Wanted to make use of what we learned about switch statements and arrays and such in #100DaysOfSwiftUI. So here to come my 6 solutions for 'Checkpoint 3' ...

100DaysOfSwitfUI - Day 6 - Checkpoint 3.swift

// Solution 1: Straightforward, I guess this is the standard way to tackle the task?!?
// Note: The order in which the conditions are checked is crucial - you see why?!?

for i in 1...100 {
        
    if i.isMultiple(of: 3) && i.isMultiple(of: 5) {
        print("FizzBuzz")
    }   else if i.isMultiple(of: 3) {
        print("Fizz")
    }   else if i.isMultiple(of: 5) {
        print("Buzz")
    }   else
    {   print("\(i)")
    }
}



// Solution 2: Instead of isMultiple you can also use the modulo operator,
// but I think the isMultiple is better to show the intend.

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



// Solution 3: We can optimize our first solution (same for the second one, if you prefer modulo) given that,
// if a number is a multiple of 3 and a multiple of 5, then it is a multiple of 15 (15, 30, 45, etc. are the
// common multiples of 3 and 5) ...

for i in 1...100 {
        
    if i.isMultiple(of: 15) {
        print("FizzBuzz")
    }   else if i.isMultiple(of: 3) {
        print("Fizz")
    }   else if i.isMultiple(of: 5) {
        print("Buzz")
    }   else
    {   print("\(i)")
    }
}



// Solution 4: Next variant - you remember that ternary operator on day 5?
// Did I mention that I like it? Less code and less comparisons
// ... but honstently, what do you think if you have to read such code?

for i in 1...100 {
        
    print( i.isMultiple(of: 3) ? "Fizz" + (i.isMultiple(of: 5) ?
        "Buzz" : "") : (i.isMultiple(of: 5) ? "Buzz" : String(i)))
}



// Solution 5: Okay, let's do it without that W?T:F magic.
// It's quite the same, but different - more readable maybe?!?

for i in 1...100 {

    var string: String

    if i.isMultiple(of: 3) {
        string = ("Fizz")
        if i.isMultiple(of: 5) {
            string.append("Buzz")
        }
    }   else {
        if i.isMultiple(of: 5) {
            string = "Buzz"
        }   else {
            string = String(i)
        }
    }
    
    print(string)
}



// Solution 6: Let's try something different. We had the switch-case on day 5 - so let's use it.
// Hint: There are 10 types of people in this world - those who understand binary and those who don't ...

for i in 1...100 {
    
    var choice = 0
    
    if i.isMultiple(of: 3) {
        choice = 1
    }
    
    if i.isMultiple(of: 5) {
        choice += 2
    }

    switch choice {
    case 1:
        print("Fizz")
    case 2:
        print("Buzz")
    case 3:
        print("FizzBuzz")
    default:
        print(String(i))
    }
}



// Solution 7: There is actually a special notation in Swift that lets us use a switch directly ...

for i in 1...100 {

    switch ( i.isMultiple(of: 3), i.isMultiple(of: 5) ) {
        case (true, true):
            print("FizzBuzz")
        case (true, false):
            print("Fizz")
        case (false, true):
            print("Buzz")
        case (false, false):
            print(String(i))
    }
}



// Solution 8: But with the approach of solution 6 we can use these arrays that we learned about on day 3 ...

var fizzBuzzChoices = ["","Fizz","Buzz","FizzBuzz"]

for i in 1...100 {
    
    fizzBuzzChoices[0] = String(i)
    
    var choice = 0
    
    if i.isMultiple(of: 3) {
        choice = 1
    }
    
    if i.isMultiple(of: 5) {
        choice += 2
    }

    print(fizzBuzzChoices[choice])
}



// Solution 9: Let's do the same spiced up with some W?T:F ... that's pretty concise, isn't it?
// But still probably hard to read and to "see" what's the intention here ...

let buzzFizzChoices = ["","Fizz","Buzz","FizzBuzz"]

for i in 1...100 {
        
    let choice = (i.isMultiple(of: 3) ? 1 : 0) + (i.isMultiple(of: 5) ? 2 : 0)
    
    print( choice == 0 ? String(i) : buzzFizzChoices[choice] )
}



// Solution 10: Talking of arrays - this solution might look a bit like "brute force" ...
// but think about it: It uses a loop as demanded, it is very very easy to understand,
// and probably it has a better performance than all the previous suggestions ...

let fizzBuzz = ["1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz", "11", "Fizz",
        "13", "14", "FizzBuzz", "16", "17", "Fizz", "19", "Buzz", "Fizz", "22", "23", "Fizz", "Buzz",
        "26", "Fizz", "28", "29", "FizzBuzz", "31", "32", "Fizz", "34", "Buzz", "Fizz", "37", "38",
        "Fizz", "Buzz", "41", "Fizz", "43", "44", "FizzBuzz", "46", "47", "Fizz", "49", "Buzz", "Fizz",
        "52", "53", "Fizz", "Buzz", "56", "Fizz", "58", "59", "FizzBuzz", "61", "62", "Fizz", "64",
        "Buzz", "Fizz", "67", "68", "Fizz", "Buzz", "71", "Fizz", "73", "74", "FizzBuzz", "76", "77",
        "Fizz", "79", "Buzz", "Fizz", "82", "83", "Fizz", "Buzz", "86", "Fizz", "88", "89", "FizzBuzz",
        "91", "92", "Fizz", "94", "Buzz", "Fizz", "97", "98", "Fizz", "Buzz",
    ]

for i in 0..<100 {
    print(fizzBuzz[i])
}



// That's it. Any other ideas? Which one do you prefer? What does your #100DaysOfSwiftUI ‘Checkpoint 3’
// solution look like?!? Would love to learn how others solved it!

Awesome work