FrogJmp

Back to Menu

tab1

Task description

A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.

Count the minimal number of jumps that the small frog must perform to reach its target.

Write a function:

class Solution { public int solution(int X, int Y, int D); }

 that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.

For example, given:
         X = 10
         Y = 85
         D = 30

the function should return 3, because the frog will be positioned as follows:

• after the first jump, at position 10 + 30 = 40 
• after the second jump, at position 10 + 30 + 30 = 70 
• after the third jump, at position 10 + 30 + 30 + 30 = 100 

Write an efficient algorithm for the following assumptions:

•  X, Y and D are integers within the range [1..1,000,000,000];
•  X ≤ Y.

Algorithm

/**
 * @author Ravi Yadav
 * blog address: www.meritmix.com 
 */
public class FrogJump {
 public static void main(String[] args) {

  System.out.println(new FrogJump().solution(10, 85, 30));
 }
 public int solution(int X, int Y, int D) {

  if(X == Y) {
   return 0;
  }

  return (Y-X)%D == 0? ((Y-X)/D) : ((Y-X)/D+1);
    }
}

Back to Menu

tab1