ORACLE SQL: I've got a crude solution to a problem using LEAD and CASE WHEN but would like a more elegant solution - loops

Question

Goal: Given a table of patient records that show entry and exit dates per facility, I want to find the Original_entrydate for each record in the table.
Consider this table, which is sorted by entry_date descending and grouped by ID:
ID
Facility
entry_date
exit_date
original_entrydate
003246
C
2022-03-22
null
2012-10-01
003246
B
2015-07-24
2022-03-22 (matches row 1 entry_date)
2012-10-01
003246
A
2012-10-01
2015-07-24 (matches row 2 entry_date)
2012-10-01
003246
D
2001-02-02
2010-04-05 (does NOT match row 3 entry_date!)
2001-02-02
Starting from the earliest record, Patient #003246
Entered Facility D on 2/2/2001 and left on 4/5/2010.
Entered Facility A on 10/1/2012 and left on 7/24/2015 - so there was a gap in service between 2010 & 2012.
Entered Facility B on 7/24/2015 and left on 3/22/2022
Entered Facility C on 3/22/2022 and remains there today.
As you can see, when the individual entered Facility A, there has been no gap in service between 10/1/2012 and today--even though the patient switched facilities, they have been a resident since 10/1/2012--therefore 10/1/2012 is the original_entrydate for the first three records in the table, but not the 4th record, since there was a gap in service between Facilities D and A.
I managed to write a query that gave me what I want, but it's extremely clunky and will fail if a resident appears more than 4 times in the table (because my LEADs only go 4 rows deep--sufficient for my test table but not necessarily for the actual data).
SELECT
t1.resident_key,
t1.lidda,
t1.facility_key,
t1.admit_dt_key,
t1.separation_dt_key,
CASE
WHEN prev_sep1 IS NULL
THEN admit_dt_key
WHEN admit_dt_key != prev_sep1
THEN admit_dt_key
WHEN admit_dt_key = prev_sep1
THEN CASE
WHEN prev_sep2 IS NULL
THEN prev_admit1
WHEN prev_admit1 != prev_sep2
THEN prev_admit1
WHEN prev_admit1 = prev_sep2
THEN CASE
WHEN prev_sep3 IS NULL
THEN prev_admit2
WHEN prev_admit2 != prev_sep3
THEN prev_admit2
WHEN prev_admit2 = prev_sep3
THEN CASE
WHEN prev_sep4 IS NULL
THEN prev_admit3
WHEN prev_admit3 != prev_sep4
THEN prev_admit3
END
END
END
END
as Original_AdmDt
FROM
(SELECT my_table.*,
LEAD (separation_dt_key, 1) OVER (
PARTITION BY resident_key
ORDER BY admit_dt_key DESC
) prev_sep1,
LEAD (admit_dt_key, 1) OVER (
PARTITION BY resident_key
ORDER BY admit_dt_key DESC
) prev_admit1,
LEAD (separation_dt_key, 2) OVER (
PARTITION BY resident_key
ORDER BY admit_dt_key DESC
) prev_sep2,
LEAD (admit_dt_key, 2) OVER (
PARTITION BY resident_key
ORDER BY admit_dt_key DESC
) prev_admit2,
LEAD (separation_dt_key, 3) OVER (
PARTITION BY resident_key
ORDER BY admit_dt_key DESC
) prev_sep3,
LEAD (admit_dt_key, 3) OVER (
PARTITION BY resident_key
ORDER BY admit_dt_key DESC
) prev_admit3,
LEAD (separation_dt_key, 4) OVER (
PARTITION BY resident_key
ORDER BY admit_dt_key DESC
) prev_sep4,
LEAD (admit_dt_key, 4) OVER (
PARTITION BY resident_key
ORDER BY admit_dt_key DESC
) prev_admit4
FROM my_table) t1
Can I make this iterative, to deal with however many times a unique ID may appear in the table?

Answer 1

SELECT t1.date1, t2.date2,
CASE
WHEN t1.date1 < t2.date2 THEN t1.date1
ELSE t2.date2
END AS result
FROM table1 t1
JOIN table1 t2
ON t1.id < t2.id
WHERE t1.date1 < t2.date2
ORDER BY t1.id, t2.id;