Solution to Codility lesson1-exercise2 Frog Jump problem, link to Codility problem and complete solution at the end of the post.

**Definitions:**

** Problem:**Define a function

*solution*with input

*X, Y, D*, such that

*solution*returns the minimum number of jumps needed to reach a position equal or greater than

*Y*from position X, being

*D*the distance traveled in each jump.

- Expected worst-case time complexity is
*O(1)*; - Expected worst-case space complexity is
*O(1);*

### Analysis:

To solve this one, we must establish the distance to be traveled, which is *(Y – X)*. This distance divided by *D* will give us the exact number of jumps to travel from *X *to *Y. *But, since the frog can’t jump in fractions, the result of this solution will be the ceiling of the float division *(Y – X)/D*. Without the use of python math.ceil() function, this can be achieved with the follow line of code.

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return ((Y - X)//D) + (1 if (Y - X)%D > 0 else 0) |

Since we do only a few operations regardless the size of our input, the solution time complexity is *O(1), *the same goes to the space complexity.

**Complete Solution:**

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# Name: FrogJmp# Link: https://codility.com/demo/take-sample-test/frog_jmp/ def solution(X, Y, D): # Calcs the integer part of the division, distance to be traveled by jump size, if the division leaves a rest, add 1 to the result. return ((Y - X)//D) + (1 if (Y - X)%D > 0 else 0)

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# Name: FrogJmp # Link: https://codility.com/demo/take-sample-test/frog_jmp/ def solution(X, Y, D): # Calcs the integer part of the division, distance to be traveled by jump size, if the division leaves a rest, add 1 to the result. return ((Y - X)//D) + (1 if (Y - X)%D > 0 else 0) |

*Time Complexity O(1)*

*Space Complexity O(1)*