Source: https://codility.com/demo/take-sample-test/count_div/
Task description
Write a function:
Assume that:
Solution:
using System; // you can also use other imports, for example: // using System.Collections.Generic; // you can use Console.WriteLine for debugging purposes, e.g. // Console.WriteLine("this is a debug message"); class Solution { public int solution(int A, int B, int K) { if(K==0) return 0; int left = A/K; int right = B/K; int modLeft = A%K; int modRight = B%K; if((modLeft==00 || modRight==0)) return (B-A)/K +1 ; return right-left; } }
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.:class Solution { public int solution(int A, int B, int K); }
{ 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:
Complexity:
- 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.
- expected worst-case time complexity is O(1);
- expected worst-case space complexity is O(1).
Solution:
using System; // you can also use other imports, for example: // using System.Collections.Generic; // you can use Console.WriteLine for debugging purposes, e.g. // Console.WriteLine("this is a debug message"); class Solution { public int solution(int A, int B, int K) { if(K==0) return 0; int left = A/K; int right = B/K; int modLeft = A%K; int modRight = B%K; if((modLeft==00 || modRight==0)) return (B-A)/K +1 ; return right-left; } }
