Smallest repeating substring

Question

I was looking at a Perl code golf page (don't ask why) and came across this:

Hole 3 - Smallest Repeating Pattern

Write a subroutine that accepts a string which may consist of a repeating pattern, and returns the smallest repeating substring. If the string does not consist of a repeating pattern, the subroutine should return undef or the empty string. e.g.:

input    output 
'aaaaaa' 'a' 
'ababab' 'ab' 
'aabaab' 'aab' 
'ababaa' ''

Apparently in Perl this can be expressed as sub g3 { pop=~/^(.*?)\1+\z/s&&$1 }

I don't know much Perl so I don't understand how this works. What is the best we can do in Scala? I'm more interested in elegance than the precise number of characters.

Here is my attempt, but it's pretty ugly, which is why I'm asking.

def srp(s: String) =
  s.inits.toList.tail.init.map(i => (i * (s.size / i.size), i)).
    filter(_._1 == s).map(_._2).reverse.headOption.getOrElse("")

This doesn't qualify as code golf since it doesn't have an *objective* optimization criterion. The OP is just looking for an elegant Scala solution. — Aaron Novstrup, Oct 25 '11 at 01:47

huynhjl · Accepted Answer · 2011-10-25T08:19:20.870

Elegance is subjective...

def smallestPat(input: String) = {
   (1 to input.length).view
      .map(i => input.take(i))
      .find{p => 
        Iterator.iterate(p)(_ + p)
          .takeWhile(_.length <= input.length)
          .exists(_ == input) && p != input}
      .getOrElse("")
}

List("aaaaaa", "ababab", "aabaab", "ababaa") map smallestPat
// res13: List[String] = List(a, ab, aab, "")

Edit and re-edited: slighty shorter:

def smallestPat(i: String) = {
   (1 to i.length - 1)
      .map(i.take)
      .find(p => p * (i.length / p.length) == i)
      .getOrElse("")
}

One more, using grouped:

def smallestPat(i: String) = {
  (1 to i.size/2).map(i.take)
  .find(p => i.grouped(p.size) forall(p==))
  .getOrElse("")
}

Travis Brown · Answer 2 · 2011-10-25T01:54:58.370

7

Are you willing to admit a Regex-based solution?

def srp(s: String) = {
  val M = """^(.*?)\1+$""".r
  s match {
    case M(m) => Some(m)
    case _ => None
  }
}

Or a one-liner:

val srp = """^(.*?)\1+$""".r.findFirstMatchIn(_: String).map(_.group(1))

Not as concise as the Perl, but I find both considerably more readable.

edited Oct 25 '11 at 01:54

answered Oct 25 '11 at 01:48

Travis Brown

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What does the \1 do in the regexp? – Jens Schauder Oct 25 '11 at 05:36
I never got my head around greedy/reluctant/posessive quantifiers, but now it makes perfect sense! @Jens `\1` refers to the first sub-expression, i.e. the contents of the `()` – Luigi Plinge Oct 25 '11 at 05:46
You should add an `getOrElse ""`, to return exactly the same output. – Daniel C. Sobral Oct 25 '11 at 17:35

Daniel C. Sobral · Answer 3 · 2011-10-25T18:30:26.643

2

This is the equivalent Scala one-liner:

"""(?:^(.*?)\1+$)|.*""".r replaceAllIn (_: String, "$1")

edited Oct 25 '11 at 18:30

answered Oct 25 '11 at 17:32

Daniel C. Sobral

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This always returns `""` ... is there a typo? – Luigi Plinge Oct 25 '11 at 17:55
@Luigi Yes, sorry. I pasted something I was working on (and which was not working). The above now fully accomplishes the task as proposed. – Daniel C. Sobral Oct 25 '11 at 18:32

Smallest repeating substring

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