**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; }