[codility] 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:

public func solution(_ X : Int, _ Y : Int, _ D : Int) -> Int

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.Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

 

문제 해설

개구리 초기 위치 X, 도착 지점 Y, 점프 값 D, X<=Y

 

구현 해설

초기 위치값X가 Y보다 크거나 같으면 이동하지 않아도 도착지점 예외처리

도착지점 - 초기지점 = 이동해야 할 거리 

이동해야 할 거리 / 점프 한번 당 이동거리 = 이동횟수

이동해야 할 거리 % 점프 한번 당 이동거리의 나머지 몫이 0보다 크면 한번 카운트 추가

 

import Foundation
import Glibc

// you can write to stdout for debugging purposes, e.g.
// print("this is a debug message")

public func solution(_ X : Int, _ Y : Int, _ D : Int) -> Int {
    // write your code in Swift 4.2.1 (Linux)

    if X >= Y {
        return 0
    }

    var count = (Y-X) / D
    if (Y-X) % D > 0 {
        count += 1
    }

    return count
}