Merge Intervals

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Constraints:

• 1 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 104

Intuition

If we sort the intervals by their start value, then each set of intervals that can be merged will appear as a contiguous "run" in the sorted list.

Algorithm

First, we sort the list as described. Then, we insert the first interval into our merged list and continue considering each interval in turn as follows: If the current interval begins after the previous interval ends, then they do not overlap and we can append the current interval to merged. Otherwise, they do overlap, and we merge them by updating the end of the previous interval if it is less than the end of the current interval.

A simple proof by contradiction shows that this algorithm always produces the correct answer. First, suppose that the algorithm at some point fails to merge two intervals that should be merged. This would imply that there exists some triple of indices ii, jj, and kk in a list of intervals ints such that i<j<k and (ints[i], ints[k]) can be merged, but neither (ints[i], ints[j]) nor (ints[j], ints[k]) can be merged. From this scenario follow several inequalities:

ints[i].end < ints[j].start

ints[j].end < ints[k].start

ints[i].end} ≥ ints[k].start

We can chain these inequalities (along with the following inequality, implied by the well-formedness of the intervals:ints[j].start≤ ints[j].end) to demonstrate a contradiction:

ints[i].end < ints[j].start ≤ ints[j].end < ints[k].startints[i].end ≥ ints[k].start​

Therefore, all mergeable intervals must occur in a contiguous run of the sorted list.

Complexity Analysis

• Time complexity : O(nlogn)
• Other than the sort invocation, we do a simple linear scan of the list, so the runtime is dominated by the O(nlogn) complexity of sorting.
• Space complexity : O(logN) (or O(n))

With the above technique below similar problems can be solved.

Given an array of meeting time intervals where intervals[i] = [starti, endi], determine if a person could attend all meetings.

Example 1:

Input: intervals = [[0,30],[5,10],[15,20]]
Output: false

Example 2:

Input: intervals = [[7,10],[2,4]]
Output: true