Finding “maximum” overlapping interval pair in O(nlog(n))

Question

Problem Statement

Input set of n intervals; {[s_1,t_1], [s_2,t_2], ... ,[s_n,t_n]}.

Output pair of intervals;

Answer 1

First, sort the intervals: first by left endpoint in increasing order, then — as a secondary criterion — by right endpoint in decreasing order. For the rest of this answer, I'll assume that the intervals are already in sorted order.

Now, there are two possibilities for what the maximum possible overlap might be:

• it may be between an interval and a later interval that it completely covers.
• it may be between an interval and the very next interval that it doesn't completely cover.

We can cover both cases in O(n) time by iterating over the intervals, keeping track of the following:

• the greatest overlap we've seen so far, and the relevant pair of intervals.
• the latest interval we've seen, call it L, that wasn't completely covered by any of its predecessors. (For this, the key insight is that thanks to the ordering of the intervals, we can easily tell if an interval is completely covered by any of its predecessors — and therefore if we need to update L — by simply checking if it's completely covered by the current L. So we can keep L up-to-date in O(1) time.)

and computing each interval's overlap with L.

So:

result := []
max_overlap := 0
L := sorted_intervals[1]
for interval I in sorted_intervals[2..n]:
    overlap := MIN(L.right, I.right) - I.left
    if overlap >= max_overlap:
        result := [L, I]
        max_overlap := overlap
    if I.right > L.right:
        L := I

So the total cost is the cost of sorting the intervals, which is likely to be O(n log n) time but may be O(n) if you can use bucket-sort or radix-sort or similar.