Merge Overlapping Time Intervals based on Hierarchy in SQL - sql

Question

I am trying to solve a problem where i want to merge overlapping intervals for a given column id, but i also want to merge them based on hierarchy/priority. I have start_time and stop_time for each interval and each interval has a hierarchy/priority associated with it.
These are the following columns in the table:
id, start_time, stop_time, priority
I was able to solve the problem where i do not have take into account the priority, but i am struggling with this one.
Red colour: p1 (priority 1)
Blue Colour: p2 (priority 2)
Green colour: p3 (priority 3)
Note that in the example input below we will have 9 rows having same id, and the output will have 6 rows. Please note that for some id's might have only some of the priority values or just one, the solution should take care of that.
expected input and output:

Answer 1

Below is for BigQuery Standard SQL
#standardSQL
WITH check_times AS (
SELECT id, start_time AS time FROM `project.dataset.table` UNION DISTINCT
SELECT id, stop_time AS time FROM `project.dataset.table`
), distinct_intervals AS (
SELECT id, time AS start_time, LEAD(time) OVER(PARTITION BY id ORDER BY time) stop_time
FROM check_times
), deduped_intervals AS (
SELECT a.id, a.start_time, a.stop_time, MIN(priority) priority
FROM distinct_intervals a
JOIN `project.dataset.table` b
ON a.id = b.id
AND a.start_time BETWEEN b.start_time AND b.stop_time
AND a.stop_time BETWEEN b.start_time AND b.stop_time
GROUP BY a.id, a.start_time, a.stop_time
), combined_intervals AS (
SELECT id, MIN(start_time) start_time, MAX(stop_time) stop_time, ANY_VALUE(priority) priority
FROM (
SELECT id, start_time, stop_time, priority, COUNTIF(flag) OVER(PARTITION BY id ORDER BY start_time) grp
FROM (
SELECT id, start_time, stop_time, priority,
start_time != IFNULL(LAG(stop_time) OVER(PARTITION BY id ORDER BY start_time), start_time) OR
priority != IFNULL(LAG(priority) OVER(PARTITION BY id ORDER BY start_time), -1) flag
FROM deduped_intervals
)
)
GROUP BY id, grp
)
SELECT *
FROM combined_intervals
-- ORDER BY id, start_time
If to apply to sample data from your question - result is
Can you also share a solution where we merge intervals based on just id and no priority column
I just simply slightly adjusted above query to ignore priority
#standardSQL
WITH check_times AS (
SELECT id, start_time AS TIME FROM `project.dataset.table` UNION DISTINCT
SELECT id, stop_time AS TIME FROM `project.dataset.table`
), distinct_intervals AS (
SELECT id, TIME AS start_time, LEAD(TIME) OVER(PARTITION BY id ORDER BY TIME) stop_time
FROM check_times
), deduped_intervals AS (
SELECT a.id, a.start_time, a.stop_time
FROM distinct_intervals a
JOIN `project.dataset.table` b
ON a.id = b.id
AND a.start_time BETWEEN b.start_time AND b.stop_time
AND a.stop_time BETWEEN b.start_time AND b.stop_time
GROUP BY a.id, a.start_time, a.stop_time
), combined_intervals AS (
SELECT id, MIN(start_time) start_time, MAX(stop_time) stop_time
FROM (
SELECT id, start_time, stop_time, COUNTIF(flag) OVER(PARTITION BY id ORDER BY start_time) grp
FROM (
SELECT id, start_time, stop_time,
start_time != IFNULL(LAG(stop_time) OVER(PARTITION BY id ORDER BY start_time), start_time) flag
FROM deduped_intervals
)
)
GROUP BY id, grp
)
SELECT *
FROM combined_intervals
-- ORDER BY id, start_time
with result
Row id start_time stop_time
1 1 0 36
2 1 41 47

Answer 2

This is a "combining" islands problem. One solution is to find where the islands begin and do a cumulative sum of the beginnings. You can determine the beginning by seeing where there are no overlaps:
select id, priority, min(start_time), max(stop_time)
from (select t.*,
countif(coalesce(prev_stop_time, stop_time) < stop_time) over (partition by id, priority order by start_time) as grp
from (select t.*,
max(stop_time) over (partition by id, priority order by start_time rows between unbounded preceding and 1 preceding) as prev_stop_time
from t
) t
) t
group by id, priority, grp;