Associating by a Key

Part of the Logic & Flow section of Coddy's Kotlin journey. Lesson 39 of 63.

associateBy creates a map from a computed key to an original element. If keys repeat, the last element for that key wins. Use groupBy when every matching element must survive; use associateBy when one latest element per key is intended.

Inside main:

val latest = listOf("a", "to", "b").associateBy { it.length }
println(latest[1])

Both a and b have key one, so b replaces a for that key.

associateBy keeps the last element when computed keys collide.

Challenge

Medium

Complete lastOfLength with parameters words: List<String>, length: Int. Use associateBy with word length as the key. Return the last input word of the requested length, or missing if none exists.

Unless stricter bounds are stated above, collections contain at most 100 elements at each level, and integer arguments and integer collection values are between -1000 and 1000. Text supplied for parsing can include invalid or out-of-range representations as described.

Return a value of type String. Keep the supplied input/output code. It reads scalar arguments one per line; a list starts with its count followed by its elements, and a map starts with its entry count followed by each key and value. Nested lists repeat the count-and-elements format for each row. The supplied main prints the return value followed by one newline. Lists use Kotlin's standard bracketed format; print no additional labels.

Try it yourself

fun lastOfLength(words: List<String>, length: Int): String {
    // Write your solution here.
    return ""
}
fun main() {
    val wordsCount = readln().toInt()
    val words = mutableListOf<String>()
    for (i0 in 0 until wordsCount) {
        val wordsItem = readLine().orEmpty()
        words.add(wordsItem)
    }
    val length = readln().toInt()
    println(lastOfLength(words, length))
}