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easyBasicsSometimes asked

FizzBuzz

Basics · 6 min

Given n, return a list of strings for 1..n where multiples of 3 are "Fizz", multiples of 5 are "Buzz", multiples of both are "FizzBuzz", and others are the number itself.

In plain English

Check divisibility by 15 (both) first, then 3, then 5, else the number.

Examples

Example 1

Input: n = 5
Output: ["1","2","Fizz","4","Buzz"]
Note: 3→Fizz, 5→Buzz.

Example 2

Input: n = 15
Output: […,"FizzBuzz"]
Note: 15 → FizzBuzz.

Example 3

Input: n = 3
Output: ["1","2","Fizz"]
Note: First Fizz at 3.

Constraints

1 ≤ n ≤ 10^4

Clarifying questions1First question: Return strings or print?Explore questions

Q1.Return strings or print?
I'll return a list of strings, stringifying the number case.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

For each i: if divisible by 3 AND 5 → FizzBuzz, else by 3 → Fizz, else by 5 → Buzz, else the number. Check the both-case first.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you order overlapping conditions most-specific-first.

Where candidates slip

The classic 15-before-3/5 ordering bug.

How to stand out

Offer the append-and-fallback variant that scales to more divisors.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Ordered divisibility checks

TimeO(n)SpaceO(n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Which numbers satisfy more than one rule, and why does order matter?

2Hint 2Explore hint

Test divisible-by-15 (both) before 3 and 5.

3Hint 3Explore hint

Alternative: append 'Fizz'/'Buzz' and fall back to the number if empty.

A · Core idea

The canonical screening problem. For each i in 1..n: multiples of both 3 and 5 → "FizzBuzz", multiples of 3 → "Fizz", multiples of 5 → "Buzz", otherwise the number. The entire test is branch ordering.

Why test divisible-by-15 first. Every multiple of 15 is also a multiple of 3 and of 5. If you check % 3 before % 15, then 15, 30, 45… match the Fizz branch and never produce "FizzBuzz". So the most specific condition (both) must come first. This is the trap the problem is famous for.

The append variant (no 15 check). Build the word incrementally: append "Fizz" if divisible by 3, append "Buzz" if divisible by 5; if the word is still empty, use the number. This avoids the explicit 15 test and extends naturally to more divisors (e.g. add 7 → "Bazz").

Output type. Return strings, not ints — the number case must be stringified too.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

n = 5

1,2

no divisor

State"1","2"

Final result["1","2","Fizz","4","Buzz"]

C · Code with teaching comments

solution.javaJava

Ordered divisibility checks — optimal

1 teaching note

1
public java.util.List<String> fizzBuzz(int n) {
2
    java.util.List<String> out = new java.util.ArrayList<>();
3
    for (int i = 1; i <= n; i++) {
4
        // Check the combined case first, before 3 and 5.
5
        if (i % 15 == 0) out.add("FizzBuzz");
6
        else if (i % 3 == 0) out.add("Fizz");
7
        else if (i % 5 == 0) out.add("Buzz");
8
        else out.add(String.valueOf(i));
9
    }
10
    return out;
11
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.11 lines

D · Why it works and complexityO(n) timeO(n) spaceOpen the invariant and proofExplore the proof

A clean variant builds the word by appending 'Fizz' then 'Buzz' and falls back to the number if empty — no 15 check needed.

ComplexityO(n) time, O(n) space — one constant set of divisibility checks per number across n numbers.

E · Edge cases and pitfalls5n = 15 — Index 15 → "FizzBuzz".Explore edge cases

Edge cases handled

n = 15Index 15 → "FizzBuzz".
n = 3First Fizz at 3.
n = 1["1"].

Pitfalls

Checking %3 or %5 before %15, so 15 never prints FizzBuzz.
Returning ints instead of strings.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic java.util.List<String> fizzBuzz(int n) { java.util.List<String> out = new java.util.ArrayList<>(); for (int i = 1; i <= n; i++) { if (i % 15 == 0) out.add("FizzBuzz"); else if (i % 3 == 0) out.add("Fizz");Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Wrong branch order

Testing %3 before %15 makes every multiple of 15 print 'Fizz', never 'FizzBuzz'.

Explore the code comparison

Don't write this

if i%3==0: 'Fizz'
elif i%5==0: 'Buzz'
elif i%15==0: 'FizzBuzz'  # unreachable

Write this instead

if i%15==0: 'FizzBuzz'
elif i%3==0: 'Fizz'
elif i%5==0: 'Buzz'

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Ordered branch (most-specific first)

Code pattern to memorize

check %15, then %3, then %5, else number

Rules to remember

1Test %15 before %3/%5.
2Else number as string.
3Order is the whole trick.

Use this pattern when

Overlapping conditional rules
Divisibility classification

Choose another pattern when

Checking general cases before the most specific one

Interview takeaway

When conditions overlap, test the most specific one first.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Ordered branch (most-specific first). Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Fizz Buzz
Coming soon
The LeetCode version.

easyBasicsSometimes asked

FizzBuzz

Basics · 6 min

Given n, return a list of strings for 1..n where multiples of 3 are "Fizz", multiples of 5 are "Buzz", multiples of both are "FizzBuzz", and others are the number itself.

In plain English

Check divisibility by 15 (both) first, then 3, then 5, else the number.

Examples

Example 1

Input: n = 5
Output: ["1","2","Fizz","4","Buzz"]
Note: 3→Fizz, 5→Buzz.

Example 2

Input: n = 15
Output: […,"FizzBuzz"]
Note: 15 → FizzBuzz.

Example 3

Input: n = 3
Output: ["1","2","Fizz"]
Note: First Fizz at 3.

Constraints

1 ≤ n ≤ 10^4

Clarifying questions1First question: Return strings or print?Explore questions

Q1.Return strings or print?
I'll return a list of strings, stringifying the number case.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

For each i: if divisible by 3 AND 5 → FizzBuzz, else by 3 → Fizz, else by 5 → Buzz, else the number. Check the both-case first.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you order overlapping conditions most-specific-first.

Where candidates slip

The classic 15-before-3/5 ordering bug.

How to stand out

Offer the append-and-fallback variant that scales to more divisors.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Ordered divisibility checks

TimeO(n)SpaceO(n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Which numbers satisfy more than one rule, and why does order matter?

2Hint 2Explore hint

Test divisible-by-15 (both) before 3 and 5.

3Hint 3Explore hint

Alternative: append 'Fizz'/'Buzz' and fall back to the number if empty.

A · Core idea

Output type. Return strings, not ints — the number case must be stringified too.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

n = 5

1,2

no divisor

State"1","2"

Final result["1","2","Fizz","4","Buzz"]

C · Code with teaching comments

solution.javaJava

Ordered divisibility checks — optimal

1 teaching note

1
public java.util.List<String> fizzBuzz(int n) {
2
    java.util.List<String> out = new java.util.ArrayList<>();
3
    for (int i = 1; i <= n; i++) {
4
        // Check the combined case first, before 3 and 5.
5
        if (i % 15 == 0) out.add("FizzBuzz");
6
        else if (i % 3 == 0) out.add("Fizz");
7
        else if (i % 5 == 0) out.add("Buzz");
8
        else out.add(String.valueOf(i));
9
    }
10
    return out;
11
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.11 lines

D · Why it works and complexityO(n) timeO(n) spaceOpen the invariant and proofExplore the proof

A clean variant builds the word by appending 'Fizz' then 'Buzz' and falls back to the number if empty — no 15 check needed.

ComplexityO(n) time, O(n) space — one constant set of divisibility checks per number across n numbers.

E · Edge cases and pitfalls5n = 15 — Index 15 → "FizzBuzz".Explore edge cases

Edge cases handled

n = 15Index 15 → "FizzBuzz".
n = 3First Fizz at 3.
n = 1["1"].

Pitfalls

Checking %3 or %5 before %15, so 15 never prints FizzBuzz.
Returning ints instead of strings.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Wrong branch order

Testing %3 before %15 makes every multiple of 15 print 'Fizz', never 'FizzBuzz'.

Explore the code comparison

Don't write this

if i%3==0: 'Fizz'
elif i%5==0: 'Buzz'
elif i%15==0: 'FizzBuzz'  # unreachable

Write this instead

if i%15==0: 'FizzBuzz'
elif i%3==0: 'Fizz'
elif i%5==0: 'Buzz'

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Ordered branch (most-specific first)

Code pattern to memorize

check %15, then %3, then %5, else number

Rules to remember

1Test %15 before %3/%5.
2Else number as string.
3Order is the whole trick.

Use this pattern when

Overlapping conditional rules
Divisibility classification

Choose another pattern when

Checking general cases before the most specific one

Interview takeaway

When conditions overlap, test the most specific one first.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Ordered branch (most-specific first). Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Fizz Buzz
Coming soon
The LeetCode version.