Given a list of intervals, merge all the overlapping intervals to produce a list that has only mutually exclusive intervals.
Intervals: [[1,4], [2,5], [7,9]] Output: [[1,5], [7,9]] Explanation: Since the first two intervals [1,4] and [2,5] overlap, we merged them into one [1,5].
Intervals: [[6,7], [2,4], [5,9]] Output: [[2,4], [5,9]] Explanation: Since the intervals [6,7] and [5,9] overlap, we merged them into one [5,9].
Intervals: [[1,4], [2,6], [3,5]] Output: [[1,6]] Explanation: Since all the given intervals overlap, we merged them into one.
Let’s take the example of two intervals (‘a’ and ‘b’) such that
a.start <= b.start. There are four possible scenarios:
Our goal is to merge the intervals whenever they overlap. For the above-mentioned three overlapping scenarios (2, 3, and 4), this is how we will merge them:
The diagram above clearly shows a merging approach. Our algorithm will look like this:
- Sort the intervals on the start time to ensure
a.start <= b.start.
- If ‘a’ overlaps ‘b’ (i.e.
b.start <= a.end), we need to merge them into a new interval ‘c’ such that:
c.start = a.start c.end = max(a.end, b.end)
- We will keep repeating the above two steps to merge ‘c’ with the next interval if it overlaps with ‘c’.
Here is what our algorithm will look like:
The time complexity of the above algorithm is O(N * logN), where ‘N’ is the total number of intervals. We are iterating the intervals only once which will take O(N), in the beginning though, since we need to sort the intervals, our algorithm will take O(N * logN).
The space complexity of the above algorithm will be O(N) as we need to return a list containing all the merged intervals. We will also need O(N) space for sorting. For Java, depending on its version, Collections.sort() either uses Merge sort or Timsort, and both these algorithms need O(N) space. Overall, our algorithm has a space complexity of O(N).
Problem 1: Given a set of intervals, find out if any two intervals overlap.
Intervals: [[1,4], [2,5], [7,9]]
Explanation: Intervals [1,4] and [2,5] overlap
Solution: We can follow the same approach as discussed above to find if any two intervals overlap.