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https://github.com/dhia-gharsallaoui/hackerrank-sql-solutions
The solutions of all SQL hackerrank in MYSQL
https://github.com/dhia-gharsallaoui/hackerrank-sql-solutions
hackerrank hackerrank-solutions mysql sql
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The solutions of all SQL hackerrank in MYSQL
- Host: GitHub
- URL: https://github.com/dhia-gharsallaoui/hackerrank-sql-solutions
- Owner: dhia-gharsallaoui
- Created: 2021-12-06T11:46:27.000Z (about 3 years ago)
- Default Branch: main
- Last Pushed: 2022-07-30T17:58:29.000Z (over 2 years ago)
- Last Synced: 2024-11-18T04:37:25.960Z (3 months ago)
- Topics: hackerrank, hackerrank-solutions, mysql, sql
- Language: SQL
- Homepage:
- Size: 15.6 KB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# HackerRank-SQL-Solutions
## Description
The solutions of all the SQL challenges for all easy, medium and hard challenges on HackerRank executed on MySQL environment compiled.
## Solutions
## [Basic Select Challenges](https://www.hackerrank.com/domains/sql?filters%5Bsubdomains%5D%5B%5D=select)
### Revising the Select Query I
```SQL
SELECT * FROM CITY WHERE population > 100000 AND Countrycode ="USA";
```
### Revising the Select Query II
```SQL
Select name from city where population > 120000 and Countrycode = "USA";
```
### Select All
```SQL
Select * from city;
```
### Select by ID
```SQL
select * from city where ID=1661;
```
### Japanese Cities' Attributes
```SQL
select * from city where countrycode="JPN";
```
### Japanese Cities' Names
```SQL
select name from city where CountryCode="JPN";
```
### Weather Observation Station 1
```SQL
select city,State from station;
```
### Weather Observation Station 3
```SQL
Select distinct city from station where ID%2=0;
```
### Weather Observation Station 4
```SQL
select count(city) - count(distinct city) from station;
```
### Weather Observation Station 5
```SQL
select city,length(city) from station order By length(city) asc, city asc limit 1;
select distinct(City),length(city) from station order by length(city) desc, city asc limit 1;
```
### Weather Observation Station 6
```SQL
select distinct city from station where left(city,1) in('a','e','i','o','u');
```
### Weather Observation Station 7
```SQL
select distinct city from station where right(city,1) in('a','e','i','o','u');
```
### Weather Observation Station 8
```SQL
select distinct city from station where left(city,1) in('a','e','i','o','u') and right(city,1) in('a','e','i','o','u') ;
```
### Weather Observation Station 9
```SQL
select distinct city from station where left(city,1) not in('a','e','i','o','u') ;
```
### Weather Observation Station 10
```SQL
select distinct city from station where right(city,1) not in('a','i','e','o','u') ;
```
### Weather Observation Station 11
```SQL
select distinct city from station where left(city,1) not in('a','i','e','o','u') or right(city,1) not in('a','i','e','o','u');
```
### Weather Observation Station 12
```SQL
select distinct city from station where left(city,1) not in('a','i','e','o','u') and right(city,1) not in('a','i','e','o','u') ;
```
### Higher than 75 marks
```SQL
select name from students where marks > 75 order by right(name,3),id asc;
```
### Employee Names
```SQL
select name from employee order by name;
```
### Employee Salaries
```SQL
select name from employee where salary > 2000 and months <10 order By employee_id;
```
## [ Advanced Select Challenges](https://www.hackerrank.com/domains/sql?filters%5Bsubdomains%5D%5B%5D=advanced-select)
### Type of Triangle
```SQL
Select case
when (A+B<=C or A+C<=B or B+C<=A ) then 'Not A Triangle'
when (A=B) and(B=C) then 'Equilateral'
when (A=B or B=C or A=C) then 'Isosceles'
else 'Scalene'
End as "triangle_type"
from triangles;
```
### The PADS
```SQL
select CONCAT_WS('',name,'(',left(occupation,1),')')
from occupations
order by name;
select CONCAT_WS("","There are a total of ",count(occupation)," ",LOWER(occupation),"s.")
from occupations
group by occupation
order by count(occupation), occupation;
```
### Occupations
```SQL
set @d=0, @p=0, @s=0, @a=0; -- initial counters
select min(Doctor), min(Professor), min(Singer), min(Actor)
from(
SELECT case
when Occupation='Doctor' then (@d:=@d+1)
when Occupation='Professor' then (@p:=@p+1)
when Occupation='Singer' then (@s:=@s+1)
when Occupation='Actor' then (@a:=@a+1) end as counter,
case when Occupation='Doctor' then Name end as Doctor,
case when Occupation='Professor' then Name end as Professor,
case when Occupation='Singer' then Name end as Singer,
case when Occupation='Actor' then Name end as Actor
FROM OCCUPATIONS order by name )temp
group by counter;
```
### Binary Tree Nodes
```SQL
select N,case
when P is NULL then 'Root'
WHEN EXISTS (SELECT B.P FROM BST B WHERE B.P = BT.N) THEN 'Inner'
else 'Leaf' end as node_type
from BST as BT order by N;
```
### New Companies
- With `left join`.
```SQL
SELECT
c.company_code,c.founder,
count(distinct lm.lead_manager_code),
count(distinct sm.senior_manager_code),
count(distinct m.manager_code),
count(distinct e.employee_code)
FROM
Company c
left join Lead_Manager lm on c.company_code=lm.company_code
left join Senior_Manager sm on sm.lead_manager_code=lm.lead_manager_code
left join Manager m on m.senior_Manager_code=sm.senior_Manager_code
left join Employee e on e.manager_code=m.manager_code
GROUP BY
c.company_code,c.founder
ORDER BY
c.company_code ASC
```
- Without `left join`
```SQL
SELECT
c.company_code,c.founder,
count(distinct lm.lead_manager_code),
count(distinct sm.senior_manager_code),
count(distinct m.manager_code),
count(distinct e.employee_code)
FROM
Company c, Lead_Manager lm, Senior_Manager sm, Manager m, Employee e
WHERE
c.company_code=lm.company_code AND
lm.lead_manager_code=sm.lead_manager_code AND
sm.senior_manager_code=m.senior_manager_code AND
m.manager_code=e.manager_code
GROUP BY
c.company_code,c.founder
ORDER BY
c.company_code ASC;
```
## [Aggregation Challenges](https://www.hackerrank.com/domains/sql?filters%5Bsubdomains%5D%5B%5D=aggregation)
### Revising Aggregations - The Sum Function
```SQL
select
sum(population)
from
city
where
district="California";
```
### Revising Aggregations - Averages
```SQL
select
avg(population)
from
city
where
district="California";
```
### Revising Aggregations - The Count Function
```SQL
select
count(*)
from
city
where
population>100000;
```
### Average Population
```SQL
select
floor(avg(population))
from
city;
```
### Japan Population
```SQL
select
sum(population)
from
city
where
countrycode='JPN';
```
### Population Density Difference
```SQL
select
max(population)-min(population)
from
city;
```
### The Blunder
```SQL
select
ceil(avg(salary)-avg(replace(salary,0,'')))
from
employees;
```
### Top Earners
```SQL
select
salary*months, COUNT(*)
from
employee
where
(salary*months) = (select max(salary*months) from employee)
group by salary*months;
```
### Weather Observation Station 2
```SQL
select
round(sum(lat_n),2),round(sum(long_w),2)
from
station;
```
### Weather Observation Station 13
```SQL
select
truncate(sum(lat_n),4)
from
station
where
lat_n>38.7880 and lat_n<137.2345 ;
```
### Weather Observation Station 14
```SQL
select
truncate(max(lat_n),4)
from
station
where
lat_n<137.2345 ;
```
### Weather Oservation Station 15
```SQL
select
round(long_w,4)
from
station
where
lat_n =(select max(lat_n) from station where lat_n <137.2345);
```
### Weather Oservation Station 16
```SQL
select
round(min(lat_n),4)
from
station
where
lat_n>38.7780;
```
### Weather Oservation Station 17
```SQL
select
round(long_w,4)
from
station
where
lat_n=(select min(lat_n) from station where lat_n>38.7780);
```
### Weather Oservation Station 18
```SQL
select
round(abs(max(lat_n)-min(lat_n))+abs(max(long_w)-min(long_w)),4)
from
station;
```
### Weather Oservation Station 19
```SQL
select
round(power((power(max(lat_n)-min(lat_n),2)+power(max(long_w)-min(long_w),2)),0.5),4)
from
station;
```
### Weather Oservation Station 20
```SQL
set
@rownum=0;
SELECT
ROUND(t.lat_n, 4)
FROM(
SELECT
s.LAT_N,
@rownum := @rownum + 1 as `row_number`,
@total_rows := @rownum
FROM
STATION s
ORDER BY
s.LAT_N
) as t
WHERE
t.row_number IN (FLOOR((@total_rows + 1) / 2), FLOOR((@total_rows + 2) / 2));
```
## [Basic Join Challenges](https://www.hackerrank.com/domains/sql?filters%5Bsubdomains%5D%5B%5D=join)[]()
### Population Census
```SQL
select
sum(city.population)
from
country left join city on city.countrycode=country.code
where
country.continent="Asia";
```
### African Cities
```SQL
select
city.name
from
country inner join city on city.countrycode=country.code
where
country.continent="Africa";
```
### Average Population of Each Continent
```SQL
select
country.continent, floor(avg(city.population))
from
country inner join city on city.countrycode=country.code
group by
country.continent
```
### The Report
```SQL
select
case when g.grade<8 then NULL
else s.name end as names,
g.grade,s.marks
from
students s join grades g on s.marks between g.min_mark and g.max_mark
order by
g.grade Desc,
s.name,
s.marks;
```
### Top Competitors
```SQL
select
h.hacker_id,h.name
from
submissions s join hackers h on s.hacker_id=h.hacker_id
join challenges c on c.challenge_id=s.challenge_id
join difficulty d on c.difficulty_level=d.difficulty_level and s.score=d.Score
group by
h.hacker_id,h.name
having count(h.hacker_id) > 1
order by
count(h.hacker_id) desc,
h.hacker_id ;
```
### Ollivander's Inventory
```SQL
select
w.id,wp.age,w.coins_needed,w.power
from
wands w join wands_property wp on w.code=wp.code
where
wp.is_evil=0 and
w.coins_needed=(select
min(w2.coins_needed)
from
wands w2 join wands_property wp2 on w2.code=wp2.code
where
w2.power=w.power and wp2.age=wp.age)
order by
w.power desc,
wp.age desc;
```
### Challenges
```SQL
select
h.hacker_id,h.name,count(h.hacker_id) as challenges_created
from
hackers h join challenges c on h.hacker_id=c.hacker_id
group by
h.hacker_id,
h.name
having
challenges_created=(
select
count(challenge_id) as max_challenges
from
challenges
group by
hacker_id
order by
max_challenges desc
limit 1 )
or
challenges_created in(
select
temp.max_challenges
from
(select count(challenge_id) AS max_challenges
FROM
challenges
group by
hacker_id order by max_challenges) temp
group by
temp.max_challenges
having
count(temp.max_challenges)=1)
order by
challenges_created desc,
h.hacker_id
```
### Contest Leaderboard
```SQL
select
h.hacker_id,h.name,sum(ss.max_score) as total_score
from
hackers h join (select
s.hacker_id,s.challenge_id,max(score) as max_score
from
submissions s
group by
s.hacker_id,s.challenge_id) ss
on h.hacker_id=ss.hacker_id
group by
h.hacker_id,h.name
having
total_score>0
order by
total_score desc,
h.hacker_id
```
## [Advanced Join Challenges](https://www.hackerrank.com/domains/sql?filters%5Bsubdomains%5D%5B%5D=advanced-join)
### SQL Project Planning
```sql
select
s.start_date,min(e.end_date)
from
(select
end_date
from
projects
where
end_date not in (select start_date from projects)) e,
(select
start_date
from
projects
where
start_date not in (select end_date from projects)) s
where
s.start_datep.salary
order by
pf.salary ;
```
### Symmetric Pairs
```sql
select
f1.x,f1.y
from
functions f1 ,functions f2
where
f1.x<=f1.y
and f1.x=f2.y
and f1.y=f2.x
group by
f1.x,f1.y
having
count(case when f1.x=f1.y then f1.x END)>2
or count(case when f1.x0
order by
c.contest_id
```
### 15 Days of Learning SQL
```sql
```
## [Alternative Queries Challenges](https://www.hackerrank.com/domains/sql?filters%5Bsubdomains%5D%5B%5D=alternative-queries)
### Draw The Triangle 1
```sql
set
@n_lines =21;
select
repeat('* ',@n_lines:=@n_lines-1)
from
INFORMATION_SCHEMA.TABLES
limit 20;
```
### Draw The Triangle 2
```sql
set
@n_lines =0;
select
repeat('* ',@n_lines:=@n_lines+1)
from
INFORMATION_SCHEMA.TABLES
limit 20;
```
### Print Prime Numbers
```sql
with recursive tblnums
as (
select 2 as nums
union all
select nums+1
from tblnums
where nums<1000)
select group_concat(tt.nums order by tt.nums separator '&') as nums
from tblnums tt
where not exists
( select 1 from tblnums t2
where t2.nums <= tt.nums/2 and mod(tt.nums,t2.nums)=0) ;
```