Find first and last trip with same source-destination by day - sql

Question

I have a table with source, dest, and time of the trip. I want to find list of all the sources that had same destination for the first and last trip of the day. Table looks like below:
Source Dest Trip_Time
1 2 2/1/2019 6:00
2 3 2/1/2019 7:00
4 2 2/1/2019 7:00
1 3 2/1/2019 8:00
2 1 2/1/2019 9:00
3 1 2/1/2019 9:00
4 1 2/1/2019 9:00
1 4 2/1/2019 15:00
2 1 2/1/2019 17:30
3 5 2/1/2019 17:30
4 5 2/1/2019 17:30
2 3 2/1/2019 19:45
3 1 2/1/2019 19:45
5 2 2/1/2019 19:45
1 4 2/2/2019 17:00
1 3 2/2/2019 21:00
I have figured out a query to get what I wanted, but I was wondering if there is more optimal way of achieving the result, especially the one that'll work with millions of rows.
select source, max(first_trip) ft, max(last_trip) lt from
(select source, case when (a.max) = 1 then (dest) end as first_trip,
case when (a.min) = 1 then (dest) end as last_trip from (select source, dest, time_trip,
Row_Number() Over (partition by source order by time_trip desc) as max,
Row_Number() Over (partition by source order by time_trip asc) as min from trips) a
where a.max = 1 or a.min = 1) b
group by b.source) c where ft = lt```
Expected result:
caller fc lc
2 3 3
3 1 1
5 2 2

Answer 1

One method is to use first_value() and last_value(). Date functions are notoriously dependent on the database, but you need to extract the date from the trip_time. The following illustrates the idea but the date function could differ on your database:
select distinct source, cast(trip_time as date)
from (select t.*,
first_value(dest) over (partition by source, cast(trip_time as date) order by trip_time asc) as first_dest,
first_value(dest) over (partition by source, cast(trip_time as date) order by trip_time desc) as last_dest
from trips t
) t
where first_dest = last_dest;