Pavol Pidanič - Page 9 of 27 - I can count to 1023 with 10 fingers

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 – Compute the Perimeter of a Polygon

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Calculate a length of a line between each pair of following points p, defined with x and y coordinate: (p0, p1), (p1, p2), …, (pn-1, p0) Length of a line between two points: $$l = \sqrt{{(x_1 – x_2)}^2 + {(y_1 […]

Hackerrank – Compute the Area of a Polygon

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Create matrices of point pairs as follows

|x1 x2|
|y1 y2|

1 2	\|x1 x2\| \|y1 y2\|

|x2 x3|
|y2 y3|

1 2	\|x2 x3\| \|y2 y3\|

… Calculate the determinat of each matrix: x1 * y2 – x2 * y1 And sum the determinants. In the end divide the sum with 2. Solution based on this. I […]

Hackerrank – Remove Duplicates

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created solution in: Scala All solutions are also available on my GitHub. Scala

object RemoveDuplicates extends App {
  println(Console.readLine.distinct)
}

object RemoveDuplicates extends App {

println(Console.readLine.distinct)

}

Hackerrank – Lambda Calculus – Reductions #4

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created solution in: Plain Text All solutions are also available on my GitHub. Plain Text

CAN'T REDUCE

1	CAN'T REDUCE

Hackerrank – Lambda Calculus – Reductions #3

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created solution in: Plain Text All solutions are also available on my GitHub. Plain Text

CAN'T REDUCE

1	CAN'T REDUCE

Hackerrank – Lambda Calculus – Reductions #2

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created solution in: Plain Text All solutions are also available on my GitHub. Plain Text

w

Hackerrank – Lambda Calculus – Reductions #1

Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created solution in: Plain Text All solutions are also available on my GitHub. Plain Text

y

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 […]