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FrogJmp (Codility)
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:
int solution(int X, int Y, int 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).
Solution:
int solution(int X, int Y, int D) { // write your code in C++11 return (Y-X)/D+((Y-X)%D==0?0:1);}
FrogJmp (Codility)
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