# 99 Scala Problems 04 - Find the number of elements of a list

By Leonardo Giordani - Updated on

## The problem¶

**P04** (*) Find the number of elements of a list.

Example:

```
scala> length(List(1, 1, 2, 3, 5, 8))
res0: Int = 6
```

## Initial thoughts¶

This should be straightforward both with object-oriented and functional techniques. There are no special cases except that of an empty list. There is an interesting solution that mixes functional and object-oriented techniques, which is known as folding, and I'll discuss it in detail.

As already said for problem 02, this function may help solving that problem in a pure recursive way, so we'll talk about this too.

## The procedural solution¶

```
def length[A](l:List[A]):Int = {
l.length
}
```

No special things to highlight here. The `length()`

method works with empty lists too, returning 0, so everything should be fine. As already shown in problem 01 the function may be also reduced to one single line

```
def length[A](l:List[A]):Int = l.length
```

## The recursive solution¶

Coming from procedural languages my first solution was

```
def length[A](l:List[A]):Int = {
def lengthN[A](n:Int, l:List[A]):Int = l match {
case Nil => n
case _::tail => lengthN(n + 1, tail)
}
lengthN(0,l)
}
```

which works fine. I create an helper function that calls itself on each element in the list passing an incremented accumulator and the remaining elements of the list. Then I call it passing an empty accumulator and the whole list.

There is a simpler way to express the same algorithm, which is

```
def lengthRecursive[A](l:List[A]):Int = l match {
case Nil => 0
case h::tail => 1 + lengthRecursive(tail)
}
```

The first solution is better, anyhow, because that function is tail recursive. You can find a good explanation of tail recursion here. The explanation given by "learn you some Erlang!" (here) is perhaps clearer, even if discussed for a different language.

Basically the first solution may be optimized since all local variables are useless once the recursive call is performed, thus the stack may be freed.

## Folding¶

There is a third way to solve the problem, mixing object-oriented and functional approaches, through the use of folding. This is something new for me, and it took me a little to grasp the concept.

```
def length[A](l:List[A]):Int = l.foldLeft(0) { (c,_) => c + 1 }
```

The `foldLeft()`

definition is the following

```
def foldLeft[B](z: B)(f: (B, A) ⇒ B): B
```

and the documentation states: "Applies a binary operator to a start value and all elements of this sequence, going left to right".

So basically the `foldLeft()`

method visits each element of the list from left to right and applies a function passing the result of the previous application. The first element receives the initial value.

The first strange thing that I see here is that the function is defined with two sets of brackets, the first encompassing "the start value" `z`

and the second what seems the definition of a function.

This has to do with the concepts of partially applied functions (NOT *partial functions*) and currying. I recommend two readings from Stackoverflow: this answer about the meaning of partial function, partially applied function and currying, and this answer (which also links the first one) that explores the difference between partially applied and currying. I consider that two answers to be exhaustive so I will not add any explanation about this topic.

There is another thing to note in the folding solution, however, which is the way Scala defines anonymous functions. The Scala tour has a good explanation of anonymous functions, which are pretty simple.

In the given solution for the problem, the anonymous function can be written more explicitly like

```
def count[A](c:Int, d:A):Int = {
c + 1
}
def length[A](l:List[A]):Int = l.foldLeft(0) { count }
```

which should be clear. The count function accepts an `Int`

and a value of the generic type `A`

and returns `c + 1`

. This function is applied on each element of the list by the `foldLeft()`

method, passing the result of the previous application as the first parameter `c`

.

The `count()`

function may however be simplified to an anonymous function due to its simplicity. Making use of type inference and placeholder syntax we may reduce it to the given form.

## Last nth element¶

Now that we have a pure recursive solution for counting elements in a list we may go back to problem 02 and give another solution.

```
def lastNth[A](n: Int, l:List[A]): A = {
def length[A](l:List[A]):Int = {
def lengthN[A](n:Int, l:List[A]):Int = l match {
case Nil => n
case _::tail => lengthN(n + 1, tail)
}
lengthN(0,l)
}
def findKth[A](k:Int, l:List[A]):A = (k,l) match {
case (0, h::_) => h
case (k, _::tail) if k > 0 => findKth(k - 1, tail)
case _ => throw new NoSuchElementException
}
val k = length(l) - n
findKth(k, l)
}
```

This simply takes advantage of the function `length()`

developed in this post and of the function `findKth()`

from problem 03.

## Final considerations¶

The discussion about this problem involved a lot of important topics for the Scala programmer: **tail recursion**, **folding**, **partial functions**, **partially applied functions**, **currying**, **anonymous functions**, **type inference** and **placeholder syntax**. We also reached the goal of writing a pure recursive solution to problem 02.

## Feedback¶

The GitHub issues page is the best place to submit corrections.

## Part 4 of the 99 Scala Problems series

### Previous articles

- 99 Scala Problems 01 - Find the last element of a list
- 99 Scala Problems 02 - Find the last but one element of a list
- 99 Scala Problems 03 - Find the Kth element of a list

### Next articles

- 99 Scala Problems 05 - Reverse a list
- 99 Scala Problems 06 - Find out whether a list is a palindrome
- 99 Scala Problems 07 - Flatten a nested list structure
- 99 Scala Problems 08 - Eliminate consecutive duplicates of list elements
- 99 Scala Problems 09 - Pack consecutive duplicates of list elements into sublists
- 99 Scala Problems 10 - Run-length encoding of a list.
- 99 Scala Problems 11 - Modified run-length encoding
- 99 Scala Problems 12 - Decode a run-length encoded list
- 99 Scala Problems 13 - Run-length encoding of a list (direct solution)
- 99 Scala Problems 14 - Duplicate the elements of a list
- 99 Scala Problems 15 - Duplicate the elements of a list a given number of times
- 99 Scala Problems 16-20