Archives java - Page 6 of 14

Hackerrank – Beautiful Binary String

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I replaced all occurences of “010” with an arbitrary placeholder. Then I counted the placeholders. I created solution in: Scala Java JavaScript Ruby All solutions are also available on my GitHub. Java

import java.util.*;

public class BeautifulBinaryString {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int l = Integer.parseInt(stdin.nextLine());
        String s = stdin.nextLine();
        String beautiful = s.replaceAll("010", "b");
        int changes = beautiful.replaceAll("[01]", "").length();
        System.out.println(changes);
        stdin.close();
    }
}

import java.util.*;

public class BeautifulBinaryString {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

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

String s = stdin.nextLine();

String beautiful = s.replaceAll("010", "b");

int changes = beautiful.replaceAll("[01]", "").length();

System.out.println(changes);

stdin.close();

}

JavaScript

'use strict';

const processData = input => {
    const lines = input.split('\n');
    const s = lines[1];
    const beautiful = s.replace(/010/g, "b");
    const changes = beautiful.replace(/[01]/g, "").length;
    console.log(changes);
};

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 s = lines[1];

const beautiful = s.replace(/010/g, "b");

const changes = beautiful.replace(/[01]/g, "").length;

console.log(changes);

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

let _input = "";

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

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

Scala

object BeautifulBinaryString {

  def main(args: Array[String]) {
    val sc = new java.util.Scanner (System.in)
    val n = sc.nextInt()
    val B = sc.next()

    val beautiful = B.replaceAll("010", "b")
    val changes = beautiful.count(_ == 'b')
    println(changes)
  }
}

object BeautifulBinaryString {

def main(args: Array[String]) {

val sc = new java.util.Scanner (System.in)

val n = sc.nextInt()

val B = sc.next()

val beautiful = B.replaceAll("010", "b")

val changes = beautiful.count(_ == 'b')

println(changes)

}

Ruby

l = gets.to_i
s = gets.to_s.strip

s.gsub!(/010/, "b")
s.gsub!(/[01]/, "")
puts s.length()

l = gets.to_i

s = gets.to_s.strip

s.gsub!(/010/, "b")

s.gsub!(/[01]/, "")

puts s.length()

Hackerrank – Circular Array Rotation

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution We rotate the array in circles. That why it is enough to calculate a position after all rotations using modulo. We split the array into two halves at position array_length – calculated_position. We change halves and concat them – the […]

Hackerrank – Jumping on the Clouds: Revisited

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I jumped for k positions and lowered the energy. If I reached position 0 I printed the remaining enenergy. I created solution in: Scala Java JavaScript Ruby All solutions are also available on my GitHub. Java

import java.util.*;

public class JumpingOnTheCloudsRevisited {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        int k = stdin.nextInt();
        int[] clouds = new int[n];
        for(int i = 0; i < n; i++) {
            clouds[i] = stdin.nextInt();
        }
        int energy = 100;
        int pos = k % n;
        energy -= ((clouds[pos] == 1) ? 3 : 1);
        while(pos != 0) {
            pos = (pos + k) % n;
            energy -= ((clouds[pos] == 1) ? 3 : 1);
        }
        System.out.println(energy);
        stdin.close();
    }
}

import java.util.*;

public class JumpingOnTheCloudsRevisited {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int n = stdin.nextInt();

int k = stdin.nextInt();

int[] clouds = new int[n];

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

clouds[i] = stdin.nextInt();

}

int energy = 100;

int pos = k % n;

energy -= ((clouds[pos] == 1) ? 3 : 1);

while(pos != 0) {

pos = (pos + k) % n;

energy -= ((clouds[pos] == 1) ? 3 : 1);

}

System.out.println(energy);

stdin.close();

}

JavaScript

'use strict';

const processData = input => {
    const lines = input.split('\n');
    const [n, k] = lines[0].split(' ').map(i => parseInt(i));
    const clouds = lines[1].split(' ').map(i => parseInt(i));

    let energy = 100;
    let pos = k % n;
    energy -= ((clouds[pos] == 1) ? 3 : 1);
    while(pos != 0) {
        pos = (pos + k) % n;
        energy -= ((clouds[pos] == 1) ? 3 : 1);
    }
    console.log(energy);
};

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 [n, k] = lines[0].split(' ').map(i => parseInt(i));

const clouds = lines[1].split(' ').map(i => parseInt(i));

let energy = 100;

let pos = k % n;

energy -= ((clouds[pos] == 1) ? 3 : 1);

while(pos != 0) {

pos = (pos + k) % n;

energy -= ((clouds[pos] == 1) ? 3 : 1);

}

console.log(energy);

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

let _input = "";

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

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

Scala […]

Hackerrank – Mars Exploration

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I divided the whole string into to 3-character parts. Then I checked for each triplet if first character is equal to S, second to O and third to S. And counted the changed letters. I created solution in: Scala Java […]

Hackerrank – Minimum Distances

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Create index pairs variations for all indexes that pair are unique – it means when (i, j) exists (j, i) is the same. Filter only this pairs of indexes, which have the same value. Make an absolute difference of them […]

Hackerrank – String Construction

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution The answer is the count of distinct characters. If the string does not contain a specific character, only then it costs 1 dollar. Because one character is a substring. When we are able to find it in a partially created […]

Hackerrank – Save the Prisoner

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution (((sweets % prisoners) + startId) % prisoners) – 1 Modulo finds the id of prisoner who gets the last poisoned candy after all circles distribution. Since the starting id S is not from first prisoner we make an offset. Do […]

Hackerrank – Kangaroo

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution There are two checks: I checked if is less then and immediatelly print “NO”. That means, kangaroo 1 would be never able to reach kangaroo 2. I calculated differences between kangaroo positions and their […]

Hackerrank – Beautiful Triplets

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I checked each consequent combination of (i, j, k), if it is valid according to condition. There is a catch. I had to stop an iteration if array[j] – array[i]

Hackerrank – Strange Counter

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I calculated time in the cycle when counter value is 1: last_time_next_cycle = actual_cycle_time + 2 * (last_time_actual_cycle – last_time_previous_cycle). The difference in seconds between cycle 0 and cycle 1 is 3, which is always like that. For a given […]

Pavol Pidanič

I can count to 1023 with 10 fingers

Tag: java

Hackerrank – Beautiful Binary String

Hackerrank – Circular Array Rotation

Hackerrank – Jumping on the Clouds: Revisited

Hackerrank – Mars Exploration

Hackerrank – Minimum Distances

Hackerrank – String Construction

Hackerrank – Save the Prisoner

Hackerrank – Kangaroo

Hackerrank – Beautiful Triplets

Hackerrank – Strange Counter