Contents.swift

import Foundation

public func solution(_ A : Int, _ B : Int, _ K : Int) -> Int {
    // write your code in Swift 4.2.1 (Linux)
    
    let numB = Int((Double(B)/Double(K)).rounded(.down))
    
    let numA = Int((Double(A)/Double(K)).rounded(.down))
    
    var numDiv: Int = numB - numA;
    
    var plus: Int = 0
    if (A % K == 0) {
        plus = 1
    }
    
    numDiv = numDiv + plus
    
    return numDiv
    
}

let A = 6
let B = 11
let K = 2

print(solution(A,B,K))



// MARK: - Prompt

/*
 Write a function:

 public func solution(_ A : Int, _ B : Int, _ K : Int) -> Int
 that, given three integers A, B and K, returns the number of integers within the range [A..B] that are divisible by K, i.e.:

 { i : A ≤ i ≤ B, i mod K = 0 }
 For example, for A = 6, B = 11 and K = 2, your function should return 3, because there are three numbers divisible by 2 within the range [6..11], namely 6, 8 and 10.

 Write an efficient algorithm for the following assumptions:

 A and B are integers within the range [0..2,000,000,000];
 K is an integer within the range [1..2,000,000,000];
 A ≤ B.
*/