Question

I came across a good question while preparing for some programming interviews.

Given a set of possibly overlapping intervals, you need to write a function to return all elementary intervals among them. For example: if you are given intervals as the following list of pairs: {{1,5}, {3,10}, {5,11}, {15,18}, {16,20}}, then you need to return the following:

{{1,3}, {3,5}, {5,10}, {10,11}, {15,16}, {16,18}, {18,20}}

Note the following in the above answer:

• The interval {11,15} is omitted in the answer because it is
  non-existent in the input.
• The interval {1,5} from the input has been split up into {1,3}, {3,5}
  in the answer because of the starting point “3” defined in {3,10} in
  the input that cuts the interval into two elementary intervals.

Method signature in Java:

List<Pair<Integer, Integer>> generateElementaryIntervals(List<Pair<Integer, Integer> intervals)

One of the solutions that I imagined was to separate the input into non-intersecting sets, and then a simple O(NlogN) sort over all numbers in each non-intersecting set will yield the answer. Is there a more efficient way to do it?

Answer 1

You could break this problem up into nested intervals first, then deal with each nesting separately. By nested, I mean intervals that share at least one point. For the example you gave:

{{1,5}, {3,10}, {5,11}, {15,18}, {16,20}}

there are two nestings:

{1,5}, {3,10}, {5,11}

and

{15,18}, {16,20}

In general, to determine the nestings, you can sort the intervals based on the left endpoint (as in your example), then run through and start a new nesting whenever you see {x,y}, {x',y'} with y < x'.

For a nesting, the “elementary intervals” are formed by the sorted sequence (without repeats) of values. In the example, the nestings give

(1,3,5,10,11) -> {1,3}, {3,5}, {5,10}, {10,11}

and

(15,16,18,20) -> {15,16}, {16,18}, {18,20}

So the overall algorithm may look like this:

• Sort the intervals based on the left endpoint
• Run through intervals until {x,y}, {x',y'} with y < x'
• From beginning to {x,y}, make sorted list of endpoints (no repeats), say a0,a1,...,ak
• Add elementary intervals {ai,a(i+1)} for i = 0...k-1
• Remove intervals up to {x,y} and continue from step 2