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SQL Puzzle | 2 Digit Puzzle
Here you need to write a query to get distinct digits from the table. Now here half data has been reversely inserted. e.g. For 21 we have inserted 12. So out of 12 and 21 we need 21 as a single row.
For details please check out the sample input and the expected output below.
- The challenge is to do it with T-SQL statements only.
- Reverse function is not allowed.
- Add your solution(s) in the comments section or send you solution(s) to firstname.lastname@example.org
Script | use below script to create table and insert some sample data
-- CREATE TABLE [2Digits] ( Id INT ) GO INSERT INTO [2Digits] VALUES (21), (12), (34), (43), (29), (92) --
SOLUTION – 1
-- SELECT DISTINCT CASE WHEN Id % 10 < (id / 10) THEN (id / 10)*10 + (Id % 10) ELSE (Id % 10 )*10+(id / 10) END Id FROM [2Digits] --
Add a comment if you have a solution in mind. I would love to learn it. We all need to learn.
Enjoy !!! Keep Learning
;with cte(id,a,b) as
select id, SUBSTRING(cast(id as varchar(2)), 1,1) as a,SUBSTRING(cast(id as varchar(2)),2,1) as b from [2digits]
select id, case when a>b then a else b end x,case when a>b then b else a end y from cte
select distinct x+y from cte1
Pawan Kumar Khowal said:
Can you try with number logic , maths?
Gregor Kralj said:
with cte as (
ROW_NUMBER() over (order by (select null)) as rowid,
REVERSE(Id) + Id as suma
top 1 with ties Id
order by ROW_NUMBER() over (partition by suma order by rowid)
LikeLiked by 1 person
Saurabh Singhal said:
Could be done easily using co-related subquery as below:
select id from d a where id in (select reverse(id) from d b where a.id>b.id)
select distinct case when a.id >b.id then a.id else b.id end ids from [2Digits] a inner join [2Digits] b
on a.id = reverse(b.id)