Toni Day 4

[tonilopezmr]

Toni's solution for day 4 challenge

Description

Easy day, I have created the recursive and iterative solution.
But the Recursive one, has parts with iterative loops :( any ideas??

Solution

Part 1

fun valid(passphrase: String): Boolean {
    val list = LinkedList(passphrase.split(" "))
    val element = list.poll()
    return validR(list, element)
  }

  tailrec fun validR(passphrase: LinkedList<String>, element: String): Boolean =
      if (passphrase.isEmpty()) true
      else if (passphrase.contains(element)) false
      else {
        val element = passphrase.poll()
        validR(passphrase, element)
      }

  fun countValid(passphrase: List<String>): Int = passphrase
      .fold(0) { acc, a -> if (valid(a)) acc + 1 else acc }

Part 2

fun containsInAnyOrder(passphrase: List<String>, element: String): Boolean =
          passphrase.fold(false) { bcc, b ->
            if (equalsInAnyOrder(element, b)) true
            else bcc
          }

  fun equalsInAnyOrder(a: String, b: String): Boolean = a.fold(true) { acc, it ->
    if (a.length != b.length) false else acc && b.contains(it)
  }

  fun valid(passphrase: String): Boolean {
    val list = LinkedList(passphrase.split(" "))
    val element = list.poll()
    return validR(list, element)
  }

  tailrec fun validR(passphrase: LinkedList<String>, element: String): Boolean =
      if (passphrase.isEmpty()) true
      else if (containsInAnyOrder(passphrase, element)) false
      else {
        val element = passphrase.poll()
        validR(passphrase, element)
      }

  fun countValid(passphrase: List<String>): Int = passphrase
      .fold(0) { acc, a -> if (valid(a)) acc + 1 else acc }

Tests battery

always the same hahah

Happy Advent of Code!

[Cotel]

Quite good. Using Kotlin collection operators will help you a lot 👍

[Cotel]

I suppose you are using LinkedList to get Dequeue behaviour. But did you know you could the same with an immutable list? (Just a bit uglier though, but in Kategory they have implemented methods for (x : xs) haskell notation like)

val list = listOf(1,2,3)
val (head, tail) = Pair(list.first, list.drop(1))
// head -> 1
// tail -> listOf(2,3)

[tonilopezmr]

You are droping in the same way, Yes I did with LinkedList to work like (x :xs)

[Cotel]

Except, drop is immutable. It returns a new list.

[tonilopezmr]

Yes, In kategory, we have several issues with OutOfMemory with drop in ListKW.fold for many elements 👎

[Cotel]

Why are you using tailrec? Did the IDE suggested it to you or you knew you had to use it?

[tonilopezmr]

No! tailrec is for recursion optimization, tailrec coerce to call the method only in the last operation in return https://kotlinlang.org/docs/reference/functions.html and https://medium.com/@JorgeCastilloPr/tail-recursion-and-how-to-use-it-in-kotlin-97353993e17f

[Cotel]

Interesting, I will search if Scala has a tailrec reserved word too 👍

[Cotel]

You could simplify this like this

fun countValid(passphrase: List<String>): Int = passphrase.count(::valid)

[Cotel]

I guess you could simplify this to

return words.all { word -> words.count { it == word } <= 1 }  

[tonilopezmr]

wooooot?

[Cotel]

all checks all the elements in a collection fulfill a predicate.

count counts how many items in a collection fulfill a predicate.

[tonilopezmr]

too much magic behind this methods

[Cotel]

Declarative magic 😉

[Cotel]

😏

[tonilopezmr]

💘

[Cotel]

This is okay but you can check if you strings are anagrams if you sort them and they are equal. That will simplify your solution a lot 👍

[tonilopezmr]

Yes but I don't what is better for performance? good thought

[Cotel]

I don't know what will be the cost of sort integrated method. But right now you have 4 nested loops. @JoseLlorensRipolles and me only have 2, I think, and we are using this approach.

[tonilopezmr]

Please NOT MERGE THIS PR I want see the @Cotel simplifications another day!