Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

How to find the largest element in a list of integers recursively?

Tags:

I'm trying to write a function which will recursively find the largest element in a list of integers. I know how to do this in Java, but can't understand how to do this at Scala.

Here is what I have so far, but without recursion:

  def max(xs: List[Int]): Int = {
    if (xs.isEmpty) throw new java.util.NoSuchElementException();
    else xs.max;
  }

How can we find it recursively with Scala semantic.

like image 997
catch23 Avatar asked Sep 27 '13 06:09

catch23


People also ask

How do you use the recursion to find the largest element in an array?

Take array Arr[] as input. Function recforMax(int arr[], int len) takes input array and its length and returns maximum in the array using recursion. Take the integer variable maximum. If the current index len is 1 then set maximum=arr[0] and return maximum.

How do you find the largest element in an array using recursion in C++?

Method 1(Using Recursion) :Create a recursive function say, largest_element (int n, int arr[]). Base Condition : If(n==1) return arr[0]. Else, return max(arr[n-1], largest_element(n-1, arr))

How do you find largest int array?

To find the largest element from the array, a simple way is to arrange the elements in ascending order. After sorting, the first element will represent the smallest element, the next element will be the second smallest, and going on, the last element will be the largest element of the array.


2 Answers

This is the most minimal recursive implementation of max I've ever been able to think up:

def max(xs: List[Int]): Option[Int] = xs match {
  case Nil => None
  case List(x: Int) => Some(x)
  case x :: y :: rest => max( (if (x > y) x else y) :: rest )
} 

It works by comparing the first two elements on the list, discarding the smaller (or the first, if both are equal) and then calling itself on the remaining list. Eventually, this will reduce the list to one element which must be the largest.

I return an Option to deal with the case of being given an empty list without throwing an exception - which forces the calling code to recognise the possibility and deal with it (up to the caller if they want to throw an exception).

If you want it to be more generic, it should be written like this:

def max[A <% Ordered[A]](xs: List[A]): Option[A] = xs match {
  case Nil => None
  case x :: Nil => Some(x)
  case x :: y :: rest => max( (if (x > y) x else y) :: rest )
}

Which will work with any type which either extends the Ordered trait or for which there is an implicit conversion from A to Ordered[A] in scope. So by default it works for Int, BigInt, Char, String and so on, because scala.Predef defines conversions for them.

We can become yet more generic like this:

def max[A <% Ordered[A]](xs: Seq[A]): Option[A] = xs match {
  case s if s.isEmpty || !s.hasDefiniteSize => None
  case s if s.size == 1 => Some(s(0))
  case s if s(0) <= s(1) => max(s drop 1)
  case s => max((s drop 1).updated(0, s(0)))
}

Which will work not just with lists but vectors and any other collection which extends the Seq trait. Note that I had to add a check to see if the sequence actually has a definite size - it might be an infinite stream, so we back away if that might be the case. If you are sure your stream will have a definite size, you can always force it before calling this function - it's going to work through the whole stream anyway. See notes at the end for why I really would not want to return None for an indefinite stream, though. I'm doing it here purely for simplicity.

But this doesn't work for sets and maps. What to do? The next common supertype is Iterable, but that doesn't support updated or anything equivalent. Anything we construct might be very poorly performing for the actual type. So my clean no-helper-function recursion breaks down. We could change to using a helper function but there are plenty of examples in the other answers and I'm going to stick with a one-simple-function approach. So at this point, we can to switch to reduceLeft (and while we are at it, let's go for `Traversable' and cater for all collections):

def max[A <% Ordered[A]](xs: Traversable[A]): Option[A] = {
  if (xs.hasDefiniteSize) 
    xs reduceLeftOption({(b, a) => if (a >= b) a else b}) 
  else None
}

but if you don't consider reduceLeft recursive, we can do this:

def max[A <% Ordered[A]](xs: Traversable[A]): Option[A] = xs match {
  case i if i.isEmpty => None
  case i if i.size == 1 => Some(i.head)
  case i if (i collect { case x if x > i.head => x }).isEmpty => Some(i.head)
  case _ => max(xs collect { case x if x > xs.head => x })
}

It uses the collect combinator to avoid some clumsy method of bodging a new Iterator out of xs.head and xs drop 2.

Either of these will work safely with almost any collection of anything which has an order. Examples:

scala>  max(Map(1 -> "two", 3 -> "Nine", 8 -> "carrot"))
res1: Option[(Int, String)] = Some((8,carrot))

scala> max("Supercalifragilisticexpialidocious")
res2: Option[Char] = Some(x)

I don't usually give these others as examples, because it requires more expert knowledge of Scala.

Also, do remember that the basic Traversable trait provides a max method, so this is all just for practice ;)

Note: I hope that all my examples show how careful choice of the sequence of your case expressions can make each individual case expression as simple as possible.

More Important Note: Oh, also, while I am intensely comfortable returning None for an input of Nil, in practice I'd be strongly inclined to throw an exception for hasDefiniteSize == false. Firstly, a finite stream could have a definite or non-definite size dependent purely on the sequence of evaluation and this function would effectively randomly return Option in those cases - which could take a long time to track down. Secondly, I would want people to be able to differentiate between having passed Nil and having passed truly risk input (that is, an infinite stream). I only returned Option in these demonstrations to keep the code as simple as possible.

like image 135
itsbruce Avatar answered Oct 16 '22 12:10

itsbruce


The easiest approach would be to use max function of TraversableOnce trait, as follows,

val list = (1 to 10).toList
list.max

to guard against the emptiness you can do something like this,

if(list.empty) None else Some(list.max)

Above will give you an Option[Int]

My second approach would be using foldLeft

(list foldLeft None)((o, i) => o.fold(Some(i))(j => Some(Math.max(i, j))))

or if you know a default value to be returned in case of empty list, this will become more simpler.

val default = 0
(list foldLeft default)(Math.max)

Anyway since your requirement is to do it in recursive manner, I propose following,

def recur(list:List[Int], i:Option[Int] = None):Option[Int] = list match {
  case Nil => i
  case x :: xs => recur(xs, i.fold(Some(x))(j => Some(Math.max(j, x))))
}

or as default case,

val default = 0
def recur(list:List[Int], i:Int = default):Int = list match {
  case Nil => i
  case x :: xs => recur(xs, i.fold(x)(j => Math.max(j, x)))
}

Note that, this is tail recursive. Therefore stack is also saved.

like image 35
tiran Avatar answered Oct 16 '22 10:10

tiran