 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

Given an integer n , return a string array answer ( 1-indexed ) where : answer[i] == "FizzBuzz" if i is divisible by 3 and 5 . answer[i] == "Fizz" if i is divisible by 3 . answer[i] == "Buzz" if i is divisible by 5 . answer[i] == i (as a string) if none of the above conditions are true. Example 1: Input: n = 3 Output: ["1","2","Fizz"] Example 2: Input: n = 5 Output: ["1","2","Fizz","4","Buzz"] Example 3: Input: n = 15 Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","1

Detailed Explanation

The Fizz Buzz problem is a classic coding challenge that tests a programmer's understanding of basic conditional logic and string manipulation. Given an integer 'n', the task is to generate a list of strings where each element at index 'i' (1-indexed) follows these rules: * If 'i' is divisible by both 3 and 5, the element is "FizzBuzz". * If 'i' is divisible by 3, the element is "Fizz". * If 'i' is divisible by 5, the element is "Buzz". * Otherwise, the element is the string representation of 'i'. The input is a single integer 'n' representing the upper limit of the range (1 to n). The output is an array (or list) of strings following the described rules.

Solution Approach

The solution iterates through the numbers from 1 to 'n'. For each number, it checks the divisibility conditions in the specified order: first by both 3 and 5, then by 3, then by 5. If none of these conditions are met, it converts the number to a string and adds it to the result list. The algorithm utilizes simple if-else conditional statements to determine the appropriate string to add to the result.

Step-by-Step Algorithm

Step 1: Initialize an empty list or array called 'answer' to store the resulting strings.
Step 2: Iterate from 'i' = 1 to 'n' (inclusive).
Step 3: Inside the loop, check if 'i' is divisible by both 3 and 5 (i.e., i % 3 == 0 and i % 5 == 0). If true, append "FizzBuzz" to the 'answer' list.
Step 4: If step 3 is false, check if 'i' is divisible by 3 (i.e., i % 3 == 0). If true, append "Fizz" to the 'answer' list.
Step 5: If step 4 is false, check if 'i' is divisible by 5 (i.e., i % 5 == 0). If true, append "Buzz" to the 'answer' list.
Step 6: If none of the above conditions are met, convert 'i' to a string (using the appropriate string conversion function for the programming language) and append it to the 'answer' list.
Step 7: After the loop finishes, return the 'answer' list.

Key Insights

Insight 1: The core operation is checking divisibility using the modulo operator (%).
Insight 2: Implementing the conditions in the correct order is important. Checking for divisibility by both 3 and 5 before checking individual divisibility by 3 or 5 avoids incorrect results.
Insight 3: Efficiently handling the string conversion for non-divisible numbers is crucial for performance.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Math, String, Simulation.

Companies

Asked at: Cisco, Citadel, Cloudflare, Cognizant, IBM, J.P. Morgan, LinkedIn, Mastercard, Media.net, Nvidia, tcs.

Fizz Buzz | LeetCode 412

Easy

Math String Simulation

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> result = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            if (i % 3 == 0 && i % 5 == 0) {
                result.add("FizzBuzz");
            } else if (i % 3 == 0) {
                result.add("Fizz");
            } else if (i % 5 == 0) {
                result.add("Buzz");
            } else {
                result.add(String.valueOf(i));
            }
        }
        return result;
    }
}

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Problem Info

Description

Problem 412: Fizz Buzz - Detailed explanation and solution approaches.

Topics

Math String Simulation

Companies

CiscoCitadelCloudflareCognizantIBMJ.P. MorganLinkedInMastercard

Stats

Difficulty:Easy

Languages:4

Companies:11

Problem #:412

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

7 steps

1
Step 1: Initialize an empty list or array called 'answer' to store the resulting strings.
2
Step 2: Iterate from 'i' = 1 to 'n' (inclusive).
3
Step 3: Inside the loop, check if 'i' is divisible by both 3 and 5 (i.e., i % 3 == 0 and i % 5 == 0). If true, append "FizzBuzz" to the 'answer' list.
4
Step 4: If step 3 is false, check if 'i' is divisible by 3 (i.e., i % 3 == 0). If true, append "Fizz" to the 'answer' list.
5
Step 5: If step 4 is false, check if 'i' is divisible by 5 (i.e., i % 5 == 0). If true, append "Buzz" to the 'answer' list.
6
Step 6: If none of the above conditions are met, convert 'i' to a string (using the appropriate string conversion function for the programming language) and append it to the 'answer' list.
7
Step 7: After the loop finishes, return the 'answer' list.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm iterates through the numbers from 1 to 'n' once, performing a constant amount of work for each number (checking divisibility and string concatenation). Therefore, the time complexity is directly proportional to 'n', resulting in O(n) time complexity. The space complexity is also O(n) because the algorithm stores the resulting strings in a list, which grows linearly with the input '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

String Concatenation

Instead of separate if-else blocks, a string can be built incrementally. Start with an empty string. If divisible by 3, append "Fizz". If divisible by 5, append "Buzz". If the string is still empty after both checks, append the number's string representation.

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

Trade-offs: This approach can be slightly more concise, but might be less readable for beginners. It doesn't offer significant performance advantages.

✨

Key Insights

1Insight 1: The core operation is checking divisibility using the modulo operator (%).
2Insight 2: Implementing the conditions in the correct order is important. Checking for divisibility by both 3 and 5 before checking individual divisibility by 3 or 5 avoids incorrect results.
3Insight 3: Efficiently handling the string conversion for non-divisible numbers is crucial for performance.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: n = 1. The code correctly handles this by returning ['1'].
Edge case 2: n = 15. The code will return the correct FizzBuzz sequence up to 15, including 'FizzBuzz' at the appropriate position.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect order of conditions. Checking for divisibility by 3 or 5 *before* checking for divisibility by both 3 and 5 will lead to incorrect results (e.g., 'Fizz' instead of 'FizzBuzz').

✗Common mistake 2: Forgetting to convert the number to a string when it's not divisible by 3 or 5. This will result in type errors or incorrect output.

✅

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: This problem demonstrates basic conditional logic which is fundamental in almost all software development. For example, applying different discounts based on customer criteria.

🏦Real-world scenario 2: Used as a simple test for junior developers during interviews to assess their basic coding skills.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: July 23, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Fizz Buzz"?

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 string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?