# Scala algorithm: Reverse first n elements of a queue

Published

## Algorithm goal

Scala has an immutable Queue structure. Write an algorithm to reverse the order of the first n elements of this queue, but not affect the rest.

## Test cases in Scala

``````assert(Queue.apply(1, 2, 3, 4, 5).dequeue._1 == 1)
assert(Queue.apply(1, 2, 3, 4, 5).enqueue(6).dequeue._1 == 1)
assert(
Queue.empty[Int].reverseFirst(nElements = 3) == Queue.empty[Int],
"Empty queue, reversed, is still empty"
)
assert(
Queue.apply(1, 2, 3, 4, 5).reverseFirst(nElements = 3) ==
Queue.apply(3, 2, 1, 4, 5),
"First 3 elements are reversed"
)
assert(
Queue.apply(1, 2).reverseFirst(nElements = 5) == Queue.apply(2, 1),
"Where nElements is > size of queue, the queue is fully reversed"
)
``````

## Algorithm in Scala

19 lines of Scala (compatible versions 2.13 & 3.0), showing how concise Scala can be!

## Explanation

By taking items from the front of the queue, we can reverse them into a List (a Stack, effectively).

The queue that remains after dequeuing is the final queue - and we can simply append the elements of it to the list we created prior. (this is © from www.scala-algorithms.com)

## Scala concepts & Hints

1. ### foldLeft and foldRight

A 'fold' allows you to perform the equivalent of a for-loop, but with a lot less code.

``````def foldMutable[I, O](initialState: O)(items: List[I])(f: (O, I) => O): O =
items.foldLeft(initialState)(f)
``````
2. ### Lazy List

The 'LazyList' type (previously known as 'Stream' in Scala) is used to describe a potentially infinite list that evaluates only when necessary ('lazily').

3. ### Pattern Matching

Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.

``````assert("Hello World".collect {
case character if Character.isUpperCase(character) => character.toLower
} == "hw")
``````
4. ### Stack Safety

Stack safety is present where a function cannot crash due to overflowing the limit of number of recursive calls.

This function will work for n = 5, but will not work for n = 2000 (crash with java.lang.StackOverflowError) - however there is a way to fix it :-)

In Scala Algorithms, we try to write the algorithms in a stack-safe way, where possible, so that when you use the algorithms, they will not crash on large inputs. However, stack-safe implementations are often more complex, and in some cases, overly complex, for the task at hand.

``````def sum(from: Int, until: Int): Int =
if (from == until) until else from + sum(from + 1, until)

def thisWillSucceed: Int = sum(1, 5)

def thisWillFail: Int = sum(1, 300)
``````

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