Select the first row in the last group of consecutive rows - sql

Question

How would I select the row that is the first occurrence in the last 'grouping' of consecutive rows, where a grouping is defined by the consecutive appearance of a particular column value (in the example below state).
For example, given the following table:
id
datetime
state
value_needed
1
2021-04-01 09:42:41.319000
incomplete
A
2
2021-04-04 09:42:41.319000
done
B
3
2021-04-05 09:42:41.319000
incomplete
C
4
2021-04-05 10:42:41.319000
incomplete
C
5
2021-04-07 09:42:41.319000
done
D
6
2021-04-012 09:42:41.319000
done
E
I would want the row with id=5 as it it is the first occurrence of state=done in the last (i.e. most recent) grouping of state=done.

Answer 1

Assuming all columns NOT NULL.
SELECT *
FROM tbl t1
WHERE NOT EXISTS (
SELECT FROM tbl t2
WHERE t2.state <> t1.state
AND t2.datetime > t1.datetime
)
ORDER BY datetime
LIMIT 1;
db<>fiddle here
NOT EXISTS is only true for the last group of peers. (There is no later row with a different state.)
ORDER BY datetime and take the first. Voilá.

Answer 2

Here's a window function solution that accesses your table only once (which may or may not perform better for large data sets):
SELECT *
FROM (
SELECT *,
LEAD (state) OVER (ORDER BY datetime DESC)
IS DISTINCT FROM state AS first_in_group
FROM tbl
) t
WHERE first_in_group
ORDER BY datetime DESC
LIMIT 1
A dbfiddle based on Erwin Brandstetter's. To illustrate, here's the value of first_in_group for each row:
id datetime state value_needed first_in_group
---------------------------------------------------------------------
6 2021-04-12 09:42:41.319 done E f
5 2021-04-07 09:42:41.319 done D t
4 2021-04-05 10:42:41.319 incomplete C f
3 2021-04-05 09:42:41.319 incomplete C t
2 2021-04-04 09:42:41.319 done B t
1 2021-04-01 09:42:41.319 incomplete A t