In my last post I reviewed the implementation of scala.List’s foldLeft and foldRight methods. That post included a couple of simple examples, but today I’d like to give you a whole lot more. The foldLeft method is extremely versatile. It can do thousands of jobs. Of course, it’s not the best tool for EVERY job, but when working on a list problem it’s a good idea to stop and think, “Should I be using foldLeft?”

Below, I’ll present a list of problem descriptions and solutions. I thought about listing all the problems first, and then the solutions, so the reader could work on his own solution and then scroll down to compare. But this would be very annoying for those who refuse, against my strenuous urging, to start up a Scala interpreter and try to write their own solution to each problem before reading my solution.


Write a function called ‘sum’ which takes a List[Int] and returns the sum of the Ints in the list. Don’t forget to use foldLeft.

def sum(list: List[Int]): Int = list.foldLeft(0)((r,c) => r+c)
def sum(list: List[Int]): Int = list.foldLeft(0)(_+_)

I’ll explain this first example in a bit more depth than the others, just to make sure we all know how foldLeft works.

These two definitions above are equivalent. Let’s examine the first one. The foldLeft method is called on the list parameter. The first parameter is 0. This is the starting value, and the value that will be returned if list is empty. The second parameter is a function literal. It takes parameters ‘r’ (for result) and ‘c’ (for current) and returns the sum of these two values. Scala is smart enough to figure out that since the first parameter (0) is an Int, the ‘r’ parameter must also be an Int. The initial value is always the same type as ‘r’. Scala can also tell that since ‘list’ is a List[Int] the ‘c’ parameter must also be an Int, so we don’t have to specify their types in the parameter list.

The foldLeft method takes that initial value, 0, and the function literal, and it begins to apply the function on each member of the list (parameter ‘c’), updating the result value (parameter ‘r’) each time. That result value that we call ‘r’ is sometimes called the accumulator, since it accumulates the results of the function calls.

In the first defintion, foldLeft’s second parameter (a function literal) uses explicitly named parameters. Notice that ‘r’ and ‘c’ are each referred to exactly once in the function literal, and in the same order as the parameter list. When function literal parameters are used in this way (once each, same order) you can use the shorthand demonstrated in the second definition. The first ‘_’ stands for ‘r’, and the second one stands for ‘c’.


Now that you’ve got the idea, try this one. Write a function that takes a List[Int], and returns the product (of multiplication) of all the Ints in the list. It will be similar to the ‘sum’ function, but with a couple of differences.

def product(list: List[Int]): Int = list.foldLeft(1)(_*_)

Did you get it? It’s the same as ‘sum’ with two exceptions. The initial value is now 1, and the function literal’s parameters are multiplied instead of added. If the initial value were 0, as in ‘sum’, then the function would always return 0.


This one’s a little different. Write a function that takes a List[Any] and returns the number of items in the list. Don’t just call list.length()! Implement it using foldLeft.

def count(list: List[Any]): Int =
  list.foldLeft(0)((sum,_) => sum + 1)

First, we pick our initial value. Remember that this is the value that will be returned for an empty list. An empty list has 0 elements, so we use 0. What function do we want to apply for every item in the list? We just want to increase the result value by one. We call that parameter ‘sum’ in this solution. We don’t care about the actual value of each list element, so we call the second parameter ‘_’, which means it should be discarded.


Here’s a fun one. Write a function that takes a List[Double] and returns the average of the list’s values. There are two ways to go about this one. You could combine two of the previous solutions, using two foldLeft calls, or you could combine them into a single foldLeft. Try to find both solutions.

def average(list: List[Double]): Double =
  list.foldLeft(0.0)(_+_) / list.foldLeft(0.0)((r,c) => r+1)

def average(list: List[Double]): Double = list match {
  case head :: tail => tail.foldLeft( (head,1.0) )((r,c) =>
    ((r._1 + (c/r._2)) * r._2 / (r._2+1), r._2+1) )._1
  case Nil => NaN

The first solution is pretty easy and combines the ‘sum’ and ‘count’ solutions. In real life, of course, you wouldn’t use foldLeft to find the length of the list. You’d just use the length() method. Other than that, though, this is a perfectly sensible solution.

The second solution is more complex. First, the list is matched against two patterns. It is either interpreted as a head item followed by a tail, or as an empty list (Nil). If it’s empty, the function returns the same thing as the first solution, NaN (Not a Number) because you can’t divide by 0.

If the list is not empty, we use a Pair as our initial value. A Pair is just an ordered pair of values. It’s a convenient way to bundle values together. We use it when we need to keep track of more than one accumulator value. In this case, we want to keep track of the average “so far” and also the number of values that the average represents. If the function literal were just passed the average so far, it wouldn’t know how to weight the next value. Members of a Pair are accessed using special methods called ‘_1’ and ‘_2’. You can have groupings longer than 2, also. These are named Tuple3, Tuple4, and so on. In fact, Pair is just an alias of Tuple2. Notice that we didn’t use the word Pair or Tuple2 anywhere in the code. If you enclose a comma-delimited series of values in parentheses, Scala converts that series into the appropriate TupleX.

After we have built up the result, it is a Pair containing the average and the number of items in the list. We only want to return the average so we call ‘_1’ on the result of foldLeft.


Whew! That one was a little tough. Here’s an easier one. Given a List[A] return the last value in the list. Again, no using List’s last() method.

def last[A](list: List[A]): A =
  list.foldLeft[A](list.head)((_, c) => c)

Easy! Mostly. You’ll notice that we’re using a type parameter, A, in this one. If you’re not familiar with type parameters, too bad. I can’t explain them here. Suffice it to say that our use of A here allows us to take a list of any type of contents, and return a result of just that type. So Scala knows that when this is called on a List[Int], it will return an Int. When it’s called on a List[String], it returns a String.

First, we pick an initial value. For the empty list the concept of a last item doesn’t make any sense, so forget that. We can use any value, so long as it’s of type A. list.head is convenient, so that’s our initial value. The function literal is the simplest we’ve seen. For each item in the list, it just returns that item itself. So when it gets to the end of the list, the accumulator holds the last item. We don’t use the accumulator value in the function literal, so it gets parameter name ‘_’.


Write a function called ‘penultimate’ that takes a List[A] and returns the penultimate item (i.e. the next to last item) in the list. Hint: Use a tuple.

def penultimate[A](list: List[A]): A =
  list.foldLeft( (list.head, list.tail.head) )((r, c) => (r._2, c) )._1

This one is very much like the function ‘last’, but instead of keeping just the current item it keeps a Pair containing the previous and current items. When foldLeft completes, its result is a Pair containing the next-to-last and last items. The “_1” method returns just the penultimate item.


Write a function called ‘contains’ that takes a List[A] and an item of type A, and returns true if the item is one of the members of the list, and false if it isn’t.

def contains[A](list: List[A], item: A): Boolean =
  list.foldLeft(false)(_ || _==item)

We choose an initial value of false. That is, we’ll assume the item is not in the list until we can prove otherwise. We use each of the two parameters exactly once and in the proper order, so we can use the ‘_’ shorthand in our function literal. That function literal returns the result so far (a Boolean) ORed with a comparison of the current item and the target value. If the target is ever found, the accumulator becomes true and stays true as foldLeft continues.


Write a function called ‘get’ that takes a List[A] and an index Int, and returns the list value at the index position. Throw an exception if the index is out of bounds.

def get[A](list: List[A], idx: Int): A =
  list.tail.foldLeft((list.head,0)) {
    (r,c) => if (r._2 == idx) r else (c,r._2+1)
  } match {
    case (result, index) if (idx == index) => result
    case _ => throw new Exception("Bad index")

This one has two parts. First there’s the foldLeft, and the result is pattern matched. The foldLeft is pretty easy to follow. The accumulator is a Pair containing the current item and the current index. The current item keeps updating and the current index keeps incrementing until the current index equals the passed in idx. Once the correct index is found the same accumulator is returned over and over. This works fine if idx parameter is in bounds. If it’s out of bounds, though, the foldLeft just returns a Pair containing the last item and the last index. That’s where the pattern match comes in. If the Pair contains the right index then we use the result item. Otherwise, we throw an exception.


Write a function called ‘mimicToString’ that mimics List’s own toString method. That is, it should return a String containing a comma-delimited series of string representations of the list contents with “List(” on the left and “)” on the right.

def mimicToString[A](list: List[A]): String = list match {
  case head :: tail => tail.foldLeft("List(" + head)(_ + ", " + _) + ")"
  case Nil => "List()"

This one also uses a pattern match, but this time the match happens first. The pattern match just treats the empty list as a special case. For the general case (a non-empty list) we use, of course, foldLeft. The accumulator starts out as “List(” + the head item. Then each remaining item (notice foldLeft is called on tail) is appended with a leading “, ” and a final “)” is added to the result of foldLeft.


This one’s kind of fun. Make sure to try it before you look at my solution. Write a function called ‘reverse’ that takes a List and returns the same list in reverse order.

def reverse[A](list: List[A]): List[A] =
  list.foldLeft(List[A]())((r,c) => c :: r)

A very simple solution! The initial value of the accumulator is just an empty list. We don’t use Nil, but instead spell out the List type so that Scala will know what type to make ‘r’. As I say, we start with the empty list which is sensible because the reverse of an empty list is an empty list. Then, as we go through the list, we place each item at the front of the accumulator. So the item at the front of list becomes the last item in the accumulator. This goes on until we reach the end of list, and that last member of list goes onto the front of the accumulator. It’s a really neat and tidy solution.


Write a function called ‘unique’ that takes a List and returns the same List, but with duplicated items removed.

def unique[A](list: List[A]): List[A] =
  list.foldLeft(List[A]()) { (r,c) =>
    if (r.contains(c)) r else c :: r

As usual, we start with an empty list. foldLeft looks at each list item and if it’s already contained in the accumulator then then it stays as it is. If it’s not in the accumulator then it’s appended. This code bears a striking similarity to the ‘reverse’ function we wrote earlier except for the “if (r.contains(c)) r” part. Because of this, the foldLeft result is actually the original list with duplicates removed, but in reverse order. To keep the output in the same order as the input, we add the call to reverse. We could also have chained on the foldLeft from the ‘reverse’ function, like so:

def unique[A](list: List[A]): List[A] =
  list.foldLeft(List[A]()) { (r,c) =>
    if (r.contains(c)) r else c :: r
  }.foldLeft(List[A]())((r,c) => c :: r)


Write a function called ‘toSet’ that takes a List and returns a Set containing the unique elements of the list.

def toSet[A](list: List[A]): Set[A] =
  list.foldLeft(Set[A]())( (r,c) => r + c)

Super easy one. You just start out with an empty Set, which would be the right answer for an empty List. Then you just add each list item to the accumulator. Since the accumulator is a Set, it takes care of eliminating duplicates for you.


Write a function called ‘double’ that takes a List and a new List in which each item appears twice in a row. For example double(List(1, 2, 3)) should return List(1, 1, 2, 2, 3, 3).

def double[A](list: List[A]): List[A] =
  list.foldLeft(List[A]())((r,c) => c :: c :: r).reverse

Again, pretty easy. Are you starting to see a pattern. When you use foldLeft to transform one list into another, you usually end up with the reverse of what you really want.

Alternately, you could have used the foldRight method instead. This does the same thing as foldLeft, except it accumulates its result from back to front instead of front to back. I can’t recommend using it, though, due to problems I point out in my other post on foldLeft and foldRight. But here’s what it would look like:

def double[A](list: List[A]): List[A] =
  list.foldRight(List[A]())((c,r) => c :: c :: r)


This one takes some thinking. Write a function called ‘insertionSort’ that uses foldLeft to sort the input List using the insertion sort algorithm. Try it on your own before you look at the solution.

Need a hint? Use List’s ‘span’ method.

Did you find a solution? Here’s mine:

def insertionSort[A <% Ordered[A]](list: List[A]): List[A] =
  list.foldLeft(List[A]()) { (r,c) =>
    val (front, back) = r.span(_ < c)
    front ::: c :: back

First, the type parameter ensures that we have elements that can be arranged in order. We start, predictably, with an empty list as our initial accumulator. Then, for each item we assume the accumulator is in order (which it always will be), and use span to split it into two sub-lists: all already-sorted items less than the current item, and all already-sorted items greater than or equal to the current item. We put the current item in between these two and the accumulator remains sorted. This is, of course, not the fastest way to sort a list. But it’s a neat foldLeft trick.


Speaking of sorting, you can implement part of quicksort with foldLeft, the pivot. Write a function called ‘pivot’ that takes a List, and returns a Tuple3 containing: (1) a list of all elements less than the original list’s first element, (2) the first element, and (3) a List of all elements greater than or equal to the first element.

def pivot[A <% Ordered[A]](list: List[A]): (List[A],A,List[A]) =
  list.tail.foldLeft[(List[A],A,List[A])]( (Nil, list.head, Nil) ) {
    (result, item) =>
    val (r1, pivot, r2) = result
    if (item < pivot) (item :: r1, pivot, r2) else (r1, pivot, item :: r2)

We’re using the first element, head, as the pivot value, so we skip the head and call foldLeft on list.tail. We initialize the accumulator to a Tuple3 containing the head element with an empty list on either side. Then for each item in the list we just pick which of the two lists to add to based on a comparison with the pivot value.

If you take the additional step of turning this into a recursive call, you can implement a quicksort algorithm. It probably won’t be a very efficient one because it will involve a lot of building and rebuilding lists. Give it a try if you like, and then look at my solution:

def quicksort[A <% Ordered[A]](list: List[A]): List[A] = list match {
  case head :: _ :: _ =>
    list.foldLeft[(List[A],List[A],List[A])]( (Nil, Nil, Nil) ) {
      (result, item) =>
      val (r1, r2, r3) = result
      if      (item < head) (item :: r1, r2, r3)
      else if (item > head) (r1, r2, item :: r3)
      else                  (r1, item :: r2, r3)
    } match {
      case (list1, list2, list3) =>
        quicksort(list1) ::: list2  ::: quicksort(list3)
  case _ => list

Basically, for all lists that have more than 1 element the function chooses the head element as the pivot value, uses foldLeft to divide the list into three (less than, equal to, and greater than the pivot), recursively sorts the less-than and greater-than lists, and knits the three together.


Ok, we got a little into the weeds with that last one. Here’s a simpler one. Write a function called ‘encode’ that takes a List and returns a list of Pairs containing the original values and the number of times they are repeated. So passing List(1, 2, 2, 2, 2, 2, 3, 2, 2) to encode will return List((1, 1), (2, 5), (3, 1), (2, 2)).

def encode[A](list: List[A]): List[(A,Int)] =
list.foldLeft(List[(A,Int)]()){ (r,c) =>
    r match {
      case (value, count) :: tail =>
        if (value == c) (c, count+1) :: tail
        else            (c, 1) :: r
      case Nil =>
        (c, 1) :: r


You knew this was coming. Write a function called ‘decode’ that does the opposite of encode. Calling ‘decode(encode(list))’ should return the original list.

def decode[A](list: List[(A,Int)]): List[A] =
list.foldLeft(List[A]()){ (r,c) =>
    var result = r
    for (_ <- 1 to c._2) result = c._1 :: result

Encode and decode could both have been written by using foldRight and dropping the call to reverse.


One last example. Write a function called ‘group’ that takes a List and an Int size that groups elements into sublists of the specified sizes. So calling “group( List(1, 2, 3, 4, 5, 6, 7), 3)” should return List(List(1, 2, 3), List(4, 5, 6), List(7)). Don’t forget to make sure list items are in the right order. Try it yourself before you look at the solution below.

def group[A](list: List[A], size: Int): List[List[A]] =
  list.foldLeft( (List[List[A]](),0) ) { (r,c) => r match {
    case (head :: tail, num) =>
      if (num < size)  ( (c :: head) :: tail , num + 1 )
      else             ( List(c) :: head :: tail , 1 )
    case (Nil, num) => (List(List(c)), 1)
  }._1.foldLeft(List[List[A]]())( (r,c) => c.reverse :: r)

This code uses the first foldLeft to group the items in a way that’s convenient to list operations, and that last foldLeft to fix the order, which would otherwise be wrong in both the outer and inner lists.

The End!

That’s all for now. If you know of any neat foldLeft tricks, please do leave a comment. I’d be interested to hear about it.