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FizzBuzz – Solution & Complexity

Interview solutions
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FizzBuzz

1. Understand the Replacement Rules

  • For every number from 1 through n, return one string.
  • Multiples of 3 become Fizz, multiples of 5 become Buzz.
  • Multiples of both must become FizzBuzz, not just one of the two words.

2. Check the Most Specific Case First

  • A number divisible by both 3 and 5 is divisible by 15.
  • Check this case before checking only 3 or only 5.
  • Otherwise 15 would incorrectly be labeled as just Fizz or just Buzz.
if num % 15 == 0:
    result.append("FizzBuzz")

3. Loop Through the Range

  • Use range(1, n + 1) because the sequence includes both endpoints: 1 and n.
  • Append the correct string for each number.
  • Convert normal numbers with str(num).
for num in range(1, n + 1):
    if num % 15 == 0:
        result.append("FizzBuzz")
    elif num % 3 == 0:
        result.append("Fizz")

4. Handle Input Robustly

  • The problem describes n as an integer, but JSON input can sometimes provide numeric-looking values as strings.
  • Converting with int(n) makes the function tolerant of either form.
  • The output remains a list of strings.

5. Complete Solution and Complexity

  • The final solution makes one pass from 1 to n and applies the divisibility rules.
  • Time complexity is O(n).
  • Space complexity is O(n) for the returned list.
def fizz_buzz(n: int) -> list[str]:
    n = int(n)
    result = []

    for num in range(1, n + 1):
        if num % 15 == 0:
            result.append("FizzBuzz")
        elif num % 3 == 0:
            result.append("Fizz")
        elif num % 5 == 0:
            result.append("Buzz")
        else:
            result.append(str(num))

    return result

FAQ

See also:

Practice FizzBuzz online (run code in the browser)Roman to Integer solution (easy)Climbing Stairs solution (easy)Count Vowels solution (easy)Edit Distance solution (hard)All coding interview problems
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