[Codility] Lesson 3 - Time Complexity : FrogJmp

Lesson 3 - Time Complexity : FrogJmp

📌 문제

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.

📝 풀이

• 단순하게 X 의 값이 Y 보다 크거나 같을 때 까지 D 만큼 더해주고 더해준 횟수를 구하면 값을 얻을 수 있다. 그러나 그렇게 할 경우 효율성에서 TIME OUT 이 발생한다.
• 위의 방식이 아닌 다른 방법으로는 Y 에서 X 를 뺀 거리만큼을 D가 몇 개 필요한지 식을 세워서 답을 얻어낸다.

💻 구현

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