Fastest way to detect overlapping periods (or time range)

Question

Assuming I have lots of time periods identified by starting and ending timestamp. What would be the quickest way to detect which period overlaps on which periods?

Here is an example:

9 different periods, delimited by a starting (from) and ending (to) timestamp.

A = [ from : 7s , to : 11s]
B = [ from : 1s, to : 8s]
C = [ from : 9s, to : 12s]
D = [ from : 4s, to : 7s]
E = [ from 10s, to: 15s]
F = [ from 0s, to : 5s]
G (oops I skipped it when drawing the image!)
H = [ from: 5s, to: 9s]
I = [ from: 11s, to: 13s]
J = [ from: 7s, to: 14s]

How to retrieve all overlapping periods as quick as possible to get the following result?

[[A,B],[A,C],[A,E],[A,H],[A,J],[B,D],[B,F],[B,H],[B,J],[C,E],[C,I],[C,J],[D,F],[D,H],[D,J],[E,I],[E,J],[H,J],[I,J]]

JSFiddle of my own solution here

And an other similar jsfiddle but this time with real timestamps, from January to March 2017 between 8 am to 18 pm EDT, and there is a lot of them.

JSFiddle with lots of timestamps

If someone can find a quicker way to proceed, that would be great! Each milliseconds is precious for me.

What is more important "quickest" on average or "quickest" worst case? — chux - Reinstate Monica, Dec 21 '17 at 20:50
Do periods C & H overlap? (Are the end-points inclusive? Or is the first point inclusive and the 2nd point exclusive? Or is it a don't care - whatever is fastest) "[[A,B],[A,C],[A,E],[A,H],[A,..." does not seem to imply that is the goal, yet mathematically they do. — chux - Reinstate Monica, Jan 02 '18 at 17:09
Starting and ending does not inclusive. As soon as a starting / ending overlaps, they are in conflict, in my case.. — Simmoniz, Jan 03 '18 at 19:28

score 2 · Answer 1 · answered Dec 21 '17 at 20:45

Sort all the times together, marking each one with the segment it belongs to, and a start/end bit.

Then, keep a list saying which segments you are in (initially empty).

Iterate through the list of times. If a time belongs to segment X, then if it is a start time add X to the second list. If it is an end time, remove X from the second list. At all times the second list tells you which segments are overlapping.

If there are enough segments to care about big-O, the initial sort is O(N Log N). The iteration is O(N).

Of course, don't count on big-O to make you fast. There are still constant factors. Here's how I do it.

I like this approach. I will try to implement it then post the result here. Thank you ! — Simmoniz, Jan 03 '18 at 19:45

score 0 · Answer 2 · answered Dec 21 '17 at 22:19

Python:

from pprint import pprint as pp

intervals = [
    (7 , 11, 'A'),
    (1, 8, 'B'),
    (9, 12, 'C'),
    (4, 7, 'D'),
    (10, 15, 'E'),
    (0, 5, 'F'),
    (5, 9, 'H'),
    (11, 13, 'I'),
    (7, 14, 'J'),
]
intervals.sort()  # sort on interval start, then end if starts are equal
ans = []
for n0, (_, end0, name0) in enumerate(intervals[:-1]):
    for n1, (start1, _, name1) in enumerate(intervals[n0 + 1:]):
        if start1 < end0:
            ans.append(sorted((name0, name1)))
        else:
            break
pp(sorted(ans))

Output:

[['A', 'B'],
 ['A', 'C'],
 ['A', 'E'],
 ['A', 'H'],
 ['A', 'J'],
 ['B', 'D'],
 ['B', 'F'],
 ['B', 'H'],
 ['B', 'J'],
 ['C', 'E'],
 ['C', 'I'],
 ['C', 'J'],
 ['D', 'F'],
 ['D', 'H'],
 ['E', 'I'],
 ['E', 'J'],
 ['H', 'J'],
 ['I', 'J']]

Fastest way to detect overlapping periods (or time range)

2 Answers2