


1. You can assume that all the meetings will happen on the same day.
2. Also, as soon as a meeting gets over if some other meeting is scheduled to start at that moment, they can then be allocated that room.
Try to solve the problem in linear time complexity.
Consider there are three meetings scheduled with timings:
1pm - 4pm
3pm - 5pm
4pm - 6pm
At the start of time, meeting 1 will be allotted room 1, which will be occupied till 4 pm hence for meeting 2 we’ll have to provide another room. At 4 pm, meeting 3 can be organized in room 1 because by that time, meeting 1 would have ended. Hence we’ll require two rooms for holding all three meetings.
The first line of input contains an integer 'T' representing the number of the test cases. Then the test case follows.
The first line of each test case contains an integer ‘N’ representing the number of meetings scheduled.
The second line of each test case contains N space-separated integers representing the start time for each meeting.
The third line of each test case contains N space-separated integers representing the end time for each meeting.
For each test case, print the minimum number of conference rooms required.
The output of each test case should be printed in a separate line.
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 <= T <= 100
1 <= N <= 1000
0000 <= INTERVAL[start][end] <= 2359
where 'T' is the number of test cases, 'N' is the number of meetings, and 'INTERVAL[start][end]' represents the starting time and ending time of a meeting.
Time limit: 1 second
Create ARRIVAL and DEPARTURE arrays from given array INTERVALS.