# codility 3-2. FrogJmp

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문제출처

## 문제

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(X, Y, 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).


## 풀이코드

• Detected time complexity: O(1)
def solution(X, Y, D):
return (Y - X) // D if (Y - X) % D == 0 else (Y - X) // D + 1