FizzBuzz

≈ 20 min · python-loops python-conditionals · Open in WolfCode

The famous interview screener — every Ch. 3 loop skill plus the order-of-branches trap from Ch. 2 in one exercise.

The rules

Divisible by both 3 and 5 → "FizzBuzz"
Divisible by 3 (only) → "Fizz"
Divisible by 5 (only) → "Buzz"
Otherwise → str(i)

The shape

Python · runnable

def fizzbuzz(n):
    result = []
    for i in range(1, n + 1):
        if i % 15 == 0:                  # MOST specific first
            result.append("FizzBuzz")
        elif i % 3 == 0:
            result.append("Fizz")
        elif i % 5 == 0:
            result.append("Buzz")
        else:
            result.append(str(i))        # don't forget str(...)
    return result

for line in fizzbuzz(15):
    print(line)

Loading Python runtime (~5 MB, one-time)…

Check yourself

In the BROKEN ordering (3-branch first), what does fizzbuzz(15)[14] return?

Why `str(i)` not just `i`?

The TypeError-on-join trap

def fizzbuzz_TYPE_BUG(n):
    result = []
    for i in range(1, n + 1):
        if i % 15 == 0:
            result.append("FizzBuzz")
        elif i % 3 == 0:
            result.append("Fizz")
        elif i % 5 == 0:
            result.append("Buzz")
        else:
            result.append(i)        # ✗ int, not str
    return result

out = fizzbuzz_TYPE_BUG(5)
try:
    print(", ".join(out))
except TypeError as e:
    print("Error:", e)

Loading Python runtime (~5 MB, one-time)…

Now you try

Open 02-fizzbuzz.py. The autograder runs 6 tests including a type-consistency check.

FizzBuzz

The rules

The shape

Why str(i) not just i?

Now you try

Why `str(i)` not just `i`?