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) otherwise.

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","13","14","FizzBuzz"]

Constraints

• 1 <= n <= 10^4

Hints for Fizz Buzz

1. Hint 1

   Check divisibility by 15 first (or by both 3 and 5) — otherwise "Fizz" or "Buzz" will shadow "FizzBuzz".

2. Hint 2

   Building the string by concatenating "Fizz" then "Buzz" when each divides avoids the 15 case entirely.

How to solve Fizz Buzz

Walk 1…n and classify each value, testing the combined case first so a multiple of 15 is not captured by the Fizz branch.

Approach

1. For each i from 1 to n, check i % 15 == 0 → "FizzBuzz".
2. Else i % 3 == 0 → "Fizz", else i % 5 == 0 → "Buzz".
3. Otherwise append the decimal string of i.

Why it works

Divisibility by both 3 and 5 is equivalent to divisibility by 15 because they are coprime, and ordering the branches most-specific first guarantees each i lands in exactly one.

Complexity

• Time — O(n)
• Space — O(n) for the output

Pitfalls

• Testing % 3 before % 15 silently turns every FizzBuzz into Fizz.
• The array is 1-indexed in the statement; the loop must start at 1, not 0.

Reference solution

Python

def fizzBuzz(n: int):
    out = []
    for i in range(1, n + 1):
        if i % 15 == 0:
            out.append("FizzBuzz")
        elif i % 3 == 0:
            out.append("Fizz")
        elif i % 5 == 0:
            out.append("Buzz")
        else:
            out.append(str(i))
    return out

Java

public static String[] fizzBuzz(int n) {
    String[] out = new String[n];
    for (int i = 1; i <= n; i++) {
        if (i % 15 == 0) out[i - 1] = "FizzBuzz";
        else if (i % 3 == 0) out[i - 1] = "Fizz";
        else if (i % 5 == 0) out[i - 1] = "Buzz";
        else out[i - 1] = String.valueOf(i);
    }
    return out;
}

C++

vector<string> fizzBuzz(int n) {
    vector<string> out;
    for (int i = 1; i <= n; i++) {
        if (i % 15 == 0) out.push_back("FizzBuzz");
        else if (i % 3 == 0) out.push_back("Fizz");
        else if (i % 5 == 0) out.push_back("Buzz");
        else out.push_back(to_string(i));
    }
    return out;
}

JavaScript

var fizzBuzz = function(n) {
    const out = [];
    for (let i = 1; i <= n; i++) {
        if (i % 15 === 0) out.push("FizzBuzz");
        else if (i % 3 === 0) out.push("Fizz");
        else if (i % 5 === 0) out.push("Buzz");
        else out.push(String(i));
    }
    return out;
};

Also on the editorial tab: C, C#, Go, Kotlin, PHP, Ruby, Rust, Swift, TypeScript.

More strings problems

• Multiply Two Linked Lists Easy
• Convert Binary Number in a Linked List to Integer Easy
• Check if Any Element Has Prime Frequency Easy
• Convert Date to Binary Easy
• Count Ways to Tile a 2 x N Floor Easy
• Find Nth Root of M Easy