Hackerrank - Kangaroo - Pavol Pidanič

Hackerrank – Problem Statement

A description of the problem can be found on Hackerrank.

Solution

There are two checks:

I checked if
$v_1$

is less then

$v_2$

and immediatelly print "NO". That means, kangaroo 1 would be never able to reach kangaroo 2.
I calculated differences between kangaroo positions
$x_d$

and their jumps

$v_d$

. If

$x_d % v_d == 0$

than kangaroo 1 will reach kangaroo 2. Else "NO".

I created solution in:

Scala
Java
JavaScript
Ruby

All solutions are also available on my GitHub.

Java

import java.util.*;

public class Kangaroo {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int x1 = stdin.nextInt();
        int v1 = stdin.nextInt();
        int x2 = stdin.nextInt();
        int v2 = stdin.nextInt();

        System.out.println(isSameLocation(x1, v1, x2, v2));
        stdin.close();
    }

    private static String isSameLocation(int x1, int v1, int x2, int v2) {
        if(v1 < v2) return "NO";
        if(x1 < x2 && v1 > v2) {
            int xDiff = Math.abs(x1 - x2);
            int vDiff = Math.abs(v1 - v2);
            if(xDiff % vDiff == 0) {
                return "YES";
            } else {
                return "NO";
            }
        }
        return "NO";
    }
}

import java.util.*;

public class Kangaroo {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int x1 = stdin.nextInt();

int v1 = stdin.nextInt();

int x2 = stdin.nextInt();

int v2 = stdin.nextInt();

System.out.println(isSameLocation(x1, v1, x2, v2));

stdin.close();

}

private static String isSameLocation(int x1, int v1, int x2, int v2) {

if(v1 < v2) return "NO";

if(x1 < x2 && v1 > v2) {

int xDiff = Math.abs(x1 - x2);

int vDiff = Math.abs(v1 - v2);

if(xDiff % vDiff == 0) {

return "YES";

} else {

return "NO";

}

return "NO";

}

JavaScript

'use strict';

const processData = input => {
    const lines = input.split('\n');
    const [x1, v1, x2, v2] = lines[0].split(' ').map(i => parseInt(i));
    console.log(sameLocation(x1, v1, x2, v2));
};

const sameLocation = (x1, v1, x2, v2) => {
    if(v1 < v2) return "NO";
    if(x1 < x2 && v1 > v2) {
        const xDiff = Math.abs(x1 - x2);
        const vDiff = Math.abs(v1 - v2);
        if(xDiff % vDiff == 0) {
            return "YES";
        } else {
            return "NO";
        }
    }
    return "NO";
};

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 [x1, v1, x2, v2] = lines[0].split(' ').map(i => parseInt(i));

console.log(sameLocation(x1, v1, x2, v2));

};

const sameLocation = (x1, v1, x2, v2) => {

if(v1 < v2) return "NO";

if(x1 < x2 && v1 > v2) {

const xDiff = Math.abs(x1 - x2);

const vDiff = Math.abs(v1 - v2);

if(xDiff % vDiff == 0) {

return "YES";

} else {

return "NO";

}

return "NO";

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

let _input = "";

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

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

Scala

object Kangaroo extends App {
  val lines = Source.stdin.getLines().toList
  val numbers = lines.head.split(" ").map(_.toInt)
  val kangaroo1 = (numbers(0), numbers(1))
  val kangaroo2 = (numbers(2), numbers(3))

  println(if(sameLocation(kangaroo1, kangaroo2)) "YES" else "NO")

  def sameLocation(k1: (Int, Int), k2: (Int, Int)): Boolean = {
    val x1 = k1._1
    val x2 = k2._1
    val v1 = k1._2
    val v2 = k2._2
    if(v1 < v2) return false
    if(x1 < x2 && v1 > v2) {
      val xDiff = (x1 - x2).abs
      val vDiff = (v1 - v2).abs
      if(xDiff % vDiff == 0) {
        return true
      } else {
        return false
      }
    }
    false
  }
}

object Kangaroo extends App {

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

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

val kangaroo1 = (numbers(0), numbers(1))

val kangaroo2 = (numbers(2), numbers(3))

println(if(sameLocation(kangaroo1, kangaroo2)) "YES" else "NO")

def sameLocation(k1: (Int, Int), k2: (Int, Int)): Boolean = {

val x1 = k1._1

val x2 = k2._1

val v1 = k1._2

val v2 = k2._2

if(v1 < v2) return false

if(x1 < x2 && v1 > v2) {

val xDiff = (x1 - x2).abs

val vDiff = (v1 - v2).abs

if(xDiff % vDiff == 0) {

return true

} else {

return false

}

false

}

Ruby

def same_location(x1, v1, x2, v2)
  return "NO" if(v1 < v2)
  if(x1 < x2 && v1 > v2)
    xDiff = (x1 - x2).abs
    vDiff = (v1 - v2).abs
    if(xDiff % vDiff == 0)
      return "YES"
    else
      return "NO"
    end
  end
  "NO"
end

(x1, v1, x2, v2) = gets.strip.split(" ").map {|i| i.to_i}
puts same_location(x1, v1, x2, v2)

def same_location(x1, v1, x2, v2)

return "NO" if(v1 < v2)

if(x1 < x2 && v1 > v2)

xDiff = (x1 - x2).abs

vDiff = (v1 - v2).abs

if(xDiff % vDiff == 0)

return "YES"

else

return "NO"

end

"NO"

end

(x1, v1, x2, v2) = gets.strip.split(" ").map {|i| i.to_i}

puts same_location(x1, v1, x2, v2)

First. I updated the description because I made a mistake in modulo. It was v_d % x_d == 0. Now it is x_d % v_d == 0.
Second, this modulo equation is saying that I will find if two kangaroos would eventually meet on an arbitrary position x. You could find the same result calculating all next kangaroos positions. But with it you are able to find infinite kangaroo jumps, when they always jump next to each other without meeting on the same position

For example, imagine that kangaroo1 starts at x1 = 1 with v1 = 4 and kangaroo2 starts at x2 = 4 with v2 = 2. Their x_d = 3 and v_d = 2. We see that v_d % x_d != 0 => 3 % 2 = 1, so they should do not met.
Let’s see “brute force” solution:
First jumps x1 = 5, x2 = 6
Second jumps x1 = 9, x2 = 8
Kangaroo1 is already before Kangaroo2 (they did not meet) and it is faster than kangaroo2. They will never meet.

If you change that kangaroo1 starts at x1 = 1 with v1 = 4 and kangaroo2 starts at x2 = 5 with v2 = 2 using same calculations, you will see that it proves v_d % x_d == 0

2 thoughts on “Hackerrank – Kangaroo”

Keanu says:

19. July 2017 at 13:12

Hi, i tried a variation of your code as such:
//
function kangaroo(x1, v1, x2, v2) {
// Complete this function
var count = 0;
if(x2>x1 && v2 > v1){
return “NO”;
}

if(x1v2){
var bedaX = Math.abs(x1-x2);
var bedaV = Math.abs(v1-v2);
if(bedaX % bedaV == 0){
return “YES”;
}else{
return “NO”;
}
}else{
return “NO”;
}
}
//
and it works, but may I ask how what do you mean by “if vd” in your description above, and why does the 1st kangaroo catch up with the 2nd one if the modulus between the difference in X and V is 0?

1. Pavol Pidanič says:
  
  23. July 2017 at 7:26
  
  First. I updated the description because I made a mistake in modulo. It was v_d % x_d == 0. Now it is x_d % v_d == 0.
  Second, this modulo equation is saying that I will find if two kangaroos would eventually meet on an arbitrary position x. You could find the same result calculating all next kangaroos positions. But with it you are able to find infinite kangaroo jumps, when they always jump next to each other without meeting on the same position
  
  For example, imagine that kangaroo1 starts at x1 = 1 with v1 = 4 and kangaroo2 starts at x2 = 4 with v2 = 2. Their x_d = 3 and v_d = 2. We see that v_d % x_d != 0 => 3 % 2 = 1, so they should do not met.
  Let’s see “brute force” solution:
  First jumps x1 = 5, x2 = 6
  Second jumps x1 = 9, x2 = 8
  Kangaroo1 is already before Kangaroo2 (they did not meet) and it is faster than kangaroo2. They will never meet.
  
  If you change that kangaroo1 starts at x1 = 1 with v1 = 4 and kangaroo2 starts at x2 = 5 with v2 = 2 using same calculations, you will see that it proves v_d % x_d == 0

Pavol Pidanič

I can count to 1023 with 10 fingers

Hackerrank – Kangaroo

Hackerrank – Problem Statement

Solution

Java

JavaScript

Scala

Ruby

2 thoughts on “Hackerrank – Kangaroo”

Leave a Reply Cancel reply