count-div.js

// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

// O(1)
// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(A, B, K) {
    // write your code in JavaScript (Node.js 8.9.4)
    let b = B/K;
    let a = (A > 0 ? (A - 1)/K : 0);
    
    if (A == 0) {
        b++;
    }
    return b - a;
}


// O(n)
function solution(A, B, K) {
    // write your code in JavaScript (Node.js 8.9.4)
    
    let i = A;
    let divisible = [];
    
    if (K == 2000000000) return 0;
    
    while (A <= i && B >= i) {
        if (i == 2000000000) {
            return 0;
            break;
        }
        // console.log(i)
        if (i % K == 0) {
            // divisible++
            divisible.push(i);
        }
        i++
    }
    
    // console.log(divisible)
    return divisible.length;
}

/*

Write a function:

function solution(A, B, K);

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.

*/