CODILITY - LESSON 3 - divCount

PROBLEM:

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).

MY SOLUTION (THINKING IN CODE) (100% correctness, 0%performance!, time complexity O(B-A)
class Solution { public int solution(int A, int B, int K) { // write your code in C# 6.0 with .NET 4.5 (Mono) int result = 0; for (int i=A; i<=B; i++)
{
if (A[i]%K == 0) count++;
}
return result; }


MATHEMATICAL SOLUTION (100%, time complexity O(1))
class Solution { public int solution(int A, int B, int K) { // write your code in C# 6.0 with .NET 4.5 (Mono) int result = 0; if (A%K == 0) { result = ((B-A)/K)+1; } else { result = (B/K - ((A/K)+1))+1; } return result; }

The for loop is the 1st thing that comes to mind while thinking code, but it comes with a expensive time complexity, while the second one is barely a coding challenge but a purely mathematical problem.