Sorting by a Key

Part of the Logic & Flow section of Coddy's Kotlin journey. Lesson 36 of 63.

sortedBy computes a comparison key for each element and returns a new list in ascending key order. sortedByDescending reverses the key order. Equal keys retain their original relative order because Kotlin's sorting is stable.

Inside main:

println(listOf("pear", "a", "kiwi", "to").sortedBy { it.length })

Words are ordered by length; pear remains before kiwi because their lengths tie.

sortedBy preserves original order among elements with equal keys.

Challenge

Medium

Complete shortWordsFirst with parameters words: List<String>. Return words in increasing length order. Preserve the original order for ties.

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