Merge two sorted lists of intervals

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]

I gave you the most effective linear algorithm for intersections in your another topic. Union could be made by analogy — MBo, Mar 31 '18 at 03:06
@MBo: You should probably move your answer from there to here (delete it there, insert copy here). — user unknown, Mar 31 '18 at 03:08
@user unknown I was too hurry. Not exact duplicate - here he wants union, there - intersection, but approach is the same. — MBo, Mar 31 '18 at 03:13
What happens to elements not unitable like [25,26]? Are they preserved or discarded? — user unknown, Mar 31 '18 at 03:41
@userunknown I explained what I did myself to try to solve the problem in the second paragraph. Elements like [25,26] are preserved. — user9577088, Mar 31 '18 at 04:40

score 2 · Answer 1 · answered Mar 31 '18 at 03:12

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

user unknown · Answer 2 · 2018-03-31T14:21:46.093

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))

score 1 · Accepted Answer · answered Mar 31 '18 at 05:26

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.

Merge two sorted lists of intervals

3 Answers3