Hackerrank - Alternating Characters

Problem Statement
A description of the problem can be found on Hackerrank.

Solution
Iterate through all string characters. If actual character i is different then character i-1 then use character i for next comparison. If they are not equal increment a deletion counter. Print the value of the counter.

I created solution in:

Java
JavaScript
Scala
Ruby

All solutions are also available on my GitHub.

Java

import java.util.Scanner;

public class AlternatingCharacters 
{
    public static void main(String[] args)
    {
        Scanner stdin = new Scanner(System.in);
        int tests = Integer.parseInt(stdin.nextLine());
        for(int i = 0; i < tests; i++)
        {
            String string = stdin.nextLine();
            int deletions = 0;
            char c = string.charAt(0);
            for(int j = 1; j < string.length(); j++)
            {
                if(c == string.charAt(j))
                {
                    deletions++;
                }
                else
                {
                    c = string.charAt(j);
                }
            }
            System.out.println(deletions);
        }
        
        stdin.close();
    }
}

import java.util.Scanner;

public class AlternatingCharacters

{

public static void main(String[] args)

{

Scanner stdin = new Scanner(System.in);

int tests = Integer.parseInt(stdin.nextLine());

for(int i = 0; i < tests; i++)

{

String string = stdin.nextLine();

int deletions = 0;

char c = string.charAt(0);

for(int j = 1; j < string.length(); j++)

{

if(c == string.charAt(j))

{

deletions++;

}

else

{

c = string.charAt(j);

}

System.out.println(deletions);

}

stdin.close();

}

JavaScript

function processData(input) {
    var lines = input.split("\n");
    console.log(lines.slice(1, lines.length).map(deletions).join("\n"));
}

function deletions(s) {
    var last = s.charAt(0);
    var dels = 0;
    for(var i = 1; i < s.length; i++) {
        if(last === s.charAt(i)) {
            dels++;
        } else {
            last = s.charAt(i);
        }
    }
    return dels;
}

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
    processData(_input);
});

function processData(input) {

var lines = input.split("\n");

console.log(lines.slice(1, lines.length).map(deletions).join("\n"));

}

function deletions(s) {

var last = s.charAt(0);

var dels = 0;

for(var i = 1; i < s.length; i++) {

if(last === s.charAt(i)) {

dels++;

} else {

last = s.charAt(i);

}

return dels;

}

process.stdin.resume();

process.stdin.setEncoding("ascii");

_input = "";

process.stdin.on("data", function (input) {

_input += input;

});

process.stdin.on("end", function () {

processData(_input);

});

Scala

import scala.io.Source

object AlternatingCharacters extends App {
  val strings = Source.stdin.getLines().drop(1)

  println(strings.map(deletions).mkString("\n"))

  def deletions(s: String):Int = {
    var last = s.charAt(0)
    var dels = 0
    s.drop(1).foreach {
      c => {
        if (last == c) {
          dels += 1
        } else {
          last = c
        }
      }
    }
    dels
  }
}

import scala.io.Source

object AlternatingCharacters extends App {

val strings = Source.stdin.getLines().drop(1)

println(strings.map(deletions).mkString("\n"))

def deletions(s: String):Int = {

var last = s.charAt(0)

var dels = 0

s.drop(1).foreach {

c => {

if (last == c) {

dels += 1

} else {

last = c

}

dels

}

Ruby

def deletions(s)
  last = s[0]
  dels = 0
  for i in 1..(s.length) do
    if last == s[i]
      dels += 1
    else
      last = s[i]
    end
  end
  dels
end

var tests = gets.to_i
tests.times do
  s = gets
  puts deletions(s)
end

def deletions(s)

last = s[0]

dels = 0

for i in 1..(s.length) do

if last == s[i]

dels += 1

else

last = s[i]

end

dels

end

var tests = gets.to_i

tests.times do

s = gets

puts deletions(s)

end

Pavol Pidanič

I can count to 1023 with 10 fingers

Hackerrank – Alternating Characters