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CountDiv.md

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Write a function:

class Solution { public int solution(int A, int B, int 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.

Assume that:

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. Complexity:

expected worst-case time complexity is O(1); expected worst-case space complexity is O(1).