Find the combination of strings in all orders and delete the duplicate records - Amazing Interview Question

This was a question posted to me in the in-the premises round of a leading E-commerce company.

Problem; Find the count of strings  and their combinations in a table, and count all the combination of letters as duplicates.

If AB is a string, then BA is a duplicate string. And count(AB) should be 2.


The Solution attempted by me is as follows:

create table table1
(
value1 varchar(30));

Now , in the list below, AB and BA are treated as duplicates.
Similarly, 'CD' and 'DC' are duplicates.
We need to find the count of all combinations possible for 'AB' and 'CD' .


insert into table1 values ('AB');
insert into table1 values ('BA');
insert into table1 values ('CD');
insert into table1 values ('DC');

select * from table1;

select A.value1,count(*) from table1 as t1
inner join
(
select
case when value1 > value_ref then value_ref else value1 end as value1
, value_ref
, value1 as val1

from
(
select value1, concat(substr(value1,2,1),substr(value1,1,1)) as value_ref from table1
group by value_ref
) as A) as A
 on t1.value1 = A.value_ref
group by A.value1;

The trick is to create a lookup with all the values in the table.
The inner query does this trick as shown below:

Lookup table of values and their combination

Result set will be:

Count of all combinations possible

Write your solution in the comment section.