Hello coders, today we are going to solve The PADS HackerRank Solution in SQL.
Problem
Generate the following two result sets:
- Query an alphabetically ordered list of all names in OCCUPATIONS, immediately followed by the first letter of each profession as a parenthetical (i.e.: enclosed in parentheses). For example:
AnActorName(A),ADoctorName(D),AProfessorName(P), andASingerName(S). - Query the number of ocurrences of each occupation in OCCUPATIONS. Sort the occurrences in ascending order, and output them in the following format:
There are a total of [occupation_count] [occupation]s.where [occupation_count] is the number of occurrences of an occupation in OCCUPATIONS and [occupation] is the lowercase occupation name. If more than one Occupation has the same [occupation_count], they should be ordered alphabetically.
Note: There will be at least two entries in the table for each type of occupation.
Input Format
| Column | Type |
|---|---|
| Name | String |
| Occupation | String |
The OCCUPATIONS table is described as follows:
Occupation will only contain one of the following values: Doctor, Professor, Singer or Actor.
Sample Input
An OCCUPATIONS table that contains the following records:
| Name | Occupation |
|---|---|
| Samantha | Doctor |
| Julia | Actor |
| Maria | Actor |
| Meera | Singer |
| Ashley | Professor |
| Ketty | Professor |
| Christeen | Professor |
| Jane | Actor |
| Jenny | Doctor |
| Priya | Singer |
Sample Output
Ashely(P)
Christeen(P)
Jane(A)
Jenny(D)
Julia(A)
Ketty(P)
Maria(A)
Meera(S)
Priya(S)
Samantha(D)
There are a total of 2 doctors.
There are a total of 2 singers.
There are a total of 3 actors.
There are a total of 3 professors.Explanation
The results of the first query are formatted to the problem description’s specifications.
The results of the second query are ascendingly ordered first by number of names corresponding to each profession (2 <= 2 <= 3 <= 3), and then alphabetically by profession (doctor <= singer, and actor <= professor).
Solution – The PADS in SQL
MySQL
SELECT concat(NAME,concat("(",concat(substr(OCCUPATION,1,1),")"))) FROM OCCUPATIONS ORDER BY NAME ASC;
SELECT "There are a total of ", count(OCCUPATION), concat(lower(occupation),"s.") FROM OCCUPATIONS GROUP BY OCCUPATION ORDER BY count(OCCUPATION), OCCUPATION ASC
Disclaimer: The above Problem (The PADS) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.
this works with MSSQL and it does not matter how many occupations
with x as (
select occupation, count(*) v from occupations group by occupation
)
,oc as (
select concat(name ,'(‘,substring(occupation,1,1),’)’) x
from occupations)
,oc2 as
(
select *
,DENSE_RANK() OVER (ORDER BY x asc) AS R
from oc
)
,x2 as (
select
concat(‘There are a total of ‘, v, ‘ ‘ ,lower(occupation) , ‘s.’) v,
(select max(r) from oc2) +
DENSE_RANK() OVER (ORDER BY v,occupation asc) AS R
from x)
select x from
(select x,r from oc2
union all
select v,r from x2) x
order by r