Sequence full of colors – HackerRank Solution

In this post, we will solve Sequence full of colors HackerRank Solution. This problem (Sequence full of colors) is a part of HackerRank Functional Programming series.

Task

You are given a sequence of N balls in 4 colors: red, green, yellow and blue. The sequence is full of colors if and only if all of the following conditions are true:

• There are as many red balls as green balls.
• There are as many yellow balls as blue balls.
• Difference between the number of red balls and green balls in every prefix of the sequence is at most 1.
• Difference between the number of yellow balls and blue balls in every prefix of the sequence is at most 1.

Your task is to write a program, which for a given sequence prints True if it is full of colors, otherwise it prints False.

Input

In the first line there is one number T denoting the number of tests cases.
T lines follow. In each of them there is a sequence of letters {R, G, Y, B} denoting the input sequence (R – red, G– green, Y– yellow, B – blue).

Output

For each test case, print True if this is a sequence full of colors, otherwise print False.

Constraints

• 1 <= T <= 10
• Sequence will only consists of letters {R, G, Y, B}.
• Sum of length of all sequences will not exceed 10⁶.

Sample Input

4
RGGR
RYBG
RYRB
YGYGRBRB

Sample Output

True
True
False
False

Explanation

In the first two test cases, all four conditions are satisfied.
In the third test case, condition #1 fails as there are more red balls than green balls and condition #3 also fails for prefix “RYR” as the difference between the number of red and green balls is more than 1. In the fourth test, for a prefix “YGYG” condition 4^th fails.

Solution – Sequence full of colors – HackerRank Solution

Scala

import java.util.Scanner

object Solution {

  def main(args: Array[String]): Unit = {
    val sc = new Scanner(System.in)

    val t = sc.nextInt
    sc.nextLine

    val colors = "RGYB"

    (0 until t).foreach(_ => {
      val s = sc.nextLine

      val counts = colors.map(c => c -> s.count(_ == c)).toMap

      println(if (counts('R') == counts('G') && counts('Y') == counts('B') && prefix(s.toList, 'R', 'G') && prefix(s.toList, 'R', 'G')) "True" else "False")
    })

    sc.close()
  }

  def prefix(s: List[Char], c0: Char, c1: Char): Boolean = {
    @scala.annotation.tailrec
    def inner(s: List[Char], delta: Int): Boolean = s match {
      case Nil => true
      case x :: xs =>
        val nextDelta = delta + (x match {
          case `c0` => -1
          case `c1` => 1
          case _ => 0
        })

        math.abs(nextDelta) <= 1 && inner(xs, nextDelta)
    }

    inner(s, 0)
  }
}

Note: This problem (Sequence full of colors) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.