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SOLVED: Checkpoint 3 Nested loop efficiency question

Forums > 100 Days of SwiftUI

Hi all, Just started this course and on Day 6 now. I had a question regarding the FizzBuzz checkpoint. I know I can solve it with a if/else if/else if/else condition, specifically by checking the && condition in the first if statement, but is the below approach not prefered for any reason? Is it less efficient to have a while loop in there to check for the && multiple condition and then break out of the loop?

Note: super beginner and just trying to understand if both approaches are a matter of preference or is one just not recommended as its not as efficient and bad practice

for i in 1...100{
    while i % 3 == 0 && i % 5 == 00{
        print ("\(i) FizzBuzz")
    if i % 3 == 0 {
        print("\(i) Fizz")
    } else if i % 5 == 0{
        print("\(i) Buzz")
    } else {
        print("\(i) ...")


I don't know about efficiency, but semantically it doesn't make much sense to do it that way.

You are checking one number (whatever the current value of i is), why would you use a loop to do so?


@roosterboy. I suspect you have not watched the checkpoint 3 video? The very first part says "Your goal is to loop from 1 to 100". That would be a good reason for OP to use a loop in his logic?


The for loop, yes. That is going over the numbers from 1 to 100.

But not the inner while loop. That is looping over the value of i each time through the for loop, so it's "looping" over... 1 number.


Oh I see My appologies, thought you were talking about the for loop. I was like, I hope not lol otherwise I got it wrong too. Yeah I agree the while loop is redundant.


Yeah makes sense. Thanks guys. I guess I was caught up in the simple solution being 'too simple' but it seems to be the right thing to do.


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