:question: What's the cleanest way to test if all ...

# getting-started

[voddan]

❓ What's the cleanest way to test if all elements in a list are distinct?
The best I could come up with was list.size == list.distinct().size, which is efficient, but ugly

[hho]

Unless the list in question is huge this is the best solution. I don't think it's that ugly, either – just put it in an extension:

fun Collection<*> allDistinct(): Boolean = this.size == this.distinct().size

[sitepodmatt]

distinct converts it to a mutableset then toList()

[sitepodmatt]

which should be fine, but if the list is large then you false-fast when you find the first duplicate

[sitepodmatt]

fun <T> List<T>.isDistinct() = this.fold(mutableMapOf<T,Int>()) { acc, i ->
    acc.also {
        it[i] = ((it[i] ?: 0) + 1).also { count ->
            if(count > 1) {
                return false
            }
        }
    }
}.let { true }

[sitepodmatt]

you would have to benchmark to see if there was any real performance gains to be had

[sitepodmatt]

if its sorted there much quicker ways as you only have to compare the previous element

[marstran]

Could also go for a procedural approach that stops immediately when it finds a duplicate:

fun <T> List<T>.isDistinct(): Boolean {
    val mutSet = mutableSetOf<T>()
    for (elem in this) {
        if (elem in mutSet) {
            return false
        }
        mutSet.add(elem)
    }
    return true
}

[Pavlo Liapota]

Or just

fun <T> List<T>.isDistinct(): Boolean {
    val mutSet = mutableSetOf<T>()
    return all { mutSet.add(it) }
}

[voddan]

Nice optimizations! Feels like such a method could be in the stdlib. Do you think I should open an issue?

[voddan]

Here is the issue https://youtrack.jetbrains.com/issue/KT-30270
@Pavlo Liapota I hope you don't mind I posted your code there