Merge two sorted lists of intervals - algorithm

Question

Given A and B, which are two interval lists. A has no overlap inside A and B has no overlap inside B. In A, the intervals are sorted by their starting points. In B, the intervals are sorted by their starting points. How do you merge the two interval lists and output the result with no overlap?
One method is to concatenate the two lists, sort by the starting point, and apply merge intervals as discussed at https://www.geeksforgeeks.org/merging-intervals/. Is there a more efficient method?
Here is an example:
A: [1,5], [10,14], [16,18]
B: [2,6], [8,10], [11,20]
The output:
[1,6], [8, 20]

Answer 1

So you have two sorted lists with events - entering interval and leaving interval.
Merge these lists keeping current state as integer 0, 1, 2 (active interval count)
Get the next coordinate from both lists
If it is entering event
Increment state
If state becomes 1, start new output interval
If it is closing event
Decrement state
If state becomes 0, close current output interval
Note that this algo is similar to intersection finding there

Answer 2

Here is a different approach, in the spirit of the answer to the question of overlaps.
<!--code lang=scala-->
def findUnite (l1: List[Interval], l2: List[Interval]): List[Interval] = (l1, l2) match {
case (Nil, Nil) => Nil
case (as, Nil) => as
case (Nil, bs) => bs
case (a :: as, b :: bs) => {
if (a.lower > b.upper) b :: findUnite (l1, bs)
else if (a.upper < b.lower) a :: findUnite (as, l2)
else if (a.upper > b.upper) findUnite (a.union (b).get :: as, bs)
else findUnite (as, a.union (b).get :: bs)
}
}
If both lists are empty - return the empty list.
If only one is empty, return the other.
If the upper bound of one list is below the lower bound of the other, there is no unification possible, so return the other and proceed with the rest.
If they overlap, don't return, but call the method recursively, the unification on the side of the more far reaching interval and without the consumed less far reaching interval.
The union method looks similar to the one which does the overlap:
<!--code scala-->
case class Interval (lower: Int, upper: Int) {
// from former question, to compare
def overlap (other: Interval) : Option [Interval] = {
if (lower > other.upper || upper < other.lower) None else
Some (Interval (Math.max (lower, other.lower), Math.min (upper, other.upper)))
}
def union (other: Interval) : Option [Interval] = {
if (lower > other.upper || upper < other.lower) None else
Some (Interval (Math.min (lower, other.lower), Math.max (upper, other.upper)))
}
}
The test for non overlap is the same. But min and max have changed places.
So for (2, 4) (3, 5) the overlap is (3, 4), the union is (2, 5).
lower upper
_____________
2 4
3 5
_____________
min 2 4
max 3 5
Table of min/max lower/upper.
<!--code lang='scala'-->
val e = List (Interval (0, 4), Interval (7, 12))
val f = List (Interval (1, 3), Interval (6, 8), Interval (9, 11))
findUnite (e, f)
// res3: List[Interval] = List(Interval(0,4), Interval(6,12))
Now for the tricky or unclear case from above:
val e = List (Interval (0, 4), Interval (7, 12))
val f = List (Interval (1, 3), Interval (5, 8), Interval (9, 11))
findUnite (e, f)
// res6: List[Interval] = List(Interval(0,4), Interval(5,12))
0-4 and 5-8 don't overlap, so they form two different results which don't get merged.

Answer 3

A simple solution could be, to deflate all elements, put them into a set, sort it, then iterate to transform adjectant elements to Intervals.
A similar approach could be chosen for your other question, just eliminating all distinct values to get the overlaps.
But - there is a problem with that approach.
Lets define a class Interval:
case class Interval (lower: Int, upper: Int) {
def deflate () : List [Int] = {(lower to upper).toList}
}
and use it:
val e = List (Interval (0, 4), Interval (7, 12))
val f = List (Interval (1, 3), Interval (6, 8), Interval (9, 11))
deflating:
e.map (_.deflate)
// res26: List[List[Int]] = List(List(0, 1, 2, 3, 4), List(7, 8, 9, 10, 11, 12))
f.map (_.deflate)
// res27: List[List[Int]] = List(List(1, 2, 3), List(6, 7, 8), List(9, 10, 11))
The ::: combines two Lists, here two Lists of Lists, which is why we have to flatten the result, to make one big List:
(res26 ::: res27).flatten
// res28: List[Int] = List(0, 1, 2, 3, 4, 7, 8, 9, 10, 11, 12, 1, 2, 3, 6, 7, 8, 9, 10, 11)
With distinct, we remove duplicates:
(res26 ::: res27).flatten.distinct
// res29: List[Int] = List(0, 1, 2, 3, 4, 7, 8, 9, 10, 11, 12, 6)
And then we sort it:
(res26 ::: res27).flatten.distinct.sorted
// res30: List[Int] = List(0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12)
All in one command chain:
val united = ((e.map (_.deflate) ::: f.map (_.deflate)).flatten.distinct).sorted
// united: List[Int] = List(0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12)
// ^ (Gap)
Now we have to find the gaps like the one between 4 and 6 and return two distinct Lists.
We go recursively through the input list l, and if the element is from the sofar collected elements 1 bigger than the last, we collect that element into this sofar-list. Else we return the sofar collected list as partial result, followed by splitting of the rest with a List of just the current element as new sofar-collection. In the beginning, sofar is empty, so we can start right with adding the first element into that list and splitting the tail with that.
def split (l: List [Int], sofar: List[Int]): List[List[Int]] = l match {
case Nil => List (sofar)
case h :: t => if (sofar.isEmpty) split (t, List (h)) else
if (h == sofar.head + 1) split (t, h :: sofar)
else sofar :: split (t, List (h))
}
// Nil is the empty list, we hand in for initialization
split (united, Nil)
// List(List(4, 3, 2, 1, 0), List(12, 11, 10, 9, 8, 7, 6))
Converting the Lists into intervals would be a trivial task - take the first and last element, and voila!
But there is a problem with that approach. Maybe you recognized, that I redefined your A: and B: (from the former question). In B, I redefined the second element from 5-8 to 6-8. Because else, it would merge with the 0-4 from A because 4 and 5 are direct neighbors, so why not combine them to a big interval?
But maybe it is supposed to work this way? For the above data:
split (united, Nil)
// List(List(6, 5, 4, 3, 2, 1), List(20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8))