Archívy Hackerrank - Stránka 7 z 26

Hackerrank – Java Stdin and Stdout II

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

import java.util.Scanner;

public class Solution {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int i = scan.nextInt();
        double d = scan.nextDouble();
        scan.nextLine();
        String s = scan.nextLine();

        System.out.println("String: " + s);
        System.out.println("Double: " + d);
        System.out.println("Int: " + i);
    }
}

import java.util.Scanner;

public class Solution {

public static void main(String[] args) {

Scanner scan = new Scanner(System.in);

int i = scan.nextInt();

double d = scan.nextDouble();

scan.nextLine();

String s = scan.nextLine();

System.out.println("String: " + s);

System.out.println("Double: " + d);

System.out.println("Int: " + i);

}

Hackerrank – Mini-Max Sum

Hackerrank – Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Ak budú prvky usporiadané vzostupne – najmenší súčet bude mať prvých n-1 prvkov – najväčší súčet je medzi poslednými n-1 prvkami Vytvoril som riešenie v týchto programovacích jazykoch: Scala Java JavaScript Ruby Všetky riešenia sú dostupné aj na mojom GitHub profile. Java

import java.util.*;

public class MiniMaxSum {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        Object[] numbers = Arrays.stream(stdin.nextLine().split(" "))
                .map(Long::parseLong).toArray();
        Arrays.sort(numbers);
        long minSum = 0;
        long maxSum = 0;
        for(int i = 0; i < numbers.length - 1; i++) {
            minSum += (long) numbers[i];
            maxSum += (long) numbers[i + 1];
        }

        System.out.println(minSum + " " + maxSum);
        stdin.close();
    }
}

import java.util.*;

public class MiniMaxSum {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

Object[] numbers = Arrays.stream(stdin.nextLine().split(" "))

.map(Long::parseLong).toArray();

Arrays.sort(numbers);

long minSum = 0;

long maxSum = 0;

for(int i = 0; i < numbers.length - 1; i++) {

minSum += (long) numbers[i];

maxSum += (long) numbers[i + 1];

}

System.out.println(minSum + " " + maxSum);

stdin.close();

}

[…]

Hackerrank – Bon Appétit

Hackerrank – Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Vypočítame správnu sumu peňazí, ktoré je potrebné zaplatiť – (food_sum – index_of_allergic) / 2. Potom stačí porovnať, či sa vypočítaná suma zhoduje s množstom zadaným pri popise problému. Vytvoril som riešenie v týchto programovacích jazykoch: Scala Java JavaScript Ruby Všetky riešenia sú dostupné […]

Hackerrank – Sequence full of colors

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: Scala Všetky riešenia sú dostupné aj na mojom GitHub profile. Scala

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)
  }
}

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)

}

Hackerrank – Jumping Bunnies

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

import scala.io.Source

object JumpingBunnies extends App {
  val lines = Source.stdin.getLines().toList
  val bunnies = lines(1).split(" ").map(_.toInt)

  val commonPoint = bunnies.foldLeft(BigInt(1))(lcm(_, _))

  println(commonPoint)

  def gcd(x: BigInt, y: BigInt): BigInt = {
    if(y == 0) x else gcd(y, x % y)
  }

  def lcm(x: BigInt, y: BigInt): BigInt = {
    (x * y) / gcd(x, y)
  }
}

import scala.io.Source

object JumpingBunnies extends App {

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

val bunnies = lines(1).split(" ").map(_.toInt)

val commonPoint = bunnies.foldLeft(BigInt(1))(lcm(_, _))

println(commonPoint)

def gcd(x: BigInt, y: BigInt): BigInt = {

if(y == 0) x else gcd(y, x % y)

}

def lcm(x: BigInt, y: BigInt): BigInt = {

(x * y) / gcd(x, y)

}

Hackerrank – Repeated String

Hackerrank – Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Dĺžka slova opakovaného niekoľkokrát za sebou by prekročila časové limity pre toto zadanie. Musíme si ho nejako zjednodušiť. Máme zadaný reťazec – s. Spočítame výskyt znaku „a“ a označíme si ho ako – c Získaj dĺžku výsledného reťazca (zadané na druhom riadku) – […]

Hackerrank – Viral Advertising

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

import java.util.*;

public class ViralAdvertising {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int n = Integer.parseInt(stdin.nextLine());
        int totalLiked = 0;
        int dayStartPeople = 5;
        for(int i = 0; i < n; i++) {
            int dayLiked = dayStartPeople / 2;
            int received = dayLiked * 3;
            totalLiked += dayLiked;
            dayStartPeople = received;
        }
        System.out.println(totalLiked);
        stdin.close();
    }
}

import java.util.*;

public class ViralAdvertising {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

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

int totalLiked = 0;

int dayStartPeople = 5;

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

int dayLiked = dayStartPeople / 2;

int received = dayLiked * 3;

totalLiked += dayLiked;

dayStartPeople = received;

}

System.out.println(totalLiked);

stdin.close();

}

JavaScript

'use strict';

const processData = input => {
    const n = parseInt(input)

    let totalLiked = 0;
    let dayStartPeople = 5;
    for(let i = 0; i < n; i++) {
        const dayLiked = parseInt(dayStartPeople / 2);
        const received = dayLiked * 3;
        totalLiked += dayLiked;
        dayStartPeople = received;
    }
    console.log(totalLiked);
};

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

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

'use strict';

const processData = input => {

const n = parseInt(input)

let totalLiked = 0;

let dayStartPeople = 5;

for(let i = 0; i < n; i++) {

const dayLiked = parseInt(dayStartPeople / 2);

const received = dayLiked * 3;

totalLiked += dayLiked;

dayStartPeople = received;

}

console.log(totalLiked);

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

let _input = "";

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

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

Scala

import scala.io.Source

object ViralAdvertising extends App {
  val n = Source.stdin.bufferedReader().readLine().toInt

  println(like(5).take(n).sum)


  def like(start: Long): Stream[Long] = {
    val liked = start / 2
    val received = liked * 3
    liked #:: like(received)
  }
}

import scala.io.Source

object ViralAdvertising extends App {

val n = Source.stdin.bufferedReader().readLine().toInt

println(like(5).take(n).sum)

def like(start: Long): Stream[Long] = {

val liked = start / 2

val received = liked * 3

liked #:: like(received)

}

Ruby

n = gets.to_i
    
total_liked = 0
day_start_people = 5
for i in 0...n do
  day_liked = day_start_people / 2
  received = day_liked * 3
  total_liked = total_liked + day_liked
  day_start_people = received
end

puts total_liked

n = gets.to_i

total_liked = 0

day_start_people = 5

for i in 0...n do

day_liked = day_start_people / 2

received = day_liked * 3

total_liked = total_liked + day_liked

day_start_people = received

end

puts total_liked

Hackerrank – Apple and Orange

Hackerrank – Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Musíme spočítať, koľko jabĺk alebo pomarančov sa nachádza medzi zadanými bodmi na priamke. Použijeme túto rovnicu. Vytvoril som riešenie v týchto programovacích jazykoch: Scala Java JavaScript Ruby Všetky riešenia sú dostupné aj na mojom GitHub profile. Java

import java.util.*;

public class AppleAndOrange {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int s = stdin.nextInt();
        int t = stdin.nextInt();
        int a = stdin.nextInt();
        int b = stdin.nextInt();
        int m = stdin.nextInt();
        int n = stdin.nextInt();
        int[] apples = new int[m];
        for(int i = 0; i < m; i++) {
            apples[i] = stdin.nextInt();
        }
        int[] oranges = new int[n];
        for(int i = 0; i < n; i++) {
            oranges[i] = stdin.nextInt();
        }

        int fallenApples = (int) Arrays.stream(apples).filter(ap -> (a + ap >= s) && (a + ap <= t)).count();
        int fallenOranges = (int) Arrays.stream(oranges).filter(o -> (b + o >= s) && (b + o <= t)).count();

        System.out.println(fallenApples);
        System.out.println(fallenOranges);

        stdin.close();
    }
}

import java.util.*;

public class AppleAndOrange {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int s = stdin.nextInt();

int t = stdin.nextInt();

int a = stdin.nextInt();

int b = stdin.nextInt();

int m = stdin.nextInt();

int n = stdin.nextInt();

int[] apples = new int[m];

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

apples[i] = stdin.nextInt();

}

int[] oranges = new int[n];

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

oranges[i] = stdin.nextInt();

}

int fallenApples = (int) Arrays.stream(apples).filter(ap -> (a + ap >= s) && (a + ap <= t)).count();

int fallenOranges = (int) Arrays.stream(oranges).filter(o -> (b + o >= s) && (b + o <= t)).count();

System.out.println(fallenApples);

System.out.println(fallenOranges);

stdin.close();

}

JavaScript

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

var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;

process.stdin.on('data', function (data) {
    input_stdin += data;
});

process.stdin.on('end', function () {
    input_stdin_array = input_stdin.split("\n");
    main();    
});

function readLine() {
    return input_stdin_array[input_currentline++];
}

function main() {
    var s_temp = readLine().split(' ');
    var s = parseInt(s_temp[0]);
    var t = parseInt(s_temp[1]);
    var a_temp = readLine().split(' ');
    var a = parseInt(a_temp[0]);
    var b = parseInt(a_temp[1]);
    var m_temp = readLine().split(' ');
    var m = parseInt(m_temp[0]);
    var n = parseInt(m_temp[1]);
    apple = readLine().split(' ');
    apple = apple.map(Number);
    orange = readLine().split(' ');
    orange = orange.map(Number);

    var fallenApples = apple.filter(function(ap) {
        return (a + ap >= s) && (a + ap <= t);
    });

    var fallenOranges = orange.filter(function(o) {
        return (b + o >= s) && (b + o <= t);
    });

    console.log(fallenApples.length);
    console.log(fallenOranges.length);
}

process.stdin.resume();

process.stdin.setEncoding('ascii');

var input_stdin = "";

var input_stdin_array = "";

var input_currentline = 0;

process.stdin.on('data', function (data) {

input_stdin += data;

});

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

input_stdin_array = input_stdin.split("\n");

main();

});

function readLine() {

return input_stdin_array[input_currentline++];

}

function main() {

var s_temp = readLine().split(' ');

var s = parseInt(s_temp[0]);

var t = parseInt(s_temp[1]);

var a_temp = readLine().split(' ');

var a = parseInt(a_temp[0]);

var b = parseInt(a_temp[1]);

var m_temp = readLine().split(' ');

var m = parseInt(m_temp[0]);

var n = parseInt(m_temp[1]);

apple = readLine().split(' ');

apple = apple.map(Number);

orange = readLine().split(' ');

orange = orange.map(Number);

var fallenApples = apple.filter(function(ap) {

return (a + ap >= s) && (a + ap <= t);

});

var fallenOranges = orange.filter(function(o) {

return (b + o >= s) && (b + o <= t);

});

console.log(fallenApples.length);

console.log(fallenOranges.length);

}

Scala […]

Hackerrank – Equalize the Array

Hackerrank – Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Vytvoríme skupiny čísel, ktoré budú obsahovať rovnaké čísla. Výsledok je veľkosť poľa bez veľkosti najväčšej skupiny. Vytvoril som riešenie v týchto programovacích jazykoch: Scala Java JavaScript Ruby Všetky riešenia sú dostupné aj na mojom GitHub profile. Java

import java.util.*;

public class EqualizeTheArray {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        int[] array = new int[n];
        for(int i = 0; i < n; i++) {
            array[i] = stdin.nextInt();
        }
        Map<Integer, Integer> numCount = new HashMap<>();
        for(int i = 0; i < n; i++) {
            int count = numCount.getOrDefault(array[i], 0);
            numCount.put(array[i], count + 1);
        }
        int maxCount = numCount.values().stream().max(
                (Integer o1, Integer o2) -> o1 - o2
        ).orElse(0);

        System.out.println(n - maxCount);
        stdin.close();
    }
}

import java.util.*;

public class EqualizeTheArray {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int n = stdin.nextInt();

int[] array = new int[n];

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

array[i] = stdin.nextInt();

}

Map<Integer, Integer> numCount = new HashMap<>();

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

int count = numCount.getOrDefault(array[i], 0);

numCount.put(array[i], count + 1);

}

int maxCount = numCount.values().stream().max(

(Integer o1, Integer o2) -> o1 - o2

).orElse(0);

System.out.println(n - maxCount);

stdin.close();

}

JavaScript

'use strict';

const processData = input => {
    const lines = input.split('\n');
    const array = lines[1].split(' ');
    const numCounts = {};
    for(let i = 0; i < array.length; i++) {
        let count = numCounts[array[i]] || 0;
        numCounts[array[i]] = count + 1;
    }
    let maxCount = 0;
    for(let p in numCounts) {
        if(numCounts.hasOwnProperty(p) && numCounts[p] > maxCount) {
            maxCount = numCounts[p];
        }
    }
    console.log(array.length - maxCount);
};

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

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

'use strict';

const processData = input => {

const lines = input.split('\n');

const array = lines[1].split(' ');

const numCounts = {};

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

let count = numCounts[array[i]] || 0;

numCounts[array[i]] = count + 1;

}

let maxCount = 0;

for(let p in numCounts) {

if(numCounts.hasOwnProperty(p) && numCounts[p] > maxCount) {

maxCount = numCounts[p];

}

console.log(array.length - maxCount);

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

let _input = "";

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

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

Scala

import scala.io.Source

object EqualizeTheArray extends App {
  val lines = Source.stdin.getLines().toList
  val array = lines(1).split(" ").map(_.toInt)

  val maxOccuredNumber = array.groupBy(n => n).map(_._2.length).max
  val toRemove = array.length - maxOccuredNumber
  println(toRemove)
}

import scala.io.Source

object EqualizeTheArray extends App {

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

val array = lines(1).split(" ").map(_.toInt)

val maxOccuredNumber = array.groupBy(n => n).map(_._2.length).max

val toRemove = array.length - maxOccuredNumber

println(toRemove)

}

Ruby […]

Hackerrank – Sock Merchant

Hackerrank – Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Vytvoril som množiny ponožiek rovnakej farby. Pre každú množinu som vytvoril počet párov: number_of_socks / 2. Výsledok je celkový počet párov. Vytvoril som riešenie v týchto programovacích jazykoch: Scala Java JavaScript Ruby Všetky riešenia sú dostupné aj na mojom GitHub profile. Java

import java.util.*;

public class SockMerchant {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int sockCount = stdin.nextInt();
        Map<Integer, Integer> sockGroupsCount = new HashMap<>();
        for(int i = 0; i < sockCount; i++) {
            int color = stdin.nextInt();
            if(sockGroupsCount.containsKey(color)) {
                int count = sockGroupsCount.get(color);
                sockGroupsCount.put(color, count + 1);
            } else {
                sockGroupsCount.put(color, 1);
            }
        }
        int pairs = 0;
        for(int count : sockGroupsCount.values()) {
            pairs += count / 2;
        }

        System.out.println(pairs);
        stdin.close();
    }
}

import java.util.*;

public class SockMerchant {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int sockCount = stdin.nextInt();

Map<Integer, Integer> sockGroupsCount = new HashMap<>();

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

int color = stdin.nextInt();

if(sockGroupsCount.containsKey(color)) {

int count = sockGroupsCount.get(color);

sockGroupsCount.put(color, count + 1);

} else {

sockGroupsCount.put(color, 1);

}

int pairs = 0;

for(int count : sockGroupsCount.values()) {

pairs += count / 2;

}

System.out.println(pairs);

stdin.close();

}

[…]

Pavol Pidanič

Dokážem na 10 prstoch napočítať do 1023

Kategória: Hackerrank

Hackerrank – Java Stdin and Stdout II

Hackerrank – Mini-Max Sum

Hackerrank – Bon Appétit

Hackerrank – Sequence full of colors

Hackerrank – Jumping Bunnies

Hackerrank – Repeated String

Hackerrank – Viral Advertising

Hackerrank – Apple and Orange

Hackerrank – Equalize the Array

Hackerrank – Sock Merchant