Sharing an answer code of mine about FrogJmp problem of Codility lesson 3.

## Lesson 3: FrogJmp

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:

```
def solution(A, K)
```

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:

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).

## Example answer code in Python 2.7

```
def solution(X, Y, D):
# write your code in Python 2.7
if X >= Y: # does not have to jump
return 0;
else: # have to jump
range = Y-X
if range % D == 0: # When the position equal to Y after jumps
return range / D
else: # When the position is greater than Y after jumps
return (range / D) + 1
```

- Time complexity: O(1)