(Time Complexity) Frog Jump

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).

Here is my solution

public class FrogJmp{
	public static int solution(int x, int y, int distance){
		int diffDistance = y - x;
		if(diffDistance%distance == 0) 
			return diffDistance/distance;
		else
			return diffDistance/distance + 1;
	}
	
	public static void main(String[] args){
		System.out.println("#Jump must made:"+solution(10,110,30));
	}	
}