Problem:

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

        // Frog problem
	int frogMinimumCount(int X, int Y, int D) {
	    int numberOfJumps = 0;
	    
	    if((Y-X)<D){
	    	numberOfJumps = 1;
	    }
		if((Y-X)%D == 0){
			numberOfJumps = (Y-X)/D;
	    }else{
	    	numberOfJumps = ((Y-X)/D)+1;
	    }
		
		return numberOfJumps;
	}