Archives ruby - Page 2 of 10

Hackerrank – Designer PDF Viewer

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Find a max height of given characters. Square is max height multiplied with length of the string. I created solution in: Scala Java JavaScript Ruby All solutions are also available on my GitHub. Java

import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        Object[] heights = Arrays.stream(stdin.nextLine().split(" "))
                .map(Integer::parseInt).toArray();
        String word = stdin.nextLine();
        int maxHeight = 0;
        for(int i = 0; i < word.length(); i++) {
            int charIndex = word.charAt(i) - 'a';
            int height = (int) heights[charIndex];
            if(height > maxHeight) {
                maxHeight = height;
            }
        }

        long square = word.length() * maxHeight;
        System.out.println(square);
        stdin.close();
    }
}

import java.util.*;

public class Solution {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

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

.map(Integer::parseInt).toArray();

String word = stdin.nextLine();

int maxHeight = 0;

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

int charIndex = word.charAt(i) - 'a';

int height = (int) heights[charIndex];

if(height > maxHeight) {

maxHeight = height;

}

long square = word.length() * maxHeight;

System.out.println(square);

stdin.close();

}

JavaScript

'use strict';

const processData = input => {
    const lines = input.split('\n');
    const heights = lines[0].split(' ').map(i => parseInt(i));
    const word = lines[1];
    let maxHeight = 0;
    for(let i = 0; i < word.length; i++) {
        const charIndex = word[i].codePointAt() - 97;
        const height = heights[charIndex];
        if(height > maxHeight) {
            maxHeight = height;
        }
    }

    const square = word.length * maxHeight;
    console.log(square);
};

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 heights = lines[0].split(' ').map(i => parseInt(i));

const word = lines[1];

let maxHeight = 0;

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

const charIndex = word[i].codePointAt() - 97;

const height = heights[charIndex];

if(height > maxHeight) {

maxHeight = height;

}

const square = word.length * maxHeight;

console.log(square);

};

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 DesignerPdfViewer extends App {
  val lines = Source.stdin.getLines().toList
  val heights = lines.head.split(" ").map(_.toInt)
  val word = lines(1)
  val characterIndexes = word.map(_ - 'a')
  val maxHeight = characterIndexes.map(heights(_)).max

  val square = word.length * maxHeight
  println(square)
}
}

import scala.io.Source

object DesignerPdfViewer extends App {

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

val heights = lines.head.split(" ").map(_.toInt)

val word = lines(1)

val characterIndexes = word.map(_ - 'a')

val maxHeight = characterIndexes.map(heights(_)).max

val square = word.length * maxHeight

println(square)

}

Ruby […]

Hackerrank – Sum vs XOR

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution A mathematical trick. just count total numbers of zero present in binary number of given n, and answer will be the 2 to the power of (total num of zero) Look at the truth table of XOR:

0 ^ 0 = 0
0 ^ 1 = 1
1 ^ 0 = 1
1 ^ 1 = 0

0 ^ 0 = 0

0 ^ 1 = 1

1 ^ 0 = 1

1 ^ 1 = 0

Now look […]

Hackerrank – Flatland Space Stations

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Fist I sorted the Space station, because they are not always ordered. Then I found out how many cities are between each pair of space stations i and i+1: Then I calculated the maximum distance for each city […]

Hackerrank – Beautiful Days at the Movies

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution For each number in the given interval between i and j count the numbers with valid given condition. I created solution in: Scala Java JavaScript Ruby All solutions are also available on my GitHub. Java

import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int start = stdin.nextInt();
        int end = stdin.nextInt();
        int k = stdin.nextInt();
        int count = 0;
        for(int i = start; i <= end; i++) {
            String s = String.valueOf(i);
            String reversed = new StringBuilder(s).reverse().toString();
            int diff = Math.abs(Integer.parseInt(s) - Integer.parseInt(reversed));
            if(diff % k == 0) {
                count++;
            }
        }
        System.out.println(count);
        stdin.close();
    }
}

import java.util.*;

public class Solution {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int start = stdin.nextInt();

int end = stdin.nextInt();

int k = stdin.nextInt();

int count = 0;

for(int i = start; i <= end; i++) {

String s = String.valueOf(i);

String reversed = new StringBuilder(s).reverse().toString();

int diff = Math.abs(Integer.parseInt(s) - Integer.parseInt(reversed));

if(diff % k == 0) {

count++;

}

System.out.println(count);

stdin.close();

}

JavaScript

'use strict';

const reverse = (s) => s.split("").reverse().join("");

const processData = input => {
    const [start, end, d] = input.split(' ').map(i => parseInt(i));
    let count = 0;
    for(let i = start; i <= end; i++) {
        const s = i.toString();
        const reversed = reverse(s);
        const diff = Math.abs(parseInt(s) - parseInt(reversed));
        if(diff % d == 0) {
            count++;
        }
    }
    console.log(count);
};

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 reverse = (s) => s.split("").reverse().join("");

const processData = input => {

const [start, end, d] = input.split(' ').map(i => parseInt(i));

let count = 0;

for(let i = start; i <= end; i++) {

const s = i.toString();

const reversed = reverse(s);

const diff = Math.abs(parseInt(s) - parseInt(reversed));

if(diff % d == 0) {

count++;

}

console.log(count);

};

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 BeautifulDaysAtTheMovies extends App {
  val nums = Source.stdin.bufferedReader().readLine().split(" ").map(_.toInt)
  val start = nums(0)
  val end = nums(1)
  val d = nums(2)

  val beautifulNum = (start to end).count(num => {
    val reversed = num.toString.reverse.toInt
    val diff = (num - reversed).abs
    diff % d == 0
  })

  println(beautifulNum)
}

import scala.io.Source

object BeautifulDaysAtTheMovies extends App {

val nums = Source.stdin.bufferedReader().readLine().split(" ").map(_.toInt)

val start = nums(0)

val end = nums(1)

val d = nums(2)

val beautifulNum = (start to end).count(num => {

val reversed = num.toString.reverse.toInt

val diff = (num - reversed).abs

diff % d == 0

})

println(beautifulNum)

}

[…]

Hackerrank – Super Reduced String

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created an accumulator with initial value as empty string. I iterated through each character and updated the accumulator if an actual character is equals to last accumulator character, delete this character from the accumulator else append the character to […]

Hackerrank – Jumping on the Clouds

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created a pointer for an actual position which starts at index 0. I always tried to jump at position i + 2. If it is not possible because of thundercloud I step at i + 1 position. I do […]

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 […]

Pavol Pidanič

I can count to 1023 with 10 fingers

Tag: ruby

Hackerrank – Designer PDF Viewer

Hackerrank – Sum vs XOR

Hackerrank – Flatland Space Stations

Hackerrank – Beautiful Days at the Movies

Hackerrank – Super Reduced String

Hackerrank – Jumping on the Clouds

Hackerrank – Beautiful Binary String

Hackerrank – Circular Array Rotation

Hackerrank – Jumping on the Clouds: Revisited

Hackerrank – Mars Exploration