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I did it pretty diffrent but it works

My code is somewhat diffrent but works.. is it still a valid way to do the challange or am i messing up ?

name = input("Please enter your name: ") number = int(input("Please enter a number: ")) print("Hi {} Hope you enjoy the game!".format(name)) print("The number you selected was {}".format(number))

def is_fizz(number): if number%3==0: return True else: return False

def is_buzz(number): if number%5==0: return True else: return False

if is_buzz(number) == True and is_fizz(number) == True: print("its a Fizzbuzz number") elif is_buzz(number) == True: print("its a buzz number") elif is_fizz(number) == True: print("its a fizz number") else: print("is neither a fizzy or a buzzy number")

1 Answer

Steven Parker
Steven Parker
229,982 Points

It's typical that as programs get more complex, there will be more ways to achieve the same result. There is rarely a single "right" way. But in addition to doing the job, you can always shoot for making the code easy to read and maintain, efficient in execution, and concise and compact.

This code looks fine as-is (good job! :+1:), but here's some tips for making it a bit more compact:

You never need to compare a boolean with "True". For example:

if is_buzz(number) == True:  # comparison is not needed
if is_buzz(number):          # just test directly!

You also don't need to test a comparison just to return a True or False:

def is_buzz(number):
    return number % 5 == 0   # just return the comparison itself

Keep up the good work, and happy coding!