With this line of code, I have a situation where if there is one "A" in `letterList.value` all the ...

# getting-started

[Nat Strangerweather]

With this line of code, I have a situation where if there is one "A" in letterList.value all the "A"s in boardLetterList are removed. Is there a way of getting only one "A" filtered out if I only have one in my letterList?

val difference = boardLetterList.filterNot{ letterList.value!!.contains(it) }

[Youssef Shoaib [MOD]]

MutableCollection.remove() specifies:

Removes a single instance of the specified element from this collection, if it is present.

and so:

val difference = buildList(boardLetterList.size) {
    addAll(boardLetterList)
    for (letter in letterList.value!!){
        remove(letter)
    }
}

[Nat Strangerweather]

Thanks, this works nicely! 😊

[ephemient]

if you want to do this more efficiently, you can use a multiset/counting map

[ephemient]

e.g.

fun <T> Iterable<T>.removeAllOnce(elements: Iterable<T>): List<T> {
    val removeCounts = elements.groupingBy { it }.eachCountTo(mutableMapOf())
    return this.filter { element ->
        val count = removeCounts[element] ?: return@filter true
        if (count > 0) {
            removeCounts[element] = count - 1
            false
        } else true
    }
}

[Youssef Shoaib [MOD]]

Is that that much efficient though? groupingBy is iterating once over the removal elements (O(n) for speed) and adding them all into a map (O(n) for space) and then you're iterating over the original elements (O(m) for speed) and adding only the valid elements (O(m - n) for space)
Oh right, remove has to iterate over the whole list to find the element to remove. So yes, OP, the removeAllOnce solution with the multiset is significantly faster!

[ephemient]

yep, you're comparing O(m + n) < O(m * n). stdlib's Iterable.removeAll does something similar but without counting