Longest Common Subsequence in Kotlin

Welcome to the Longest Common Subsequence in Kotlin page! Here, you'll find the source code for this program as well as a description of how the program works.

Current Solution

private fun lcsOf(a: MutableList<String>, b: MutableList<String>, indexA: Int, indexB: Int): MutableList<String> {
    return if (indexA < 0 || indexB < 0) {
        mutableListOf()
    } else if (a[indexA] == b[indexB]) {
        // get the best subsequence of the rest, then add this one at the end (prevents needing to reverse at the end)
        lcsOf(a, b, indexA - 1, indexB - 1).also { it.add(a[indexA] )}
    } else {
        // compare both subsequences and return the one that has more element
        val subA = lcsOf(a, b, indexA, indexB - 1)
        val subB = lcsOf(a, b, indexA - 1, indexB)
        if (subA.size >= subB.size) subA else subB
    }
}

// only require consumers to pass in the lists... we'll handle the indices ourselves
fun lcsOf(a: List<String>, b: List<String>) = lcsOf(a.toMutableList(), b.toMutableList(), a.size - 1, b.size - 1)

fun main(args: Array<String>) {
    if (args.size != 2 || args[0].isBlank() || args[1].isBlank()) {
        // print and exit if we don't have the correct number of arguments
        println("Usage: please provide two lists in the format \"1, 2, 3, 4, 5\"")
        return
    }

    // split each argument by comma, remove whitespace around each element, and pack them all in a list
    val seqA = args[0].split(",").map { it.trim() }
    val seqB = args[1].split(",").map { it.trim() }

    lcsOf(seqA, seqB).joinToString(", ").also { println(it) }
}

Longest Common Subsequence in Kotlin was written by:

• Blake.Ke
• David Phillips

If you see anything you'd like to change or update, please consider contributing.

How to Implement the Solution

No 'How to Implement the Solution' section available. Please consider contributing.

How to Run the Solution

No 'How to Run the Solution' section available. Please consider contributing.