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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.
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:
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Note: The solution shown above is the current solution in the Sample Programs repository as of Oct 09 2020 23:02:08. The solution was first committed on Oct 05 2020 00:20:57. As a result, documentation below may be outdated.
No 'How to Implement the Solution' section available. Please consider contributing.
No 'How to Run the Solution' section available. Please consider contributing.