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!
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
foldLeft and foldRight
A 'fold' allows you to perform the equivalent of a for-loop, but with a lot less code.
The 'LazyList' type (previously known as 'Stream' in Scala) is used to describe a potentially infinite list that evaluates only when necessary ('lazily').
Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.
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.