Hello, I just started learning Kotlin. I am trying...

# getting-started

[Arun M]

Hello, I just started learning Kotlin. I am trying to remove duplicates in a character mutable list. I did not find appropriate idiom / function to achieve this. Could anyone knowledgeable about this, please help?
var charList = ('a', 'a', 'b', 'c', 'c') ----> Expected o/p ('a', 'b', 'c')

[Sam]

My first thought would be the distinct function

[Sam]

That will create a copy of the list. I’m not sure about mutating one. You’ll find that many of the Kotlin idioms and stdlib functions prefer immutability.

[CLOVIS]

For naming conventions: if they're in the past tense, they create a new immutable value, if they're in the present tense, they mutate the current value:

list.sort()   // mutates
list.sorted() // immutable copy

[CLOVIS]

Specifically for this case, there is no efficient way to find duplicates in a List. The efficient data structure to avoid duplicates is a Set .

.disctint() actually creates a Set , adds all elements to it (thus removing duplicates), then creates a new List and adds back all elements to it. Depending on what you want to do, it may be a better idea to use a Set from the start and completely avoid List here

[Klitos Kyriacou]

Just to be a bit nerdy on naming conventions: they create a new immutable value if they are adjectives. Names like sorted are past participle adjectives. Names like distinct are just plain vanilla adjectives.

[CLOVIS]

Ahah, thanks for the precision

[ephemient]

if it's a mutable collection,

charList.retainAll(mutableSetOf<Char>()::add)

will remove all duplicates

[CLOVIS]

Wow that's more condensed that I thought possible 😅 I didn't know it was allowed to simultaneously create an instance and use it as a method reference directly, clever

[ephemient]

this is equivalent to

val set = mutableSetOf<Char>()
charList.retainAll { set.add(it) }

which is equivalent to

val set = mutableSetOf<Char>()
val iterator = charList.iterator()
while (iterator.hasNext()) {
    if (!set.add(iterator.next() {
        iterator.remove()
    }
}