HackerRank SQL Practice ①

Basic SELECT

Revising the Select Query I

Query all columns for all American cities in the CITY table with populations larger than 100000. The CountryCode for America is USA.

The CITY table is described as follows:

Solution

SELECT *
FROM city
WHERE countrycode = “USA” AND population > 100000

Revising the Select Query II

Query the NAME field for all American cities in the CITY table with populations larger than 120000. The CountryCode for America is USA.

The CITY table is described as follows:

Solution

SELECT name
FROM city
WHERE countrycode = “USA” AND population > 120000

Select All

Query all columns (attributes) for every row in the CITY table.

The CITY table is described as follows:

Solution

SELECT *
FROM city

Select By ID

Query all columns for a city in CITY with the ID 1661.

The CITY table is described as follows:

Solution

SELECT *
FROM city
WHERE id = 1661 -- beware of data type(not string, number)

Japanese Cities’ Attributes

Query all attributes of every Japanese city in the CITY table. The COUNTRYCODE for Japan is JPN.

The CITY table is described as follows:

Solution

SELECT *
FROM city
WHERE countrycode = “JPN”

Japanese Cities’ Names

Query the names of all the Japanese cities in the CITY table. The COUNTRYCODE for Japan is JPN.
The CITY table is described as follows:

Solution

SELECT name
FROM city
WHERE countrycode = “JPN”

Weather Observation Station 1

Query a list of CITY and STATE from the STATION table.
The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Solution

SELECT city, state
FROM station

Weather Observation Station 3

Query a list of CITY names from STATION for cities that have an even ID number. Print the results in any order, but exclude duplicates from the answer.
The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Solution

SELECT DISTINCT city
FROM station
WHERE id % 2 = 0

OR

WHERE MOD(id, 2) = 0

The MOD() function returns the remainder of a number divided by another number.

Weather Observation Station 4

Find the difference between the total number of CITY entries in the table and the number of distinct CITY entries in the table.
The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

For example, if there are three records in the table with CITY values ‘New York’, ‘New York’, ‘Bengalaru’, there are 2 different city names: ‘New York’ and ‘Bengalaru’. The query returns 1, because

Solution

SELECT COUNT(city) — COUNT(DISTINCT city)
FROM station

Weather Observation Station 5

Query the two cities in STATION with the shortest and longest CITY names, as well as their respective lengths (i.e.: number of characters in the name). If there is more than one smallest or largest city, choose the one that comes first when ordered alphabetically.
The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Note
You can write two separate queries to get the desired output. It need not be a single query.

Solution

(SELECT city, LENGTH(city)
FROM station
ORDER BY LENGTH(city), city
LIMIT 1)
UNION
(SELECT city, LENGTH(city)
FROM station
ORDER BY LENGTH(city) DESC, city
LIMIT 1)

Weather Observation 6

Query the list of CITY names starting with vowels (i.e., a, e, i, o, or u) from STATION. Your result cannot contain duplicates.

Input Format

The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Solution

SELECT DISTINCT city 
FROM station 
WHERE city REGEXP ‘^[AEIOUaeiou]’

Weather Observation Station 7

Query the list of CITY names ending with vowels (a, e, i, o, u) from STATION. Your result cannot contain duplicates.

Input Format

The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Solution

SELECT DISTINCT city 
FROM station 
WHERE city REGEXP ‘^.*[aeiouAEIOU]$’

Weather Observation Station 8

Query the list of CITY names from STATION which have vowels (i.e., a, e, i, o, and u) as both their first and last characters. Your result cannot contain duplicates.

Input Format

The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Soultion

SELECT DISTINCT city 
FROM station 
WHERE city REGEXP ‘^[aeiouAEIOU].*[aeiouAEIOU]$’

Weather Observation Station 9

Query the list of CITY names from STATION that do not start with vowels. Your result cannot contain duplicates.

Input Format

The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Solution

SELECT DISTINCT city 
FROM station 
WHERE city REGEXP ‘^[^aeiouAEIOU]’

Weather Observation Station 10

Query the list of CITY names from STATION that do not end with vowels. Your result cannot contain duplicates.

Input Format

The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Solution

SELECT DISTINCT city 
FROM station 
WHERE city REGEXP ‘[^aeiouAEIOU]$’

Weather Observation Station 11

Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. Your result cannot contain duplicates.

Input Format

The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Solution

SELECT DISTINCT city 
FROM station 
WHERE city REGEXP '^[^aeiouAEIOU]' OR city REGEXP '[^aeiouAEIOU]$'

Weather Observation Station 12

Query the list of CITY names from STATION that do not start with vowels and do not end with vowels. Your result cannot contain duplicates.

Input Format

The STATION table is described as follows:

where LAT_N is the northern latitude and LONG_W is the western longitude.

Solution

SELECT DISTINCT city 
FROM station 
WHERE city REGEXP ‘^[^aeiouAEIOU].*[^aeiouAEIOU]$’

Higher Than 75 Marks

Query the Name of any student in STUDENTS who scored higher than 75 Marks. Order your output by the last three characters of each name. If two or more students both have names ending in the same last three characters (i.e.: Bobby, Robby, etc.), secondary sort them by ascending ID.

Input Format

The STUDENTS table is described as follows:

The Name column only contains uppercase (A-Z) and lowercase (a-z) letters.

Solution

SELECT name 
FROM students 
WHERE marks > 75 
ORDER BY RIGHT(name, 3), id

Employee Names

Write a query that prints a list of employee names (i.e.: the name attribute) from the Employee table in alphabetical order.

Input Format

The Employee table containing employee data for a company is described as follows:

where employee_id is an employee’s ID number, name is their name, months is the total number of months they’ve been working for the company, and salary is their monthly salary.

Solution

SELECT name 
FROM employee 
ORDER BY name

Employee Salaries

Write a query that prints a list of employee names (i.e.: the name attribute) for employees in Employee having a salary greater than per month who have been employees for less than 10 months. Sort your result by ascending employee_id.

Input Format

The Employee table containing employee data for a company is described as follows:

where employee_id is an employee’s ID number, name is their name, months is the total number of months they’ve been working for the company, and salary is the their monthly salary.

Solution

SELECT name 
FROM employee 
WHERE months < 10 AND salary > 2000 
ORDER BY employee_id

Asian Population

Given the CITY and COUNTRY tables, query the sum of the populations of all cities where the CONTINENT is ‘Asia’.

Note: CITY.CountryCode and COUNTRY.Code are matching key columns.

Input Format

The CITY and COUNTRY tables are described as follows:

Solution

SELECT SUM(city.population) 
FROM city 
LEFT JOIN country ON city.countrycode = country.code  
WHERE country.continent = 'Asia'

African Cities

Given the CITY and COUNTRY tables, query the names of all cities where the CONTINENT is ‘Africa’.

Note: CITY.CountryCode and COUNTRY.Code are matching key columns.

Input Format

The CITY and COUNTRY tables are described as follows:

Solution

SELECT city.name  
FROM city 
LEFT JOIN country ON city.countrycode = country.code 
WHERE country.continent = 'Africa'

Average Population of Each Continent

Given the CITY and COUNTRY tables, query the names of all the continents (COUNTRY.Continent) and their respective average city populations (CITY.Population) rounded down to the nearest integer.

Note: CITY.CountryCode and COUNTRY.Code are matching key columns.

Input Format

The CITY and COUNTRY tables are described as follows:

Solution

SELECT country.continent, FLOOR(AVG(city.population)) 
FROM country     
INNER JOIN city ON city.countrycode = country.code  
GROUP BY country.continent

The Report

You are given two tables: Students and Grades. Students contains three columns ID, Name and Marks.

Grades contains the following data:

Ketty gives Eve a task to generate a report containing three columns: Name, Grade and Mark. Ketty doesn’t want the NAMES of those students who received a grade lower than 8. The report must be in descending order by grade — i.e. higher grades are entered first. If there is more than one student with the same grade (8–10) assigned to them, order those particular students by their name alphabetically. Finally, if the grade is lower than 8, use “NULL” as their name and list them by their grades in descending order. If there is more than one student with the same grade (1–7) assigned to them, order those particular students by their marks in ascending order.

Write a query to help Eve.

Note

Print “NULL” as the name if the grade is less than 8.

Solution

SELECT      
CASE WHEN g.grade >= 8 THEN s.name     
ELSE null     
END name,     
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, name, s.marks

Top Competitors

Julia just finished conducting a coding contest, and she needs your help assembling the leaderboard! Write a query to print the respective hacker_id and name of hackers who achieved full scores for more than one challenge. Order your output in descending order by the total number of challenges in which the hacker earned a full score. If more than one hacker received full scores in same number of challenges, then sort them by ascending hacker_id.

Input Format

The following tables contain contest data:

• Hackers: The hacker_id is the id of the hacker, and name is the name of the hacker.

• Difficulty: The difficult_level is the level of difficulty of the challenge, and score is the score of the challenge for the difficulty level.

• Challenges: The challenge_id is the id of the challenge, the hacker_id is the id of the hacker who created the challenge, and difficulty_level is the level of difficulty of the challenge.

• Submissions: The submission_id is the id of the submission, hacker_id is the id of the hacker who made the submission, challenge_id is the id of the challenge that the submission belongs to, and score is the score of the submission.

Solution

SELECT hacker_id, name 
FROM 
(SELECT h.hacker_id hacker_id, h.name name, c.challenge_id challenge_id 
FROM hackers h 
JOIN submissions s ON s.hacker_id = h.hacker_id  
JOIN challenges c ON c.challenge_id = s.challenge_id 
JOIN difficulty d ON d.difficulty_level = c.difficulty_level 
WHERE s.score = d.score AND c.difficulty_level = d.difficulty_level) sub 
GROUP BY hacker_id, name 
HAVING COUNT(challenge_id) > 1 
ORDER BY COUNT(challenge_id) DESC, hacker_id