Oracle self join starting with minimum value for each partition - sql

Question

I have this table:
COL1 COL2 COL3
--------------------
A 1 VAL1
A 2 VAL2
A 4 VAL3
B 2 VAL4
B 4 VAL5
B 5 VAL6
And I would like to obtain this output:
COL1 COL2 COL3
--------------------
A 1 VAL1
A 2 VAL2
A 3 NULL
B 2 VAL4
B 3 NULL
B 4 VAL6
Logic:
with the smallest COL2 value for each partition of COL1, take the following 3 numbers and, if the combination COL1 and COL2 present in the first table, show COL3 and NULL otherwise.

Answer 1

Your question is a good example of what PARTITIONED OUTER JOIN was created for: DBFiddle
with top3 as (
select *
from (
select
col1, col2, col3
,min(col2)over(partition by col1) min_col2
,col2 - min(col2)over(partition by col1) + 1 as rn
from t
)
where col2 < min_col2 + 3
)
select
top3.col1
,r3.n as col2
,top3.col3
from
top3
partition by (col1)
right join
(select level n from dual connect by level<=3) r3
on r3.n=top3.rn;
As you can see, the first step is to get top3 and then just use partition by (col1) right join r3, where r3 is just generator of 3 rows.
Results:
COL1 COL2 COL3
----- ---------- ----
A 1 VAL1
A 2 VAL2
A 3
B 1 VAL4
B 2
B 3 VAL5
6 rows selected.
Note, this approach allows you to scan your table just once!

Answer 2

Let's see. Here is the table
select * from t order by col1, col2;
COL1 COL2 COL3
----- ---------- -----
A 1 VAL1
A 2 VAL2
A 4 VAL3
B 2 VAL4
B 4 VAL5
B 5 VAL6
6 rows selected
and now let's try to apply the described logic
with offsets as
(select level - 1 offset from dual connect by level <= 3),
smallest_col2 as
(select col1, min(col2) min_col2 from t group by col1)
select sc2.col1, sc2.min_col2 + o.offset col2, t.col3
from smallest_col2 sc2
cross join offsets o
left join t
on t.col1 = sc2.col1
and t.col2 = sc2.min_col2 + o.offset
order by 1, 2;
COL1 COL2 COL3
----- ---------- -----
A 1 VAL1
A 2 VAL2
A 3
B 2 VAL4
B 3
B 4 VAL5
6 rows selected

Answer 3

Use a recursive CTE to get the COL2s from the min of each COL1 up to the next 2 and then a left join to the table:
WITH cte(COL1, COL2, max_col2) AS (
SELECT COL1, MIN(COL2), MIN(COL2) + 2
FROM tablename
GROUP BY COL1
UNION ALL
SELECT COL1, COL2 + 1, max_col2
FROM cte
WHERE COL2 < max_col2
)
SELECT c.COL1, c.COL2, t.COL3
FROM cte c LEFT JOIN tablename t
ON t.COL1 = c.COL1 AND t.COL2 = c.COL2
ORDER BY c.COL1, c.COL2
See the demo.

Answer 4

The partitioned outer join, already demonstrated in Sayan's answer, is probably the best approach for that part of the assignment (data densification).
For the first part, in Oracle 12.1 and higher you can use the match_recognize clause:
select col1, col2, col3
from this_table
match_recognize(
partition by col1
order by col2
measures col2 - a.col2 + 1 as rn
all rows per match
pattern ( ^ a b* )
define b as col2 <= a.col2 + 2
)
partition by (col1)
right outer join
(select level as rn from dual connect by level <= 3) using (rn)
;

Answer 5

Another solution with the "recursive WITH clause"
With rws_numbered (COL1, COL2, COL3, rn) as (
select COL1, COL2, COL3
, row_number()over(order by col1, col3) rn
from Your_table
)
, cte ( COL1, COL2, COL3, rn ) as (
select COL1, COL2, COL3, rn
from rws_numbered
where rn = 1
union all
select
t.COL1
, case when t.col1 = c.col1 then c.col2 + 1 else t.col2 end COL2
, t.COL3
, t.rn
from rws_numbered t
join cte c
on c.rn + 1 = t.rn
)
select COL1, COL2, case when exists (select null from Your_table t where t.COL1 = cte.COL1 and t.COL2 = cte.COL2) then COL3 else null end COL3
from cte
order by 1, 2
;
db<>fiddle