[Codility] Time Complexity - FrogJmp 풀이

예문

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.

해석

  • 개구리는 길 건너편으로 가길 원함.
  • X : 현재 위치
  • Y : 타겟 위치
  • 개구리는 현재 위치 X 에서 Y 보다 크거나 같은 위치로 가길 원함
  • 개구리는 항상 고정된 거리 D 만큼 점프
  • 개구리가 Y에 도달하는 최소 횟수를 리턴하라!

풀이

  • (Y - X) / D
  • 소수점은 올림 처리

제약사항

  • X, Y, D의 범위는 정수 [1..1,000,000,000]
  • X ≤ Y

코드

public int solution(int X, int Y, int D) {
  double divideNum = (Y - X) * 1.0 / D; //double 형 변환을 위해 1.0 곱함.
      return (int) Math.ceil(divideNum);  //올림 처리
}

결과