Codility solutions - lesson1, exercise: FrogJmp

Content of Codility excercise:


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).

Solution:

    public int solution(int X, int Y, int D) {
        if ((Y-X)%D == 0){
            return (Y-X)/D;
        }else {
            return (Y-X)/D + 1;
        }
    }