How do you approach FizzBuzz?

Check divisibility by 15 first (since 15 = 3 * 5).

What is the time complexity of FizzBuzz?

Time complexity: O(n). Space complexity: O(n) for the result array.

What are common mistakes on FizzBuzz?

Overflowing at an intermediate step even though the final answer fits. Assuming a modulo is non-negative. In most languages -1 % n is negative, which breaks cyclic indexing. Testing floating-point values for exact equality where integer arithmetic would have been exact.

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FizzBuzz

EasyAlgorithms

Description

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

Constraints

- 1 <= n <= 10^4

FizzBuzz: approach, complexity and pitfalls

An easy math problem, graded against 5 test cases (2 of them hidden).

The pattern: Math

Number theory, digit manipulation, and closed-form reasoning.

Reach for it when you see: Primes, digits, base conversion, or a problem with a formula hiding behind it.

More Mathproblems →

If you are stuck

Show the 3-step hint ladder

Check divisibility by 15 first (since 15 = 3 * 5).
Or build the string by checking 3 and 5 separately.
Remember to convert numbers to strings.

The optimal approach

Show the approach: String Concatenation

Build the output string by checking divisibility conditions.

Time: O(n)
Space: O(n) for the result array

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

Common Math mistakes

These apply to the pattern as a whole, not just this problem.

Overflowing at an intermediate step even though the final answer fits.
Assuming a modulo is non-negative. In most languages `-1 % n` is negative, which breaks cyclic indexing.
Testing floating-point values for exact equality where integer arithmetic would have been exact.