Count Divisors : Compute number of integers divisible by k in range [a..b].

Task description 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). class Solution { public int solution(int A, int B, int K) { // write your code in Java SE 8 long a=A; long b=B; long k=K; long count=0; long i=a%k; //System.out.println(a+" "+b+" "+k); if(i!=0){ a=a+Math.abs(i-k); } long j=b%k; if(j!=0){ b=b-(Math.abs(j)); } if(a!=b){ if(a%k==0) count=1; count=count+(b-a)/k; }else{ if(a==0 && b==0){ count=1; }else if(a%k==0){ count=(b/a); }else{ count=0; } } return (int)count; } } 100%                               100 out of 100 points Analysis summary The solution obtained perfect score. Analysis Detected time complexity: O(1) expand allExample tests ▶ example  A = 6, B = 11, K = 2 ✔ OK expand allCorrectness tests ▶ simple  A = 11, B = 345, K = 17 ✔ OK ▶ minimal  A = B in {0,1}, K = 11 ✔ OK ▶ extreme_ifempty  A = 10, B = 10, K in {5,7,20} ✔ OK ▶ extreme_endpoints  verify handling of range endpoints, multiple runs ✔ OK expand allPerformance tests ▶ big_values  A = 100, B=123M+, K=2 ✔ OK ▶ big_values2  A = 101, B = 123M+, K = 10K ✔ OK ▶ big_values3  A = 0, B = MAXINT, K in {1,MAXINT} ✔ OK ▶ big_values4  A, B, K in {1,MAXINT} ✔