increasing code performance of codility

Question

today i heard about this website called codility where a user can give various programming test to check their code's performance.

When I started, they presented me with this sample test,

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

Assume that: X, Y and D are integers within the range

[1..1,000,000,000]; X ≤ Y. Complexity: expected worst-case time

complexity is O(1); expected worst-case space complexity is O(1).

The question was pretty straight forward and it took me like 2 minutes to write the solution, which is following,

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

        int p = 0;
        while (X < Y){
            p++;
            X = X + D;
        }
    return p;
    }
}

However, the test result shows that the performance of my code is just 20% and I scored just 55%,

Here is the link to result, https://codility.com/demo/results/demo66WP2H-K25/

That was so simple code, where I have just used a single while loop, how could it possibly be make much faster ?

Answer 1

Here is the 100% total score Python solution:

def solution(X, Y, D):
    # write your code in Python 3.6
    s = (Y-X)/D
    return int(-(-s // 1))

Answer 2

Javascript solution, 100/100, and shorter than the existing answer:

function solution(Y, Y, D) {
 return Math.ceil((Y - X) / D);
}

Answer 3

Java(One Line), Correctness 100%, Performance 100%, Task score 100%

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int X, int Y, int D) {
      return (int) Math.ceil((double) (Y - X) / (double) D);
    }
}