The problem

P13 (**) Run-length encoding of a list (direct solution). Implement the so-called run-length encoding data compression method directly. I.e. don't use other methods you've written (like P09's pack); do all the work directly.

Example:

scala> encodeDirect(List('a, 'a, 'a, 'a, 'b, 'c, 'c, 'a, 'a, 'd, 'e, 'e, 'e, 'e))
res0: List[(Int, Symbol)] = List((4,'a), (1,'b), (2,'c), (2,'a), (1,'d), (4,'e))

Initial thoughts

Obviously the issue shouldn't be solved just copying all stuff from problem 09 and problem 10 inside a single file. There shall be a way to build a single-function solution.

Spanning

Scala lists provide a span() method that splits the list in two. It scans all elements in order, storing them into a list while the given predicate (a function) is true. When the predicate is false span() stops and returns the two resulting lists.

It seems to be the perfect candidate for our job. We just have to recursively apply it obtaining two lists. The first list is made by equal elements, and can be reduced to a tuple, the second list is given to the recursive call.

def encodeDirect[A](l: List[A]):List[(Int, A)] = {
    def _encodeDirect(res: List[(Int, A)], rem: List[A]):List[(Int, A)] = rem match {
        case Nil => res
        case ls => {
            val (s, r) = rem span { _ == rem.head }
            _encodeDirect(res:::List((s.length, s.head)), r)
        }
    }
    _encodeDirect(List(), l)
} 

The predicate given to span() is _ == rem.head, an anonymous function that tests if the current element is equal to the first element. While this is always true for the first element, by definition, other elements can be equal or not, and the job of span() is to find where to split the list.

So example given in the problem runs through the following steps

scala> var rem = List('a, 'a, 'a, 'a, 'b, 'c, 'c, 'a, 'a, 'd, 'e, 'e, 'e, 'e)
rem: List[Symbol] = List('a, 'a, 'a, 'a, 'b, 'c, 'c, 'a, 'a, 'd, 'e, 'e, 'e, 'e)

scala> var (res, rem1) = rem span { _ == rem.head }
res: List[Symbol] = List('a, 'a, 'a, 'a)
rem1: List[Symbol] = List('b, 'c, 'c, 'a, 'a, 'd, 'e, 'e, 'e, 'e)

scala> var (res, rem2) = rem1 span { _ == rem1.head }
res: List[Symbol] = List('b)
rem2: List[Symbol] = List('c, 'c, 'a, 'a, 'd, 'e, 'e, 'e, 'e)

scala> var (res, rem3) = rem2 span { _ == rem2.head }
res: List[Symbol] = List('c, 'c)
rem3: List[Symbol] = List('a, 'a, 'd, 'e, 'e, 'e, 'e)

scala> var (res, rem4) = rem3 span { _ == rem3.head }
res: List[Symbol] = List('a, 'a)
rem4: List[Symbol] = List('d, 'e, 'e, 'e, 'e)

scala> var (res, rem5) = rem4 span { _ == rem4.head }
res: List[Symbol] = List('d)
rem5: List[Symbol] = List('e, 'e, 'e, 'e)

scala> var (res, rem6) = rem5 span { _ == rem5.head }
res: List[Symbol] = List('e, 'e, 'e, 'e)
rem6: List[Symbol] = List()

Final considerations

I learned another very useful method to split Scala lists, span(). I also used a complex expression in a case statement, as already done in problems 09 and 10.

Feedback

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