문제

https://app.codility.com/programmers/lessons/3-time_complexity/frog_jmp/
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

• 시간복잡도 O(1)로 풀 수 있다
• 목표 위치에서 현재 위치를 빼고 갈 수 있는 거리로 나누면 된다. 나누어 떨어지지 않는다면 몫 + 1

코드

/**
개구리가 현재 x위치에 존재
y 포지션 이상으로 도달하고 싶음
언제나 고정된 거리 D만큼 이동 가능

개구리가 목표 위치로 갈 수 있는 최소한의 이동 횟수
**/
class Solution {
    public int solution(int X, int Y, int D) {
        
        int answer = 0;
        if((Y-X) % D > 0) answer = ((Y-X) / D) + 1; 
        else answer = (Y-X) / D;
        return answer;
    }
}