This problem is similar to NumberOfDiscIntersections from Codility.
The best illustration for this algorithm is on Stack Overflow: Counting Disk Intersections.
Given a list L, the item at index k is a disc on a 2D plane at position (0, i), and L[k] is the disc's radius. An intersection is if two discs at minimum touch each other.
For instance, a List of [1, 1] has just 1 intersection, whereas a list [1, 1, 1] has 2 because every disc touches every other disc.
Test cases in Scala
assert(brute(List(1)) == 0) assert(brute(List(0, 0)) == 0) assert(brute(List(0, 1)) == 1) assert(brute(List(1, 1)) == 1) assert(brute(List(1, 1, 1)) == 3) assert(brute(List(1, 0, 0, 1)) == 2) assert(brute(List(5, 0, 0, 1)) == 4) assert(efficient(List(1)) == 0) assert(efficient(List(0, 0)) == 0) assert(efficient(List(0, 1)) == 1) assert(efficient(List(1, 1)) == 1) assert(efficient(List(1, 1, 1)) == 3) assert(efficient(List(1, 0, 0, 1)) == 2) assert(efficient(List(5, 0, 0, 1)) == 4)
Algorithm in Scala
69 lines of Scala (compatible versions 2.13 & 3.0).
We begin with a brute-force solution, so that we could compare it to an optimised solution. The brute solution is highly inefficient as it traverses the list twice. We can do better by traversing it once:
We begin by counting the total range using O(n) operations, and then for each item in that range, we check how many new openings there are compared against what has been opened. When 5 new openings are added, if there are 4 existing ones, that gives us 20 extra openings, as well as the number of openings that we get from the 5 intersecting each other (that is 5 * 4 / 2, ie Binomial '(n choose 2)'). (this is © from www.scala-algorithms.com)
Note that this solution is highly verbose for the purpose of comprehension and readability. Many online solutions use convoluted logic that is quite hard to comprehend. It means: this can be written much more concisely if needed.
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 for-comprehension is highly important syntatic enhancement in functional programming languages.
Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.
(1 to n)syntax produces a "Range" which is a representation of a sequence of numbers.
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.
A state machine is the use of `sealed trait` to represent all the possible states (and transitions) of a 'machine' in a hierarchical form.
.viewsyntax creates a structure that mirrors another structure, until "forced" by an eager operation like .toList, .foreach, .forall, .count.