Archives scala - Stránka 8 z 15

Hackerrank – Two Arrays

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Usporiadl som obidve zadané polia. array1 vzostupne, array2 zostupne. Spočítaj prvky na každej i-tej pozícii, či je ich súčet menší ako zadaný súčet. Vytvoril som riešenie v týchto programovacích jazykoch: Java JavaScript Scala Ruby Všetky riešenia sú dostupné aj na mojom GitHub profile. Java

import java.util.*;

public class TwoArrays {

    private static class Descending implements Comparator<Integer> {
        @Override
        public int compare(Integer i, Integer j) {
            return j - i;
        }
    }

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int tests = stdin.nextInt();
        for(int i = 0; i < tests; i++) {
            int n = stdin.nextInt();
            int sum = stdin.nextInt();
            Integer[] array1 = new Integer[n];
            Integer[] array2 = new Integer[n];
            for(int j = 0; j < n; j++){
                array1[j] = stdin.nextInt();
            }
            for(int j = 0; j < n; j++){
                array2[j] = stdin.nextInt();
            }
            Arrays.sort(array1);
            Arrays.sort(array2, new Descending());
            System.out.println(isCorrect(array1, array2, sum));
        }
        stdin.close();
    }

    private static String isCorrect(Integer[] array1, Integer[] array2, int sum) {
        for(int i = 0; i < array1.length; i++) {
            if(array1[i] + array2[i] < sum) {
                return "NO";
            }
        }
        return "YES";
    }
}

import java.util.*;

public class TwoArrays {

private static class Descending implements Comparator<Integer> {

@Override

public int compare(Integer i, Integer j) {

return j - i;

}

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int tests = stdin.nextInt();

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

int n = stdin.nextInt();

int sum = stdin.nextInt();

Integer[] array1 = new Integer[n];

Integer[] array2 = new Integer[n];

for(int j = 0; j < n; j++){

array1[j] = stdin.nextInt();

}

for(int j = 0; j < n; j++){

array2[j] = stdin.nextInt();

}

Arrays.sort(array1);

Arrays.sort(array2, new Descending());

System.out.println(isCorrect(array1, array2, sum));

}

stdin.close();

}

private static String isCorrect(Integer[] array1, Integer[] array2, int sum) {

for(int i = 0; i < array1.length; i++) {

if(array1[i] + array2[i] < sum) {

return "NO";

}

return "YES";

}

[…]

Hackerrank – The Love-Letter Mystery

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Spočítal som sumu rozdielov ASCII kódu znaku na pozícii length(s)-i-1 a ASCII kódu znaku na pozícii i. Vytvoril som riešenie v týchto programovacích jazykoch: Java Ruby JavaScript Scala Všetky riešenia sú dostupné aj na mojom GitHub profile. Java

import java.util.Scanner;

public class LoveLetterMystery {
    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int tests = stdin.nextInt();
        for(int i = 0; i < tests; i++) {
            String s = stdin.next();
            int diff = 0;
            for(int j = 0; j < s.length() / 2; j++) {
                diff += Math.abs(Character.codePointAt(s, j) - Character.codePointAt(s, s.length() - j - 1));
            }
            System.out.println(diff);
        }
        stdin.close();
    }
}

import java.util.Scanner;

public class LoveLetterMystery {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int tests = stdin.nextInt();

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

String s = stdin.next();

int diff = 0;

for(int j = 0; j < s.length() / 2; j++) {

diff += Math.abs(Character.codePointAt(s, j) - Character.codePointAt(s, s.length() - j - 1));

}

System.out.println(diff);

}

stdin.close();

}

Ruby

tests = gets.chomp.to_i
tests.times do
  input = gets.chomp
  changes = 0
  for i in 0...input.length / 2
    char_front = input[i]
    char_back = input[input.length - 1 - i]
    diff = (char_front.ord - char_back.ord).abs
    changes += diff
  end
  puts changes
end

tests = gets.chomp.to_i

tests.times do

input = gets.chomp

changes = 0

for i in 0...input.length / 2

char_front = input[i]

char_back = input[input.length - 1 - i]

diff = (char_front.ord - char_back.ord).abs

changes += diff

end

puts changes

end

JavaScript

function processData(input) {
    var lines = input.split('\n');
    for(var i = 1; i < lines.length; i++) {
        var diff = 0;
        for(var j = 0; j < lines[i].length / 2; j++) {
            diff += Math.abs(lines[i].charCodeAt(j) - lines[i].charCodeAt(lines[i].length - j - 1));
        }
        console.log(diff);
    }
}

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');

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

var diff = 0;

for(var j = 0; j < lines[i].length / 2; j++) {

diff += Math.abs(lines[i].charCodeAt(j) - lines[i].charCodeAt(lines[i].length - j - 1));

}

console.log(diff);

}

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 TheLoveLetterMystery extends App {
  val lines = Source.stdin.getLines().drop(1)
  val diffs = lines.map(changes)
  println(diffs.mkString("\n"))

  def changes(s: String): Int = {
    val halves = s.splitAt(s.length / 2)
    val reverted2 = halves._2.reverse
    halves._1.indices.map(i => ordinalDiff(halves._1.charAt(i), reverted2.charAt(i))).sum
  }

  def ordinalDiff(ch1: Char, ch2: Char): Int = {
    (ch2 - ch1).abs
  }
}

import scala.io.Source

object TheLoveLetterMystery extends App {

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

val diffs = lines.map(changes)

println(diffs.mkString("\n"))

def changes(s: String): Int = {

val halves = s.splitAt(s.length / 2)

val reverted2 = halves._2.reverse

halves._1.indices.map(i => ordinalDiff(halves._1.charAt(i), reverted2.charAt(i))).sum

}

def ordinalDiff(ch1: Char, ch2: Char): Int = {

(ch2 - ch1).abs

}

Hackerrank – The British and American Style of Spelling

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Skontroloval som pre každé slovo vo vete, či obsahuje &guot;ze" pre americkú angličtinu alebo &guot;se" pre britskú angličtinu. Vytvoril som riešenie v týchto programovacích jazykoch: Scala Ruby JavaScript Java Všetky riešenia sú dostupné aj na mojom GitHub profile. Scala

import scala.io.Source

object AmericanEnglishSpelling extends App {
  val lines = Source.stdin.getLines().toList
  val n = lines(0).toInt
  val words = lines.slice(1, n + 1).map(_.split(" ")).flatten
  val tests = lines(n + 1).toInt
  val testWords = lines.drop(n + 2)
  val counts = testWords.map(w => words.count(isUsOrUkSpelling(_, w)))
  println(counts.mkString("\n"))

  def isUsOrUkSpelling(word: String, matching: String): Boolean = {
    word.matches(matching) || word.matches(matching.replaceAll("ze", "se"))
  }
}

import scala.io.Source

object AmericanEnglishSpelling extends App {

val lines = Source.stdin.getLines().toList

val n = lines(0).toInt

val words = lines.slice(1, n + 1).map(_.split(" ")).flatten

val tests = lines(n + 1).toInt

val testWords = lines.drop(n + 2)

val counts = testWords.map(w => words.count(isUsOrUkSpelling(_, w)))

println(counts.mkString("\n"))

def isUsOrUkSpelling(word: String, matching: String): Boolean = {

word.matches(matching) || word.matches(matching.replaceAll("ze", "se"))

}

Ruby

lines = gets.chomp.to_i
all_words = []
lines.times do
  words = gets.chomp.split.map { |s| s.to_s }
  words.each { |w| all_words << w}
end

tests = gets.chomp.to_i
tests.times do
  us_word = gets.chomp.to_s
  uk_word = us_word.sub("ze", "se")
  count = 0
  all_words.each do |word|
    count += 1 if word == us_word || word == uk_word
  end
  puts count
end

lines = gets.chomp.to_i

all_words = []

lines.times do

words = gets.chomp.split.map { |s| s.to_s }

words.each { |w| all_words << w}

end

tests = gets.chomp.to_i

tests.times do

us_word = gets.chomp.to_s

uk_word = us_word.sub("ze", "se")

count = 0

all_words.each do |word|

count += 1 if word == us_word || word == uk_word

end

puts count

end

JavaScript

function processData(input) {
    var lines = input.split("\n");
    var n = parseInt(lines[0]);
    var words = [];
    for(var i = 1; i <= n; i++) {
        var arr = lines.split(" ");
        for(var j = 0; j < arr.length; j++) {
            words.push(arr[j]);
        }
    }
    var tests = parseInt(lines[n + 1]);
    for(var t = n + 1; t <= n + tests; t++) {
        var count = 0;
        var american = lines[t];
        var british = lines[t].replace(/ze/g, "se");
        for(var w = 0; w < words.length; w++) {
            if(w.match(american) || w.match(british)) {
                count++;
            }
        }
        console.log(count);
    }
}

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");

var n = parseInt(lines[0]);

var words = [];

for(var i = 1; i <= n; i++) {

var arr = lines.split(" ");

for(var j = 0; j < arr.length; j++) {

words.push(arr[j]);

}

var tests = parseInt(lines[n + 1]);

for(var t = n + 1; t <= n + tests; t++) {

var count = 0;

var american = lines[t];

var british = lines[t].replace(/ze/g, "se");

for(var w = 0; w < words.length; w++) {

if(w.match(american) || w.match(british)) {

count++;

}

console.log(count);

}

process.stdin.resume();

process.stdin.setEncoding("ascii");

_input = "";

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

_input += input;

});

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

processData(_input);

});

Java […]

Hackerrank – Split the Phone Numbers

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Regulárny výraz potrebný na rozdelenie telefónneho čísla: \s|\- Vytvoril som riešenie v týchto programovacích jazykoch: Scala Java JavaScript Ruby Všetky riešenia sú dostupné aj na mojom GitHub profile. Scala

import scala.io.Source

object SplitPhoneNumber extends App {

  private[this] val SPLIT_PATTERN = " |\\-"

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

  numbers.foreach(num => {
    val splits = num.split(SPLIT_PATTERN)
    println(s"CountryCode=${splits(0)},LocalAreaCode=${splits(1)},Number=${splits(2)}")
  })
}

import scala.io.Source

object SplitPhoneNumber extends App {

private[this] val SPLIT_PATTERN = " |\\-"

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

numbers.foreach(num => {

val splits = num.split(SPLIT_PATTERN)

println(s"CountryCode=${splits(0)},LocalAreaCode=${splits(1)},Number=${splits(2)}")

})

}

Java

import java.util.*;

public class SplitPhoneNumbers {

    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[] splits = stdin.nextLine().split(" |\\-");
            System.out.println("CountryCode=" + splits[0]
                    + ",LocalAreaCode=" + splits[1] + ",Number=" + splits[2]);
        }
        stdin.close();
    }
}

import java.util.*;

public class SplitPhoneNumbers {

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[] splits = stdin.nextLine().split(" |\\-");

System.out.println("CountryCode=" + splits[0]

+ ",LocalAreaCode=" + splits[1] + ",Number=" + splits[2]);

}

stdin.close();

}

JavaScript

'use strict';

const processData = input => {
    let numbers = input.split('\n').slice(1);
    for(let i = 0; i < numbers.length; i++) {
        let splits = numbers[i].split(/ |\-/);
        console.log("CountryCode=" + splits[0]
            + ",LocalAreaCode=" + splits[1] + ",Number=" + splits[2]);
    }
};

process.stdin.resume();
process.stdin.setEncoding("ascii");

var _input = "";
process.stdin.on("data", input => _input += input);
process.stdin.on("end", () => processData(_input));

'use strict';

const processData = input => {

let numbers = input.split('\n').slice(1);

for(let i = 0; i < numbers.length; i++) {

let splits = numbers[i].split(/ |\-/);

console.log("CountryCode=" + splits[0]

+ ",LocalAreaCode=" + splits[1] + ",Number=" + splits[2]);

}

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

var _input = "";

process.stdin.on("data", input => _input += input);

process.stdin.on("end", () => processData(_input));

Ruby

tests = gets.chomp.to_i
tests.times do
  (country_code, local_area, number) = gets.chomp.split(/ |\-/)
  puts "CountryCode=#{country_code},LocalAreaCode=#{local_area},Number=#{number}"
end

tests = gets.chomp.to_i

tests.times do

(country_code, local_area, number) = gets.chomp.split(/ |\-/)

puts "CountryCode=#{country_code},LocalAreaCode=#{local_area},Number=#{number}"

end

Hackerrank – Sherlock and The Beast

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Vytvoril som si 2 premenné: jedna vyjadruje počet_pätiek a druhá vyjadruje počet_trojok. Na začiatku si obe premenné inicializujeme počet_pätiek hodnotou dĺžky zadaného čísla a počet_trojok hodnotou 0. Zistíme či je počet_pätiek deliteľný 3 a počet_trojok deliteľný 5. Ak neplatia podmieky, potom odčítame 5 z počet_pätiek […]

Hackerrank – Rotate String

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Vytvoril som riešenie v týchto programovacích jazykoch: Scala Všetky riešenia sú dostupné aj na mojom GitHub profile. Scala

object RotateString extends App {

   val lines = io.Source.stdin.getLines().toList
   val n = lines.head.toInt
   val strings = lines.tail

   strings.foreach(s =>
     println(rotations(s, 0).mkString(" "))
   )

   def rotate(s: String): String = {
     s.tail + s.head
   }

   def rotations(s: String, n: Int): List[String] = {
     if(n == s.length) Nil else rotate(s) :: rotations(rotate(s), n + 1)
   }
 }

object RotateString extends App {

val lines = io.Source.stdin.getLines().toList

val n = lines.head.toInt

val strings = lines.tail

strings.foreach(s =>

println(rotations(s, 0).mkString(" "))

)

def rotate(s: String): String = {

s.tail + s.head

}

def rotations(s: String, n: Int): List[String] = {

if(n == s.length) Nil else rotate(s) :: rotations(rotate(s), n + 1)

}

Hackerrank – Prefix Compression

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Porovnávaj postupne znaky v reťazcoh od ich začiatku na rovnakých pozíciách. Ak nájdeme nerovnaké, prefix zodpovedá všetkých rovnakým znakom nájdených dovtedy. Vytvoril som riešenie v týchto programovacích jazykoch: Scala Všetky riešenia sú dostupné aj na mojom GitHub profile. Scala

object PrefixCompression extends App {
  
  val lines = io.Source.stdin.getLines.toArray
  val a = lines(0)
  val b = lines(1)
  val prefix = commonPrefix(a, b, 0)
  
  println(prefix + " " + a.substring(0, prefix))
  println((a.length - prefix) + " " + a.substring(prefix))
  println((b.length - prefix) + " " + b.substring(prefix))
  
  private def commonPrefix(a : String, b: String, index: Int): Int = 
  {
    if(index < 0)
    {
      return 0
    }
    if(index == a.length)
    {
      return index
    }
    if(index == b.length)
    {
      return index
    }
    if(a.charAt(index) == b.charAt(index))
    {
      return commonPrefix(a, b, index + 1)
    }
    return index
  }
}

object PrefixCompression extends App {

val lines = io.Source.stdin.getLines.toArray

val a = lines(0)

val b = lines(1)

val prefix = commonPrefix(a, b, 0)

println(prefix + " " + a.substring(0, prefix))

println((a.length - prefix) + " " + a.substring(prefix))

println((b.length - prefix) + " " + b.substring(prefix))

private def commonPrefix(a : String, b: String, index: Int): Int =

{

if(index < 0)

{

return 0

}

if(index == a.length)

{

return index

}

if(index == b.length)

{

return index

}

if(a.charAt(index) == b.charAt(index))

{

return commonPrefix(a, b, index + 1)

}

return index

}

Hackerrank – Picking Cards

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Usporiadal som pole zostupne. Načítaj N zo vstupu. Pre každý prvok e urob rozdiel N – e. Tzn. že existuje iba jedna N – e kombinácií na vybratie dvoch kariet. Všetky kombinácie kariet je súčin všetkých N – e rozdielom pre všetky prvky. Ak je […]

Hackerrank – Pangrams

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Skontrolovať, či vstup obsahuje 26 rôznych znakov. Vytvoril som riešenie v týchto programovacích jazykoch: Java JavaScript Ruby Scala Všetky riešenia sú dostupné aj na mojom GitHub profile. Java

import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;


public class Pangrams 
{
    public static void main(String[] args)
    {
        Scanner stdin = new Scanner(System.in);
        String string = stdin.nextLine().toLowerCase();
        Set<Character> alphabet = new HashSet<>();
        for(int i = 0; i < string.length(); i++)
        {
            if(string.charAt(i) != ' ')
            {
                alphabet.add(string.charAt(i));
            }
        }
        
        if(alphabet.size() == 26)
        {
            System.out.println("pangram");
        }
        else 
        {
            System.out.println("not pangram");
        }
        stdin.close();
    }
}

import java.util.HashSet;

import java.util.Scanner;

import java.util.Set;

public class Pangrams

{

public static void main(String[] args)

{

Scanner stdin = new Scanner(System.in);

String string = stdin.nextLine().toLowerCase();

Set<Character> alphabet = new HashSet<>();

for(int i = 0; i < string.length(); i++)

{

if(string.charAt(i) != ' ')

{

alphabet.add(string.charAt(i));

}

if(alphabet.size() == 26)

{

System.out.println("pangram");

}

else

{

System.out.println("not pangram");

}

stdin.close();

}

JavaScript

function processData(input) {
    var hash = {};
    input = input.toLowerCase().replace(/\s+/g, '');
    for(var i = 0; i < input.length; i++) {
        hash[input[i]] = true;
    }

    if(isPangram(hash)) {
        console.log("pangram");
    } else {
        console.log("not pangram");
    }
}

function isPangram(hash) {
    var chars = 0;
    for(prop in hash) {
        if(hash.hasOwnProperty(prop)) {
            chars++;
        }
    }
    return chars === 26;
}

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 hash = {};

input = input.toLowerCase().replace(/\s+/g, '');

for(var i = 0; i < input.length; i++) {

hash[input[i]] = true;

}

if(isPangram(hash)) {

console.log("pangram");

} else {

console.log("not pangram");

}

function isPangram(hash) {

var chars = 0;

for(prop in hash) {

if(hash.hasOwnProperty(prop)) {

chars++;

}

return chars === 26;

}

process.stdin.resume();

process.stdin.setEncoding("ascii");

_input = "";

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

_input += input;

});

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

processData(_input);

});

Ruby

require 'set'

s = Set.new(gets.downcase.gsub(/\s+/, "").chars())
if(s.size() == 26)
  puts "pangram"
else
  puts "not pangram"
end

require 'set'

s = Set.new(gets.downcase.gsub(/\s+/, "").chars())

if(s.size() == 26)

puts "pangram"

else

puts "not pangram"

end

Scala

import scala.io.Source

object Pangrams extends App {
  println(
    if(Source.stdin.bufferedReader().readLine().toLowerCase().distinct.replaceAll("\\s+", "").distinct.length == 26)
      "pangram"
    else
      "not pangram"
  )
}

import scala.io.Source

object Pangrams extends App {

println(

if(Source.stdin.bufferedReader().readLine().toLowerCase().distinct.replaceAll("\\s+", "").distinct.length == 26)

"pangram"

else

"not pangram"

)

}

Hackerrank – Pairs

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Usporiadaj zadané pole. Vyber postupne 1 element a skontroluj postupne všetky nasledujúce prvky, či je ich rodiel rovný k. Ak je už rozdiel väčší, vyber nasledujúci. Poznámka: Riešenie pre JavaScript a Ruby používajú rovnaký algoritmus, ale niektoré testy skončili timeout-om. Vytvoril som riešenie v týchto […]

Pavol Pidanič

Dokážem na 10 prstoch napočítať do 1023

Značka: scala

Hackerrank – Two Arrays

Hackerrank – The Love-Letter Mystery

Hackerrank – The British and American Style of Spelling

Hackerrank – Split the Phone Numbers

Hackerrank – Sherlock and The Beast

Hackerrank – Rotate String

Hackerrank – Prefix Compression

Hackerrank – Picking Cards

Hackerrank – Pangrams

Hackerrank – Pairs