# Frog Problem: Count minimal number of jumps from position X to Y.

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;
}

```

## 1 thought on “Frog Problem: Count minimal number of jumps from position X to Y.”

1. Delia

Thanks for finally writing about >Frog Problem:
Count minimal number of jumps from position X to Y. | CodeSolution.org <Loved it!