[Codility] Lesson-03.1: FrogJmp
In this post, the solution for lesson 3 with reducing time complexity is handled. For accessing the problem, click the following link.
Task Description
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
Write an efficient algorithm for the following assumptions:
- X, Y and D are integers within the range [1..1,000,000,000];
- X ≤ Y.
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Key Point
- When the input X is equal or larger than Y, calculating the number of the jump is not need. Just return 0.
- If not, the required number of jumps can be simply counted and through a ‘while’ or ‘for’ loop, but it is not efficient.
Solution (using Python)
def solution(X, Y, D):
if(X >= Y): return 0
else: return (Y - X) // D + int((Y - X) % D > 0)Please use the above solution for reference.
I recommend you to write your own source code.
