Python/dynamic_programming/fizz_buzz.py
Christian Clauss 64543faa98
Make some ruff fixes (#8154)
* Make some ruff fixes

* Undo manual fix

* Undo manual fix

* Updates from ruff=0.0.251
2023-03-01 17:23:33 +01:00

65 lines
1.9 KiB
Python

# https://en.wikipedia.org/wiki/Fizz_buzz#Programming
def fizz_buzz(number: int, iterations: int) -> str:
"""
Plays FizzBuzz.
Prints Fizz if number is a multiple of 3.
Prints Buzz if its a multiple of 5.
Prints FizzBuzz if its a multiple of both 3 and 5 or 15.
Else Prints The Number Itself.
>>> fizz_buzz(1,7)
'1 2 Fizz 4 Buzz Fizz 7 '
>>> fizz_buzz(1,0)
Traceback (most recent call last):
...
ValueError: Iterations must be done more than 0 times to play FizzBuzz
>>> fizz_buzz(-5,5)
Traceback (most recent call last):
...
ValueError: starting number must be
and integer and be more than 0
>>> fizz_buzz(10,-5)
Traceback (most recent call last):
...
ValueError: Iterations must be done more than 0 times to play FizzBuzz
>>> fizz_buzz(1.5,5)
Traceback (most recent call last):
...
ValueError: starting number must be
and integer and be more than 0
>>> fizz_buzz(1,5.5)
Traceback (most recent call last):
...
ValueError: iterations must be defined as integers
"""
if not isinstance(iterations, int):
raise ValueError("iterations must be defined as integers")
if not isinstance(number, int) or not number >= 1:
raise ValueError(
"""starting number must be
and integer and be more than 0"""
)
if not iterations >= 1:
raise ValueError("Iterations must be done more than 0 times to play FizzBuzz")
out = ""
while number <= iterations:
if number % 3 == 0:
out += "Fizz"
if number % 5 == 0:
out += "Buzz"
if 0 not in (number % 3, number % 5):
out += str(number)
# print(out)
number += 1
out += " "
return out
if __name__ == "__main__":
import doctest
doctest.testmod()