# Scala algorithm: Count factors/divisors of an integer

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## 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.

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | â€¦ | 15 | 16 | Total count | |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

Divides 16? | âœ“ | âœ“ | âœ— | âœ“ | âœ— | âœ— | âœ— | âœ“ | âœ— | âœ— | âœ— | âœ“ | 5 |

Factor count so far | 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 Â© from www.scala-algorithms.com)

## Scala concepts & Hints

### Collect

'collect' allows you to use Pattern Matching, to filter and map items.

### Pattern Matching

Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.

### Range

The

`(1 to n)`

syntax produces a "Range" which is a representation of a sequence of numbers.### View

The

`.view`

syntax creates a structure that mirrors another structure, until "forced" by an eager operation like .toList, .foreach, .forall, .count.

## Algorithm in Scala

8 lines of Scala (version 2.13), showing how concise Scala can be!

## 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)
```