Very challenging SQL interview (can't use stored procedure) - sql

Question

You have a SQL table table with two columns: name and pen. Both columns are text strings.
name | pen
---------------
mike | red
mike | red
mike | blue
mike | green
steve | red
steve | yellow
anton | red
anton | blue
anton | green
anton | black
alex | black
alex | green
alex | yellow
alex | red
Person's name is given as the input argument.
Please write a SQL statement (not a stored procedure) which returns names of persons having unique set of pens which is equivalent or wider/bigger than the set of pens of given person.
Examples:
input: mike
output: anton
Mike has (red, blue, green).
Anton has more gadgets (red, blue, green) + black.
input: steve
output: alex
Steve has (red, yellow).
Alex has (red, yellow) + green+ black.
Mike, Anton aren't printed - they do not have yellow.
input: alex
output:

Answer 1

Here's one way (Online Demo), assuming input name "steve".
This can be rephrased as "Looking for all users for which there does not exist a pen owned by steve that they do not own"
SELECT DISTINCT name
FROM table t1
WHERE NOT EXISTS (SELECT *
FROM table t2
WHERE name = 'steve'
AND NOT EXISTS (SELECT *
FROM table t3
WHERE t2.pen = t3.pen
AND t1.name = t3.name))
AND t1.name <> 'steve' /*Exclude input name from results*/
See this article for other techniques

Answer 2

with test1 as
(select a.name nm, count(distinct a.pen) ct
from table a, table b
where b.pen = a.pen
and b.name = 'anton'
group by a.name
order by 2 desc),
test2 as
(select nm name1
from test1
where ct = (select max(ct) from test1))
select distinct c.name
from table c
where c.name in (select name1
from test2
where name1 not in (case
when (select count(distinct name1) from test2) > 1 then
'anton'
else
' '
end))

Answer 3

Table Format:
E_NAME E_PEN
----------------
mike green
mike blue
mike red
mike red
steve red
steve yellow
anton red
anton blue
anton green
anton black
alex black
alex green
alex yellow
alex red
Query:
SELECT A.E_NAME, A.E_PEN
FROM V_NAME A, V_NAME B
WHERE TRIM(UPPER(A.E_NAME)) = TRIM(UPPER(B.E_NAME))
AND TRIM(UPPER(A.E_NAME)) != 'MIKE'
AND TRIM(UPPER(A.E_PEN)) = TRIM(UPPER(B.E_PEN(+)))
Procedure to take Input from user:
CREATE OR REPLACE PROCEDURE E_TEST(I_NAME VARCHAR2) AS
BEGIN
EXECUTE IMMEDIATE 'CREATE TABLE E_TABLE AS
SELECT A.E_NAME,A.E_PEN FROM V_NAME A,V_NAME B
WHERE TRIM(UPPER(A.E_NAME)) = TRIM(UPPER(B.E_NAME))
AND TRIM(UPPER(A.E_NAME)) != ''' ||
I_NAME || '''
AND TRIM(UPPER(A.E_PEN))= TRIM(UPPER(B.E_PEN(+)))';
END;
Name: Nikhil Shinde
E-Mail: nikhilshinde3jun#gmail.com

Answer 4

declare #t table (name nvarchar(20), pen nvarchar(20))
insert into #t (name,pen)
values('mike','red'),
('mike','red'),
('mike','blue'),
('mike','green'),
('steve','red'),
('steve','yellow'),
('anton','red'),
('anton','blue'),
('anton','green'),
('anton','black'),
('alex','black'),
('alex','green'),
('alex','yellow'),
('alex','red')
declare #input nvarchar(20) = 'mike';
with cte_input (name, pen)
as
(
select distinct name, pen
from #t
where name = #input
)
, cte_colors (name, matches)
as
(
select name, matches = count(distinct pen)
from #t
where name != #input
and pen in (select pen from cte_input)
group by name
having count(distinct pen) = (select count(1) from cte_input)
)
select t.name
from #t t
join cte_colors m on m.name = t.name
group by t.name, m.matches
having count(distinct t.pen) > m.matches

Answer 5

We can use LEAD analytical function to achieve this solution.
The query would be
select next_name from
(select name, count(pen) CNT, LEAD(name,1) over (order by count(pen)) next_name
from table
group by name order by CNT
)
where name=input_value;
the sub query will give the results as below
name | CNT | next_name
mike | 3 | Steve
steve | 2 | anton
anton | 4 | alex
alex | 4 | (null)
Then the out query filter the row required and gives the next_name which is what we are looking for.

Answer 6

create table practise
(
name varchar (10),pens varchar (10)
)
insert practise (name,pens)
select 'mike','red' union
select 'mike','red' union
select 'mike','blue' union
select 'mike','green' union
select 'steve','red' union
select 'steve','yellow' union
select 'anton','blue' union
select 'anton','green' union
select 'anton','black' union
select 'anton','red' union
select 'alex','black' union
select 'alex','green' union
select 'alex','yellow' union
select 'alex','red'
select * into #t from practise
update #t
set Colourcode = 1 where pens = 'black'
go
update #t
set Colourcode = 2 where pens = 'blue'
go
update #t
set Colourcode = 3 where pens = 'green'
go
update #t
set Colourcode = 4 where pens = 'red'
go
update #t
set Colourcode = 5 where pens = 'yellow'
select distinct colourcode from #t
/* IMP Code /
update #t
set Number =
(
select count() from #t
where name = 'steve' and colourcode = 1
)
where name = 'steve' and colourcode = 1
/* once the above update statement is exected change colourcode = 2 and again run the statement
change colourcode = 3 and again run the statement
change colourcode = 4 and again run the statement
change colourcode = 5 and again run the statement*/
/* Follow the same process for name 'alex','anton' and 'mike' */
/* Below is the Solution*/
select name from #t
where number =
(
select distinct number from #t
where colourcode = 3 and Number =1)

Answer 7

Declare #name nvarchar(20)
set #name = 'steve';
with cte_in
as
(
select distinct Pen from tblNamePen where Name = #name
)
select Name from
(
select t1.name, count(distinct t1.Pen) ct from tblNamePen t1
join cte_in inp on t1.Name <> #name
where t1.pen = inp.Pen
group by t1.Name
) a
where a.ct >= (select count(Pen) from cte_in)