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
- X, Y and D are integers within the range [1..1,000,000,000];
- X ≤ Y.
- 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)