Contents.swift

import Foundation

public func solution(_ X : Int, _ Y : Int, _ D : Int) -> Int {
    // write your code in Swift 4.2.1 (Linux)
    
    // Handle special case
    if X >= Y {
        return 0
    }
    
    // Handle other cases
    let distanceToTravel: Int = Y - X
    let numJumps: Int = Int((Double(distanceToTravel) / Double(D)).rounded(.up))
    return numJumps

}

print(solution(1, 1000, 10000))



// MARK: - Prompt

/*
 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.
*/