Hello coders, today we are going to solve Number Line Jumps HackerRank Solution which is a Part of HackerRank Algorithm Series.
Task
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
orNO
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 12 on the number 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.
Solution – Number Line Jumps
C++
#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; }
Python
import math import os import random import re import sys # Complete the kangaroo function below. def kangaroo(x1, v1, x2, v2): if x2 > x1 and v2 > v1: return "NO" else: if v2-v1 == 0: return 'NO' else: result = (x1-x2) % (v2-v1) if result == 0: return 'YES' else: return 'NO' if __name__ == '__main__': fptr = open(os.environ['OUTPUT_PATH'], 'w') x1V1X2V2 = input().split() x1 = int(x1V1X2V2[0]) v1 = int(x1V1X2V2[1]) x2 = int(x1V1X2V2[2]) v2 = int(x1V1X2V2[3]) result = kangaroo(x1, v1, x2, v2) fptr.write(result + '\n') fptr.close()
Java
import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.regex.*; public class Solution { // Complete the kangaroo function below. static String kangaroo(int x1, int v1, int x2, int v2) { if (x1 == x2) { return "YES"; } int diff = v1 > v2 ? v1 - v2 : v2 - v1; if (diff == 0) { return "NO"; } int xdiff = x1 - x2; int vdiff = v2 - v1; if ((xdiff < 0 && vdiff < 0) || (xdiff > 0 && vdiff > 0)) { int mod = xdiff % vdiff; int mod2 = vdiff % xdiff; if (mod == 0 || mod2 == 0) { return "YES"; } } return "NO"; } private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) throws IOException { BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); String[] x1V1X2V2 = scanner.nextLine().split(" "); int x1 = Integer.parseInt(x1V1X2V2[0]); int v1 = Integer.parseInt(x1V1X2V2[1]); int x2 = Integer.parseInt(x1V1X2V2[2]); int v2 = Integer.parseInt(x1V1X2V2[3]); String result = kangaroo(x1, v1, x2, v2); bufferedWriter.write(result); bufferedWriter.newLine(); bufferedWriter.close(); scanner.close(); } }
Disclaimer: The above Problem (Number Line Jumps) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.
if x1 and x2 jump rate same but v1 v2 is different so how two kangaroos meets same location at a time ?
Say for example :
x1 = 0; v1 = 2;
x2 = 0; v2 = 5
In first jump v2 will be ahead right then how your first condition will meet my case ?
if (x1 == x2) {//say here x1 x2 same but v1 v2 different then this condition invalid
return “YES”;
}
if possible, let’s discuss about this case.