Longest common prefix

It is often useful to find the common prefix of a set of strings, that is, the longest initial portion of all strings that are identical.

Longest common prefix is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Given a set of strings, R, for a prefix S, it should hold that:

$\forall x\ \in \ R:\ S\leq$_pref $\ x$

(help here: does not express that S is the longest common prefix of x)

An example use case for this: given a set of phone numbers, identify a common dialing code. This can be accomplished by first determining the common prefix (if any), and then matching it against know dialing codes (iteratively dropping characters from rhs until a match is found, as the lcp function may match more than the dialing code).

Test cases

For a function, lcp, accepting a list of strings, the following should hold true (the empty string, $\varepsilon$, is considered a prefix of all strings):

lcp("interspecies","interstelar","interstate") = "inters"
lcp("throne","throne") = "throne"
lcp("throne","dungeon") = $\varepsilon$
lcp("cheese") = "cheese"
lcp($\varepsilon$) = $\varepsilon$
lcp($\emptyset$) = $\varepsilon$
lcp("prefix","suffix") = $\varepsilon$

Task inspired by this stackoverflow question: Find the longest common starting substring in a set of strings

Scala

<lang scala>class TestLCP extends FunSuite {

 test("shared start") {
   assert(lcp("interspecies","interstelar","interstate") === "inters")
   assert(lcp("throne","throne") === "throne")
   assert(lcp("throne","dungeon").isEmpty)
   assert(lcp("cheese") === "cheese")
   assert(lcp("").isEmpty)
   assert(lcp(Nil :_*).isEmpty)
   assert(lcp("prefix","suffix").isEmpty)
 }

 def lcp(list: String*) = list.foldLeft("")((_,_) =>
   (list.min.view,list.max.view).zipped.takeWhile(v => v._1 == v._2).unzip._1.mkString)

}</lang>

Output: