iOS repl.it - Randomisation Challenge Solution

randomisation.swift

let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

//The number of letters in alphabet equals 26

var password = alphabet[Int.random(in: 0...25)] + alphabet[Int.random(in: 0...25)] + alphabet[Int.random(in: 0...25)] + alphabet[Int.random(in: 0...25)] + alphabet[Int.random(in: 0...25)] + alphabet[Int.random(in: 0...25)]  

print(password)

I really know I am late, I've all the top solutions on this page and I really don't get what does "!" mean in alphabet.rendomElemnet()! I've been looking all over internet (in my capacity) but still haven't found a reasonable solution. I hope you guys can help me on this . Thanks

Types, say String or Int, have their variant type called Optional (Optional String written in code as String?, Optional Int written in code as Int?).
A String? can either be a String or a null value (called nil in Swift) that means "nothing".

Why do we have that? Let's say you have a program where you can define nicknames for your users, you could give it the String? type as not everybody necessarily have a nickname, so the value of a nickname variable could be either a String, either nothing (nil) if no nickname would have been defined for that user.

Now let's say you have the following function that greets your users using their nickname

func greet(personNickname: String) -> String {
    let greeting = "Hello, " + personNickname + "."
    return greeting
}

It expects the nickname as a parameter being a String and not a String?. To call your function, you could not do:

print(greet(user.nickname))

since nickname is a String? and the function expects a String, you would have a compilation error.
To effectively pass a String to your function, you would use the ! symbol that says: "I know there's a value in there, trust me and act as if it was not 'Optional'." :

print(greet(user.nickname!))

Looking at the randomElement method of the Array class in the documentation, we can see that whenever we call it, it returns a value of the type Element?:

func randomElement() -> Self.Element?

Our array alphabet being an array of String elements, calling randomElement on it would return the type String?.

Since our password is of type String and not String?, it should be composed of Strings, so using the randomElement method, we should transform the String? values into String ones by using the ! symbol/operator.

let alphabet = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"];
    var password = "";
    for i in 0...5 {
        password += alphabet.randomElement() ?? "";
    }
    print(password);

let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

//The number of letters in alphabet equals 26

var a = alphabet.randomElement()
var b = alphabet.randomElement()
var c = alphabet.randomElement()
var d = alphabet.randomElement()
var e = alphabet.randomElement()
var f = alphabet.randomElement()

var result = ""
result.append(a!)
result.append(b!)
result.append(c!)
result.append(d!)
result.append(e!)
result.append(f!)

let password = result

print(password)

let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

var password = ""
for _ in 1 ... 6 {
    password += alphabet.randomElement()!
}

print(password)

let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

let randomElements = Array(alphabet.shuffled().prefix(6))

let password = randomElements.joined()
print (password)

let alphabet = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"]
var password = ""
while password.count != 6 {
password = password + alphabet[Int.random(in: 0 ... alphabet.count - 1)]
}
print(password)

func exercise() {
let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"];

let password = alphabet.shuffled().prefix(6).joined()

print(password);

}

let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"];

let a = alphabet.randomElement() ?? "";
let b = alphabet.randomElement() ?? "";
let c = alphabet.randomElement() ?? "";
let d = alphabet.randomElement() ?? "";
let e = alphabet.randomElement() ?? "";
let f = alphabet.randomElement() ?? "";
let password = a + b + c + d + e + f;

print(password);

Hi everyone! A lot of people here using more advanced techniques to solve the challenge. It's interesting to see all the ways it can be done. I'm glad I was also exactly like the official solution to show I'm on track!

let alphabet = [
"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p",
"q", "r", "s", "t", "u", "v", "w", "x", "y", "z"
]

// The number of letters in alphabet equals 26

// a password of six random characters
var password: String = " "

for i in 1...6 {
password += alphabet.randomElement() ?? ""
}

print(password)

func exercise() {
let alphabet = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"]

// Shuffle the alphabet and pick the first 6 letters
let shuffledAlphabet = alphabet.shuffled()
let password = shuffledAlphabet.prefix(6).joined()

print(password)

}

let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

//The number of letters in alphabet equals 26

let password = alphabet.randomElement()! + alphabet.randomElement()! + alphabet.randomElement()! + alphabet.randomElement()! + alphabet.randomElement()! + alphabet.randomElement()!

print(password)

The '!' on my code refers to force unwrapping optionals. Basically what optionals are saying is, "There is a value and it is [this value] or there isn't a value at all." If there wasn't a value at all, it would be 'nil'. if tried to run with randomElement alone the code would crash as will keep running and never find a solution.

Thank you for explaining this one. I am actually trying to use the alphabet.randomElement() but I was getting compiler error -- took too long to compile. Did not know you would need the '!'.

func exercise() {
let alphabet = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"]

let password = (1...6).map { _ in alphabet.randomElement()! }.joined()

print(password)

}

let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
//The number of letters in alphabet equals 26

let password = (0..<6).compactMap { _ in
    alphabet.randomElement()}.joined()

print(password)

`func exercise() {

let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

var password = ""

for _ in 1...6{
    password += alphabet.randomElement()!
}

print(password)

}`