SQL get last column in a group based on order by and value of another column

From the table below, I am trying to create another column (last_k2X) that will show the last key2 where type was set to 'X' over time (ti).

In case X was set at the same time (ts) for multiple key2, (in same key1 partition), the new column last_k2X will have the key2 for all key2 in that partition/time.

Input:

I tried to use window functions like FIRST_VALUE() and LAG() but could not manage to have the right results. I expected the result like:

Expected output:

As you didn't specify the database system you are using, this solution is in MySQL, but can be simply converted to other database systems.

This is basically a gaps and island problem, but need to get the right partitions

CREATE TABLE data
    (key1 int, key2 varchar(1), ts varchar(2), type varchar(4))
;
    
INSERT INTO data
    (key1, key2, ts, type)
VALUES
    (1, 'A', 't0', NULL),
    (1, 'B', 't1', 'a'),
    (1, 'C', 't1', 'X'),
    (1, 'D', 't2', 'b'),
    (1, 'E', 't3', NULL),
    (1, 'F', 't4', 'c'),
    (1, 'G', 't5', 'X'),
    (1, 'H', 't5', NULL),
    (1, 'I', 't6', 'd')
;

WITH x_sel AS (SELECT key1, key2, ts,type, CASE WHEN type = 'X' then 1 ELSE 0 END rk
  FROM data
), CTE2 as
(SELECT 
x_sel.key1, x_sel.key2, x_sel.ts,x_sel.type,x_sel2.key2 k2X, SUM(rk) OVER (PARTITION BY x_sel.key1 ORDER BY x_sel.key2) s_rk  FROM x_sel
LEFT JOIN (SELECT key1, key2, ts FROM  data WHERE  type = 'X') x_sel2 
  ON x_sel.key1 = x_sel2.key1 ANd x_sel.ts = x_sel2.ts)
SELECT key1, key2, ts,type,CASE WHEN s_rk = 0 THEn k2x ELSE  COALESCE(k2x,MAX(k2X) OVER(PARTITION BY s_rk ORDER BY S_rk)) END k2x
FROM CTE2
  

fiddle