# Count factors/divisors of an integer in pure-functional immutable Scala

## Algorithm goal

A number $$y$$ is a factor of $$x$$ if $$x$$ is divisible by $$y$$. Find the number of distinct factors of a number $$x$$.

For example, 2 has two factors: $$1$$ and $$2$$. 16 has 5 factors: $$1$$, $$2$$, $$4$$, $$8$$, and $$16$$.

This problem is similar to the codility problem CountFactors - Count factors of given number n.

 Total count Divides 16? 5 Factor count so far 1 2 3 4 5 6 7 8 9 … 15 16 ✓ ✓ ✗ ✓ ✗ ✗ ✗ ✓ ✗ ✗ ✗ ✓ 1 2 2 3 3 3 3 4 4 4 4 5

## Explanation

In a brute-force approach, for number $$n$$, we can check for all numbers that are divisible, up to $$n$$.

However, there is a more efficient approach, in particular if we consider that for every factor that is under $$\sqrt{n}$$, there a corresponding factor to be counted that is above $$\sqrt{n}$$, meaning every divisor under the square root has a corresponding divisor above it - 2 divisors. (this is © from www.scala-algorithms.com)

## Test cases in Scala

assert(countFactors(1) == 1)
assert(countFactors(2) == 2)
assert(countFactors(3) == 2)
assert(countFactors(4) == 3)
assert(countFactors(5) == 2)
assert(countFactors(6) == 4)
assert(countFactors(16) == 5)
assert(countFactors(24) == 8)
assert(countFactors(36) == 9)
assert(countFactors(Int.MaxValue) == 2)

def countFactors(n: Int): Int = ???