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

Number Line Jumps

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recency

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3957 Discussions

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  • mdsanz
     + 0 comments

    My Java 8 Solution

    public static String kangaroo(int x1, int v1, int x2, int v2) {
        if ((x2 > x1 && v2 >= v1) || (x1 > x2 && v1 >= v2)) {
            return "NO";
        } else {
            int diff = Integer.MAX_VALUE, p1 = x1 + v1, p2 = x2 + v2;
            while (true) {
                if (p1 == p2) {
                    return "YES";
                } 
                
                if (Math.abs(p1 - p2) > diff) {
                    return "NO";
                }
                
                diff = Math.abs(p1 - p2);
                p1 += v1;
                p2 += v2;
            }
        }
    }

  • wobbie
     + 0 comments

    C++ solution

    string kangaroo(int x1, int v1, int x2, int v2) {
  if ((x1 < x2 && v1 <= v2) || (x2 < x1 && v2 <= v1)) {
    return "NO";
  } else if ((x1 - x2) % (v2 - v1) == 0) {
    return "YES";
  }
  
  return "NO";
}

  • sivasaichand2002
     + 0 comments

    def kangaroo(x1, v1, x2, v2): #check if both vectors are same if v1==v2: return 'NO' else: #verify if the intervel where the kangaroos are meeting is grater than 0 as the vectors are moving positively i = math.ceil(((x1-x2)/(v2-v1))) if i > 0 and (x1 + v1*i) == (x2 + i*v2) : return 'YES' else: return "NO"

  • calancea_bianca1
     + 0 comments
    def kangaroo(x1, v1, x2, v2):
    # Write your code here
    if v1!=v2:
        K=(x2-x1)/(v1-v2)
    
        if K.is_integer() and K>0:
            return "YES"
        else:
            return "NO"
    else:
        return "NO"

  • Dacheng_gu
     + 0 comments

    Bruh what is this comment search on Hackerrank. Trying to see if x1 always starts left of x2. Then you can calculate it without simulating