noahlz

/** 
  * non-performant solution to https://twitter.com/Al_Grigor/status/1357028887209902088
  * scala> partitionCount("aaaabbbcca")
  * res0: List[(Char, Int)] = List((a,4), (b,3), (c,2), (a,1))
  * PS see "stringly" for a better solution in clojure. 
  * Is there a scala equivalent to this? Feel free to comment.
  *  (->> (partition-by identity s) (map frequencies) seq))
  */
def partitionCount(s: String): List[(Char, Int)] = {
    s.toList.foldLeft(List.empty[(Char, Int)]) { (agg, c) =>