# HackerRank Number Line Jumps solution

#### ByBrokenprogrammers

Dec 12, 2022

Hello Programmers, In this post, you will know how to solve the HackerRank Number Line Jumps solution. This problem is a part of the HackerRank Algorithms Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## HackerRank Number Line Jumps Solution

You are choreographing a circus show with various animals. For one act, you are given two kangaroos on a number line ready to jump in the positive direction (i.e, toward positive infinity).

• The first kangaroo starts at location x1 and moves at a rate of v1 meters per jump.
• The second kangaroo starts at location x2 and moves at a rate of v2 meters per jump.

You have to figure out a way to get both kangaroos at the same location at the same time as part of the show. If it is possible, return `YES`, otherwise return `NO`.

Example

x1 = 2
v1 = 1
x2 = 1
v2 = 2
After one jump, they are both at x = 3, (x1 + v1 = 2 + 1, x2 + v2 = 1 + 2), so the answer is `YES`.

Function Description

Complete the function kangaroo in the editor below.

kangaroo has the following parameter(s):

• int x1, int v1: starting position and jump distance for kangaroo 1
• int x2, int v2: starting position and jump distance for kangaroo 2

Returns

• string: either `YES` or `NO`

Input Format

A single line of four space-separated integers denoting the respective values of x1, v1, x2, and v2.

Constraints

• 0 <= x1 < x2 < 10000
• 1 <= v1 < 10000
• 1 <= v2 <= 10000

Sample Input 0

```0 3 4 2
```

Sample Output 0

```YES
```

Explanation 0

The two kangaroos jump through the following sequence of locations:

From the image, it is clear that the kangaroos meet at the same location (number  on the number 12 line) after same number of jumps ( 4 jumps), and we print `YES`.

Sample Input 1

```0 2 5 3
```

Sample Output 1

```NO
```

Explanation 1

The second kangaroo has a starting location that is ahead (further to the right) of the first kangaroo’s starting location (i.e., x2 > x1). Because the second kangaroo moves at a faster rate (meaning v2 > v1) and is already ahead of the first kangaroo, the first kangaroo will never be able to catch up. Thus, we print NO.

## HackerRank Number Line Jumps solution

### Number Line Jumps Solution in C

```#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int x1;
int v1;
int x2;
int v2;
scanf("%d %d %d %d",&x1,&v1,&x2,&v2);
if(x2>x1){
if(v2<v1){
if((x2-x1)%(v1-v2)==0)
printf("YES");
else
printf("NO");
}
else
printf("NO");
}
else{
if(v1>v2){
if((x1-x2)%(v2-v1)==0)
printf("YES");
else
printf("NO");
}
else if(x1==x2&&v1==v2)
printf("YES");
else
printf("NO");
}
return 0;
}```

### Number Line Jumps Solution in Cpp

```#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

int main(){
int x1;
int v1;
int x2;
int v2;
cin >> x1 >> v1 >> x2 >> v2;
if ((v1 <= v2) || ((x2 - x1) % (v2 - v1))) {
puts("NO");
} else {
puts("YES");
}
return 0;
}```

### Number Line Jumps Solution in Java

```import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

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

if (v1>v2&&(x2-x1)%(v1-v2)==0)
System.out.println("YES");
else
System.out.println("NO");
}
}```

### Number Line Jumps Solution in Python

```#!/bin/python

import sys

x1,v1,x2,v2 = raw_input().strip().split(' ')
x1,v1,x2,v2 = [int(x1),int(v1),int(x2),int(v2)]
if v1==v2:
print 'NO'
else:
if (x1-x2+v2-v1)%(v2-v1)==0:
if (x1-x2+v2-v1)/(v2-v1)>0:
print 'YES'
else:
print 'NO'
else:
print 'NO'```

### Number Line Jumps Solution using 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();
});

return input_stdin_array[input_currentline++];
}

/////////////// ignore above this line ////////////////////

function main() {
var x1 = parseInt(x1_temp[0]);
var v1 = parseInt(x1_temp[1]);
var x2 = parseInt(x1_temp[2]);
var v2 = parseInt(x1_temp[3]);

var l1 = x1;
var l2 = x2;

for (var i=0; i< 10001; i++){
l1+=v1;
l2+=v2;

if(l1 == l2){
console.log("YES");
return;
}
}
console.log("NO");
}```

### Number Line Jumps Solution in Scala

```object Solution {

def main(args: Array[String]) {
val sc = new java.util.Scanner (System.in);
var x1 = sc.nextInt();
var v1 = sc.nextInt();
var x2 = sc.nextInt();
var v2 = sc.nextInt();

if (x1 == x2) println("YES")
else if (x1 > x2 && v1 < v2 && ((x1 - x2) % (v2 - v1) == 0)) println("YES")
else if (x1 < x2 && v1 > v2 && ((x2 - x1) % (v1 - v2) == 0)) println("YES")
else println("NO")
}
}```

### Number Line Jumps Solution in Pascal

```var
x1,x2,v1,v2:longint;

begin
if (v1>v2) AND ((x2-x1) mod (v1-v2)=0) then
writeln('YES')
else
writeln('NO');
end. ```

Disclaimer: This problem (Number Line Jumps) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.