SQL - Lag to previous distinct value

Is there a way to modify a Lag function to not only go back a previous record, but back to a previous distinct value?

Current output:

Desired output:

Lag(a1."Job Date", 1) over (partition by a1."Employee Number" Order By a1."Employee Number")

Right now I'm using the above expression, but this is returning the previous records value instead of the previous distinct/non-matching value. Is this possible with lag or perhaps a different function?

This is a nifty trick. If you use the range between with "1 PRECEDING" it starts the window at the previous distinct value. Assumes you have some key you want to partition on, but if that's not the case you can just remove the PARTITION BY clause.

Fiddle: https://dbfiddle.uk/3tPzIiDN

create table some_test_data
  (
    id integer,
    job_date date
  );

insert into some_test_data values
  (1, date '2023-06-01'),
  (1, date '2023-06-01'),
  (1, date '2023-06-01'),
  (1, date '2023-07-01'),
  (1, date '2023-07-01'),
  (1, date '2023-07-16'),
  (1, date '2023-07-16'),
  (1, date '2023-07-16');

select s.*,
       max(job_date) over (
         partition by id
         order by job_date
         range between unbounded preceding and 1 preceding)
  from some_test_data s;

Or, the equivalent with correlated subquery

select t1.*,
       ( select max(t2.job_date)
           from some_test_data t2
          where t1.id = t2.id
            and t1.job_date > t2.job_date
       ) as last_distinct_job_date
  from some_test_data t1;