Scala: Occurrences with pairs, Map and foldLeft

occurrences.scala

/**
  * This function computes for each unique character in the list `chars` the number of
  * times it occurs. times(List('a', 'b', 'a')) should return List(('a', 2), ('b', 1))
  */
def howManyTimes(chars: List[Char]): List[(Char, Int)] = {
  def incr(acc:Map[Char, Int], c:Char) = {
    val count = (acc get c).getOrElse(0) + 1 // If acc get c is none getOrElse makes it 0
    // To add values to the Map we use operator +
    // If key does not exist, a new (key, value) is inserted, otherwise substitution occurs
    acc + ((c, count)) // I Double brakets tell scala we're dealing with a pair
  }

  // Initially we have an empty map and with that initial value we perform the operation
  // starting from the right. The resulting element is an updated map (my brain hurts)
  (chars foldLeft Map[Char,Int]())(incr).iterator.toList

  // (1) foldLeft has an alternate syntax. These two are equivalent
  // xs foldLeft z
  // z /: xs
  // We could have written:
  // (Map[Char, Int]() /: chars)(incr)

  // (2)To convert a map to a list I need to use .iterator.toList
}

Great explanation, it took me a while to figure it out after my brain melted too! Thanks!