In a range \(5\) to \(15\), there are \(4\) numbers divisible by \(n = 3\): \(6\), \(9\), \(12\) and \(15\). Compute this for a generic \(n\) (all non-negative numbers)
In Codility, there is a similar problem 'CountDiv'.
Test cases in Scala
assert(countDivisibles(5 to 15, divisor = 3) == 4) assert(countDivisibles(0 to Int.MaxValue, divisor = Int.MaxValue) == 2) assert(countDivisibles(0 to 0, divisor = 3) == 1) assert(countDivisibles(1 to Int.MaxValue, divisor = Int.MaxValue / 2) == 2)
Algorithm in Scala
13 lines of Scala (compatible versions 2.13 & 3.0), showing how concise Scala can be!
A brute-force method will have a complexity that depends strictlyon the size of the input number.
Notice however that the count of numbers divisible by\(3\), up to \(15\), is \(5\) (\(3, 6, 9, 12, 15\)). If we do that with respect to a range, we need to just take the difference between counts up to the end of the range, and the counts up to the start of the range (excluding the number itself). (this is © from www.scala-algorithms.com)
Also if the number \(0\) is in range, count is increased by 1 because \(0\) is divisible by all numbers (except \(0\), of course).
Scala concepts & Hints
(1 to n)syntax produces a "Range" which is a representation of a sequence of numbers.