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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Number Line Jumps
  5. Discussions

Number Line Jumps

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3974 Discussions

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  • faidi_nabil
     + 0 comments

    def kangaroo(x1, v1, x2, v2):

    if v1==v2:
    return "NO" 
elif ((x2-x1)/(v1-v2)).is_integer() is True and ((x2-x1)/(v1-v2)) >0:
    return "YES"
else:
    return "NO"

  • neerajkalura4
     + 0 comments

    1224

  • Satyam_Agrawal
     + 0 comments

    My code:-

    public static String kangaroo(int x1, int v1, int x2, int v2) { while(!((v1>v2&&x1>x2)||(v2>v1&&x2>x1)||(v1==v2&&x1!=x2))) { if(x1==x2) return "YES"; x1+=v1; x2+=v2; } return "NO"; }

  • wmgarciaporcel
     + 0 comments

    This is my code in js:

    function kangaroo(x1, v1, x2, v2) {
    // Write your code here
    // constraint given x2 >x1
    let dist = x2-x1;
        
    while(dist>0){
    x1 += v1;
    x2 += v2;
    
    if(x1===x2) return 'YES';   
    else if ((x2-x1)>=dist)return 'NO';
    else dist = x2 - x1;
    }   
    return 'NO';
}

  • 2050yadavankit
     + 0 comments

    THIS IS MY PYTHON CODE:

    #!/bin/python3

import math
import os
import random
import re
import sys

#
# Complete the 'kangaroo' function below.
#
# The function is expected to return a STRING.
# The function accepts following parameters:
#  1. INTEGER x1
#  2. INTEGER v1
#  3. INTEGER x2
#  4. INTEGER v2
#

def kangaroo(x1, v1, x2, v2):
    if v1 == v2:
        return "YES" if x1 == x2 else "NO"
    if (x2 - x1) * (v1 - v2) < 0:
        return "NO"
    if (x2 - x1) % (v1 - v2) == 0:
        return "YES"
    return "NO"

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    first_multiple_input = input().rstrip().split()

    x1 = int(first_multiple_input[0])
    v1 = int(first_multiple_input[1])
    x2 = int(first_multiple_input[2])
    v2 = int(first_multiple_input[3])

    result = kangaroo(x1, v1, x2, v2)

    fptr.write(result + '\n')

    fptr.close()