[Codility] Lesson 3 - Time Complexity : FrogJmp

Lesson 3 - Time Complexity : FrogJmp

📌 質問する

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:

class Solution { public 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

Write an efficient algorithm for the following assumptions:

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

X ≤ Y.

📝 に答える

Xの値をY以上のDに単純に加算し、加算回数を求めることで値が得られる.しかしこれを行うと効率的にTIME OUTが発生します.

ビットの方式ではなく、他の方法はYからXの距離を減算するのにどれだけのDが必要かによって答えを出す.

💻 インプリメンテーション

static class Solution{
    
    public int solution(int X, int Y, int D){
        
        return (int) Math.ceil((double)(Y - X) / D);
    }
    
}