FizzBuzz - Print 1 to 100 with Fizz/Buzz/FizzBuzz Rules - Java

Category: Easy | Concepts used: Loops, modulo operator, conditional order

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Problem Statement

Print numbers from 1 to 100. But:

• If divisible by 3, print "Fizz" instead of the number.
• If divisible by 5, print "Buzz" instead.
• If divisible by both 3 and 5, print "FizzBuzz".

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 ...

Examples (with edge scenarios)

#	n	Output	Why
1	3	"Fizz"	Divisible by 3 only
2	5	"Buzz"	Divisible by 5 only
3	15	"FizzBuzz"	Divisible by both 3 and 5
4	7	"7"	Not divisible by either — print the number itself
5	1	"1"	Smallest value in range, not divisible by anything special

Common Fresher Mistake

Mistake	What happens	Fix
Checking %3 and %5 separately with two if blocks (not else if) BEFORE the combined check	For n=15, prints "Fizz" then "Buzz" separately, or misses "FizzBuzz" entirely	Always check the combined condition (%15==0) FIRST, before the individual %3 and %5 checks
Using %3==0 && %5==0 as a separate late check after already returning for %3 or %5	"FizzBuzz" case is never reached because %3 check already fired first	Order matters — most specific condition (both) must come first

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Before You Code: Clarify the Contract

Before choosing an algorithm, confirm whether zero and negative values are allowed, how large the input can be, and what should happen when an arithmetic result exceeds the chosen Java type. The examples use the contract stated in this article, but an interview answer should say these assumptions aloud.

Analogy: The Ticket Counter Validator

Imagine you are validating entry tickets (numbers) at a theater:

• Rule 1: If the ticket has a Green stamp (divisible by 3), give them a Fizz coupon.
• Rule 2: If the ticket has a Blue stamp (divisible by 5), give them a Buzz coupon.
• Rule 3: If the ticket has both Green and Blue stamps, give them a FizzBuzz VIP pass.
• If you check the Green stamp first and hand out a “Fizz” coupon immediately, you might let a VIP guest (with both stamps) walk away with just a basic “Fizz” coupon!
• To avoid this mistake, you must look for the combined stamps first (FizzBuzz) before evaluating individual stamps!

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Solution 1 — if-else Chain with Combined Check First (Recommended)

Intuition

A number divisible by both 3 and 5 is also divisible by 3 alone and by 5 alone — so if you check the individual conditions first, you’ll never actually reach the “both” case. The fix: always test the most restrictive condition (divisible by both, i.e., by 15) before the looser individual ones — like checking “is this a golden ticket” before checking “is this just a regular ticket.”

public class FizzBuzz {
    public static void printFizzBuzz(int limit) {
        for (int i = 1; i <= limit; i++) {
            if (i % 3 == 0 && i % 5 == 0) {
                System.out.println("FizzBuzz"); // check BOTH first
            } else if (i % 3 == 0) {
                System.out.println("Fizz");
            } else if (i % 5 == 0) {
                System.out.println("Buzz");
            } else {
                System.out.println(i);
            }
        }
    }

public static void main(String[] args) {
        printFizzBuzz(20); // demo with first 20 numbers
    }
}

Output:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
Fizz
19
Buzz

Dry Run (i = 15)

i % 3 == 0?  15%3=0 -> yes
i % 5 == 0?  15%5=0 -> yes
Both true -> print "FizzBuzz"  (never reaches the individual Fizz/Buzz checks)

Interviewer’s take

This is the expected, safe solution. The order of conditions is the entire point of this question — interviewers use FizzBuzz specifically to see if you think through condition ordering carefully, not just whether you can write a loop.

Follow-up questions you might get:

• “Why must the combined check come first?” → Because else if chains stop at the first true match — if %3==0 is checked before %15==0, a multiple of 15 would incorrectly print just "Fizz" and never reach the FizzBuzz check.
• “Can you generalize this for more divisors, like also ‘Bazz’ for divisible by 7?” → Yes — extend the combined checks and add more else if branches, always ordering from most-specific (most conditions combined) to least-specific.

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Solution 2 — Using String Concatenation (No Explicit “Both” Check Needed)

Intuition

Instead of manually handling the “both” case as a special condition, build up a result string piece by piece — append "Fizz" if divisible by 3, append "Buzz" if divisible by 5. If a number is divisible by both, both pieces naturally get appended together into "FizzBuzz" — no explicit combined check required!

public class FizzBuzzConcat {
    public static void printFizzBuzz(int limit) {
        for (int i = 1; i <= limit; i++) {
            StringBuilder output = new StringBuilder();

if (i % 3 == 0) output.append("Fizz");
            if (i % 5 == 0) output.append("Buzz");

System.out.println(output.length() == 0 ? String.valueOf(i) : output.toString());
        }
    }

public static void main(String[] args) {
        printFizzBuzz(15);
    }
}

Output:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz

Dry Run (i = 15)

output = ""
i%3==0 -> output.append("Fizz") -> output = "Fizz"
i%5==0 -> output.append("Buzz") -> output = "FizzBuzz"
output.length() != 0 -> print "FizzBuzz"

Interviewer’s take

This is actually a slightly more elegant solution — it avoids the “order of conditions” trap entirely by construction, and it scales beautifully if more rules are added later (e.g., “Bazz” for divisible by 7 — just add another if line). Many interviewers consider this the more “senior” answer since it shows awareness of extensibility.

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📊 Visual Flowchart

graph TD
    Start["Loop: i = 1 to 100"] --> CondBoth{"i % 3 == 0 AND i % 5 == 0?"}
    CondBoth -->|Yes| PrintFB["Print 'FizzBuzz'"]
    CondBoth -->|No| CondThree{"i % 3 == 0?"}
    CondThree -->|Yes| PrintF["Print 'Fizz'"]
    CondThree -->|No| CondFive{"i % 5 == 0?"}
    CondFive -->|Yes| PrintB["Print 'Buzz'"]
    CondFive -->|No| PrintNum["Print i"]
    PrintFB --> IncLoop["i++"]
    PrintF --> IncLoop
    PrintB --> IncLoop
    PrintNum --> IncLoop
    IncLoop --> Start

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Final Verdict — Which Solution Should You Give?

• Solution 1 is the standard textbook answer — correct and clearly demonstrates you understand condition ordering.
• Solution 2 is a great “next level” answer — cleaner and naturally extensible, a good one to offer if asked “can you make this cleaner/more scalable?”

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Quick Recap

Approach	Handles “both” correctly?	Extensible to more rules?	Interview Signal
if-else chain (combined check first)	Yes, if ordered correctly	Awkward — needs more else if branches	Standard, expected
String concatenation	Yes, naturally	Very easy to extend	Shows extra polish