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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}
✔

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