Find duplicate records withouyqusing group by and having

Question

I know i can delete duplicate records using an analytic function. I don't want to delete the records, I want to look at the data to try to understand why I have duplicates. I am looking at tables that don't have unique constraints (I can't do anything about it). I have some very large tables. so I am trying to find a faster way to do this.
for example
myTable
col1 number,
col2 number,
col3 number,
col4 number,
col5 number)
My key column is col1 and col2 (it is not enforced in the database). So I want to get all the records that have duplicates on these fields and put them in a table to review. This is a standard way to do it, but it requires 2 full table scans of very large tables (many, many gigabtytes. one is 150 gbs and not partitioned), a sort, and a hash join. Even if i increase sort_area_size and hash_area_size, it takes a long time to run..
create table mydup
as
select b.*
from (select col1,col2,count(*)
from myTable
group by col1, col2
having count(*) > 1) a,
myTable b
where a.col1 = b.col1
and a.col2 = b.col2
I think there is a way to do this without a join by using rank, dense_rank, or row_number or some other way. When I google this all I get is how to "delete them". I want to analyze them a nd not delete them.

Answer 1

create table mytable (col1 number,col2 number,col3 number,col4 number,col5 number);
insert into mytable values (1,2,3,4,5);
insert into mytable values (2,2,3,4,5);
insert into mytable values (3,2,3,4,5);
insert into mytable values (2,2,3,4,5);
insert into mytable values (1,2,3,4,5);
SQL> ed
Wrote file afiedt.buf
  1  select * from mytable
  2   where rowid in
  3  (select rid
  4      from
  5     (select rowid rid,
  6              row_number() over
  7              (partition by
  8                   col1,col2
  9               order by rowid) rn
10          from mytable
11      )
12    where rn <> 1
13* )
SQL> /
      COL1       COL2       COL3       COL4       COL5
         1          2          3          4          5
         2          2          3          4          5
SQL>Regards
Girish Sharma