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iOS Swift Basics (retired) Control Flow Exercise: FizzBuzz Generator

Soham Ray
Soham Ray
884 Points
Posted by Soham Ray
Soham Ray
884 Points

Your answer is inefficient, why do you need to check for a fizzbuzz, if u just do fizz as a print instead of println...

This code is more efficient I believe tell me if I'm wrong

let inputNumber = 20 if inputNumber%3==0 {print("Fizz")} if inputNumber%5==0 {println("Buzz")}

of course you can change the let in the code for different numbers

erika harvey
erika harvey
Courses Plus Student 14,540 Points

i'm not sure I totally understand your question, but in fizz buzz you have to check for fizzbuzz before fizz or buzz because you tell it to stop checking if those two conditions are met andif not then you can continue to check for fizz or buzz. (if you don't do it this way, the computer would either hit three or five and stop not checking all the way for 3 or 5)

1 Answer

Animesh Mishra
Animesh Mishra
2,393 Points
Animesh Mishra
Animesh Mishra
2,393 Points

Hi, the task is to print:

  • "FizzBuzz" if the number is divisible by BOTH 3 and 5,
  • "Fizz" if the number is divisible by 3 but NOT 5, and
  • "Buzz" if the number is divisible by 5 but NOT 3

Your solution is correct too. You are first checking if the number is divisible by 3, printing "Fizz" if it is, then checking if it is divisible by 5, printing "Buzz" in the same line followed by a newline (through println), hence making it look like "FizzBuzz". It works (although a bit hard to read for me). The only difference between yours and Amit's solution is the input that the program is expecting.

EDIT: Swift 2 has deprecated println() so your solution wouldn't work anymore.

In your algorithm, the program logic is expecting ONE single number to be checked. Amit's program logic is expecting a RANGE of numbers to be checked against the rules and output to be printed accordingly. Hence the difference in implementation. There are no inefficiencies involved. Hope this clears everything up.