Hackerrank – Popis problému
Celý popis zadania sa nacháza – Hackerrank.
Riešenie
Implementoval som rekurzívnu funkciu, ktorá berie do úvahy zadané podmienky – pozri jej implementáciu.
Vytvoril som riešenie v týchto programovacích jazykoch:
Všetky riešenia sú dostupné aj na mojom GitHub profile.
Scala
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import scala.io.Source import scala.collection.mutable.Map object SequenceFullOfColors extends App { val lines = Source.stdin.getLines().toList.tail val results = lines.map(fullOfColors).map(_.toString.capitalize) println(results.mkString("\n")) def fullOfColors(sequence: String): Boolean = { val colors = Map.empty[Char, Int] def prefixSequences(seq: String, n: Int): Boolean = { val char = seq(n) val count = colors.getOrElse(char, 0) colors.put(char, count + 1) if(n < seq.length - 1) { val greenCount = colors.getOrElse('G', 0) val redCount = colors.getOrElse('R', 0) val yellowCount = colors.getOrElse('Y', 0) val blueCount = colors.getOrElse('B', 0) if((greenCount - redCount).abs <= 1 && (yellowCount - blueCount).abs <= 1) { return prefixSequences(seq, n + 1) } return false } true } val condition = prefixSequences(sequence, 0) val greenCount = colors.getOrElse('G', 0) val redCount = colors.getOrElse('R', 0) val yellowCount = colors.getOrElse('Y', 0) val blueCount = colors.getOrElse('B', 0) condition && (greenCount == redCount) && (yellowCount == blueCount) } } |