Pavol Pidanič - Stránka 18 z 29 - Dokážem na 10 prstoch napočítať do 1023

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 – Snakes and Ladders: The Quickest Way Up

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

class Move
  attr_reader :rolls, :position
  
  def initialize(position, rolls)
    @position = position
    @rolls = rolls
  end
end

test_cases = gets.chomp.to_i
test_cases.times do
  (ladders_count, snakes_count) = gets.chomp.split(",").map { |e| e.to_i }
  ladders = gets.chomp.split
  snakes = gets.chomp.split
  ladders_hash = Hash.new
  snakes_hash = Hash.new
  ladders.each do |ladder|
    (start_p, end_p) = ladder.split(",").map { |e| e.to_i }
    ladders_hash[start_p] = end_p
  end
  
  snakes.each do |snake|
    (start_p, end_p) = snake.split(",").map { |e| e.to_i }
    snakes_hash[start_p] = end_p
  end
  
  start = Move.new(1, 0)
  moves = []
  moves.push(start)
  rolls_at_hundred = []
  while !moves.empty? do
    move = moves.shift
    actual_position = move.position
    actual_rolls = move.rolls
    
    if actual_position >= 100
      rolls_at_hundred << actual_rolls
      next 
    end
    
    for i in 1..5
      pos = ladders_hash[actual_position + i]
      if !pos.nil?
        new_move = Move.new(pos, actual_rolls + 1)
        moves.push(new_move)
      end
    end
    
    pos = ladders_hash[actual_position + 6]
    if pos.nil?
      without_ladder = Move.new(actual_position + 6, actual_rolls + 1)
      moves.push(without_ladder)
    else
      new_move = Move.new(pos, actual_rolls + 1)
      moves.push(new_move)
    end
  end

  rolls_at_hundred.sort! { |k, l| k <=> l}
  puts rolls_at_hundred[0]
end

class Move

attr_reader :rolls, :position

def initialize(position, rolls)

@position = position

@rolls = rolls

end

test_cases = gets.chomp.to_i

test_cases.times do

(ladders_count, snakes_count) = gets.chomp.split(",").map { |e| e.to_i }

ladders = gets.chomp.split

snakes = gets.chomp.split

ladders_hash = Hash.new

snakes_hash = Hash.new

ladders.each do |ladder|

(start_p, end_p) = ladder.split(",").map { |e| e.to_i }

ladders_hash[start_p] = end_p

end

snakes.each do |snake|

(start_p, end_p) = snake.split(",").map { |e| e.to_i }

snakes_hash[start_p] = end_p

end

start = Move.new(1, 0)

moves = []

moves.push(start)

rolls_at_hundred = []

while !moves.empty? do

move = moves.shift

actual_position = move.position

actual_rolls = move.rolls

if actual_position >= 100

rolls_at_hundred << actual_rolls

end

for i in 1..5

pos = ladders_hash[actual_position + i]

if !pos.nil?

new_move = Move.new(pos, actual_rolls + 1)

moves.push(new_move)

end

pos = ladders_hash[actual_position + 6]

if pos.nil?

without_ladder = Move.new(actual_position + 6, actual_rolls + 1)

moves.push(without_ladder)

else

new_move = Move.new(pos, actual_rolls + 1)

moves.push(new_move)

end

rolls_at_hundred.sort! { |k, l| k <=> l}

puts rolls_at_hundred[0]

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 […]

Hackerrank – Missing Numbers

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Urči počty výskytov čísel v oboch poliach A a B. Zisti, či sa nachádza každé číslo z B v A a aj jeho výskyt menší. Vtedy je chýbajúce. Vytvoril som riešenie v týchto programovacích jazykoch: Java JavaScript Scala Ruby Všetky riešenia sú dostupné aj na […]