Solution
Approach 1: Merge Intervals
Intuition
In an interval [a, b]
, call b
the "endpoint".
Among the given intervals, consider the interval A[0]
with the smallest endpoint. (Without loss of generality, this interval occurs in array A
.)
Then, among the intervals in array B
, A[0]
can only intersect one such interval in array B
. (If two intervals in B
intersect A[0]
, then they both share the endpoint of A[0]
 but intervals in B
are disjoint, which is a contradiction.)
Algorithm
If A[0]
has the smallest endpoint, it can only intersect B[0]
. After, we can discard A[0]
since it cannot intersect anything else.
Similarly, if B[0]
has the smallest endpoint, it can only intersect A[0]
, and we can discard B[0]
after since it cannot intersect anything else.
We use two pointers, i
and j
, to virtually manage "discarding" A[0]
or B[0]
repeatedly.
Complexity Analysis

Time Complexity: , where are the lengths of
A
andB
respectively. 
Space Complexity: , the maximum size of the answer.
Analysis written by: @awice.