HackerRank MySQL Prepare

Average Population of Each Continent

📆 08/08/2024 🚀

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:

CITY:

Field	Type
ID	NUMBER
NAME	VARCHAR2(17)
COUNTRYCODE	VARCHAR2(3)
DISTRICT	VARCHAR2(20)
POPULATION	NUMBER

COUNTRY:

Field	Type
CODE	VARCHAR2(3)
NAME	VARCHAR2(44)
CONTINENT	VARCHAR2(13)
REGION	VARCHAR2(25)
SURFACEAREA	NUMBER
INDEPYEAR	VARCHAR2(2)
POPULATION	NUMBER
LIFEEXPECTANCY	VARCHAR2(4)
GNP	NUMBER
GNPOLD	VARCHAR2(9)
LOCALNAME	VARCHAR2(44)
GOVERNMENTFORM	VARCHAR2(44)
HEADOFSTATE	VARCHAR2(32)
CAPITAL	VARCHAR2(4)
CODE2	VARCHAR2(2)

SELECT
    co.Continent,
    FLOOR(AVG(ci.Population)) AS AvgPopulation
FROM
    CITY ci
JOIN
    COUNTRY co ON ci.CountryCode = co.Code
GROUP BY
    co.Continent;

Weather Observation Station 18

📆 08/08/2024 🚀

Consider P1(a, b) and P2(c, d) to be two points on a 2D plane.

a happens to equal the minimum value in Northern Latitude (LAT_N in STATION).
b happens to equal the minimum value in Western Longitude (LONG_W in STATION).
c happens to equal the maximum value in Northern Latitude (LAT_N in STATION).
d happens to equal the maximum value in Western Longitude (LONG_W in STATION).

Query the Manhattan Distance between points and and round it to a scale of decimal places.

Input Format

The STATION table is described as follows:

Field	Type
ID	NUMBER
CITY	VARCHAR2(21)
STATE	VARCHAR2(2)
LAT_N	NUMBER
LONG_W	NUMBER

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

SELECT
    ROUND(
        ABS(MIN(LAT_N) - MAX(LAT_N)) + ABS(MIN(LONG_W) - MAX(LONG_W)),
        4
    ) AS manhattan_distance
FROM
    STATION;

New Companies

📆 08/08/2024 🚀

Amber's conglomerate corporation just acquired some new companies. Each of the companies follows this hierarchy:

Founder
|/
|
Lead Manager
|/
|
Senior Manager
|/
|
Manager
|/
|
Employee

Given the table schemas below, write a query to print the company_code, founder name, total number of lead managers, total number of senior managers, total number of managers, and total number of employees. Order your output by ascending company_code.

Note:

The tables may contain duplicate records.
The company_code is string, so the sorting should not be numeric. For example, if the company_codes are C_1, C_2, and C_10, then the ascending company_codes will be C_1, C_10, and C_2.

Input Format

The following tables contain company data:

Company: The company_code is the code of the company and founder is the founder of the company.

Column	Type
company_code	String
founder	String

Lead_Manager: The lead_manager_code is the code of the lead manager, and the company_code is the code of the working company.

Column	Type
lead_manager_code	String
company_code	String

Senior_Manager: The senior_manager_code is the code of the senior manager, the lead_manager_code is the code of its lead manager, and the company_code is the code of the working company.

Column	Type
senior_manager_code	String
lead_manager_code	String
company_code	String

Manager: The manager_code is the code of the manager, the senior_manager_code is the code of its senior manager, the lead_manager_code is the code of its lead manager, and the company_code is the code of the working company.

Column	Type
manager_code	String
senior_manager_code	String
lead_manager_code	String
company_code	String

Employee: The employee_code is the code of the employee, the manager_code is the code of its manager, the senior_manager_code is the code of its senior manager, the lead_manager_code is the code of its lead manager, and the company_code is the code of the working company.

Column	Type
employee_code	String
manager_code	String
senior_manager_code	String
lead_manager_code	String
company_code	String

Sample Input

Company Table:

company_code	founder
C1	Monika
C2	Samantha

Lead_Manager Table:

lead_manager_code	company_code
LM1	C1
LM2	C2

Senior_Manager Table:

senior_manager_code	lead_manager_code	company_code
SM1	LM1	C1
SM2	LM2	C2
SM3	LM2	C2

Manager Table:

manager_code	senior_manager_code	lead_manager_code	company_code
M1	SM1	LM1	C1
M2	SM3	LM2	C2
M3	SM3	LM2	C2

Employee Table:

manager_code	manager_code	senior_manager_code	lead_manager_code	company_code
E1	M1	SM1	LM1	C1
E2	M1	SM1	LM1	C1
E3	M2	SM3	LM2	C2
E4	M3	SM3	LM2	C2

Sample Output

C1 Monika 1 2 1 2
C2 Samantha 1 1 2 2

Explanation

In company C1, the only lead manager is LM1. There are two senior managers, SM1 and SM2, under LM1. There is one manager, M1, under senior manager SM1. There are two employees, E1 and E2, under manager M1.

In company C2, the only lead manager is LM2. There is one senior manager, SM3, under LM2. There are two managers, M2 and M3, under senior manager SM3. There is one employee, E3, under manager M2, and another employee, E4, under manager, M3.

SELECT
    c.company_code,
    c.founder,
    COALESCE(lm.lead_manager_count, 0) AS lead_manager_count,
    COALESCE(sm.senior_manager_count, 0) AS senior_manager_count,
    COALESCE(m.manager_count, 0) AS manager_count,
    COALESCE(e.employee_count, 0) AS employee_count
FROM
    Company c
    LEFT JOIN (
        SELECT company_code, COUNT(DISTINCT lead_manager_code) AS lead_manager_count
        FROM Lead_Manager
        GROUP BY company_code
    ) lm ON c.company_code = lm.company_code
    LEFT JOIN (
        SELECT company_code, COUNT(DISTINCT senior_manager_code) AS senior_manager_count
        FROM Senior_Manager
        GROUP BY company_code
    ) sm ON c.company_code = sm.company_code
    LEFT JOIN (
        SELECT company_code, COUNT(DISTINCT manager_code) AS manager_count
        FROM Manager
        GROUP BY company_code
    ) m ON c.company_code = m.company_code
    LEFT JOIN (
        SELECT company_code, COUNT(DISTINCT employee_code) AS employee_count
        FROM Employee
        GROUP BY company_code
    ) e ON c.company_code = e.company_code
ORDER BY
    c.company_code;

Binary Tree Nodes

📆 08/08/2024 🚀

You are given a table, BST, containing two columns: N and P, where N represents the value of a node in Binary Tree, and P is the parent of N.

Column	Type
N	Integer
P	Integer

Write a query to find the node type of Binary Tree ordered by the value of the node. Output one of the following for each node:

Root: If node is root node.
Leaf: If node is leaf node.
Inner: If node is neither root nor leaf node.

Sample Input

N	P
1	2
3	2
6	8
9	8
2	5
8	5
5	null

Sample Output

1 Leaf
2 Inner
3 Leaf
5 Root
6 Leaf
8 Inner
9 Leaf

Explanation

The Binary Tree below illustrates the sample:

                      5️⃣
                _______|_______
               /               \
             2️⃣                8️⃣
         _____|_____       _____|_____
        /           \     /           \
      1️⃣            3️⃣ 6️⃣            9️⃣

SELECT 
    N,
    CASE 
        WHEN P IS NULL THEN 'Root'
        WHEN N NOT IN (SELECT DISTINCT P FROM BST WHERE P IS NOT NULL) THEN 'Leaf'
        ELSE 'Inner'
    END AS NodeType
FROM 
    BST
ORDER BY 
    N;

Occupations

📆 08/08/2024 🚀

Pivot the Occupation column in OCCUPATIONS so that each Name is sorted alphabetically and displayed underneath its corresponding Occupation. The output column headers should be Doctor, Professor, Singer, and Actor, respectively.

Note: Print NULL when there are no more names corresponding to an occupation.

Input Format

The OCCUPATIONS table is described as follows:

Column	Type
Name	String
Occupation	String

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
Ashely	Professor
Ketty	Professor
Christeen	Professor
Jane	Actor
Jenny	Doctor
Priya	Singer

Sample Output

Jenny Ashley Meera Jane
Samantha Christeen Priya Julia
NULL Ketty NULL Maria

Explanation

The first column is an alphabetically ordered list of Doctor names. The second column is an alphabetically ordered list of Professor names. The third column is an alphabetically ordered list of Singer names. The fourth column is an alphabetically ordered list of Actor names. The empty cell data for columns with less than the maximum number of names per occupation (in this case, the Professor and Actor columns) are filled with NULL values.

SELECT
    MAX(CASE WHEN Occupation = 'Doctor' THEN Name ELSE NULL END) AS Doctor,
    MAX(CASE WHEN Occupation = 'Professor' THEN Name ELSE NULL END) AS Professor,
    MAX(CASE WHEN Occupation = 'Singer' THEN Name ELSE NULL END) AS Singer,
    MAX(CASE WHEN Occupation = 'Actor' THEN Name ELSE NULL END) AS Actor
FROM (
    SELECT
        Name,
        Occupation,
        ROW_NUMBER() OVER (PARTITION BY Occupation ORDER BY Name) AS RowNum
    FROM OCCUPATIONS
) AS Subquery
GROUP BY RowNum
HAVING
    MAX(CASE WHEN Occupation = 'Doctor' THEN Name ELSE NULL END) IS NOT NULL OR
    MAX(CASE WHEN Occupation = 'Professor' THEN Name ELSE NULL END) IS NOT NULL OR
    MAX(CASE WHEN Occupation = 'Singer' THEN Name ELSE NULL END) IS NOT NULL OR
    MAX(CASE WHEN Occupation = 'Actor' THEN Name ELSE NULL END) IS NOT NULL
ORDER BY RowNum;

The PADS

📆 08/08/2024 🚀

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), and ASingerName(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

The OCCUPATIONS table is described as follows:

Column	Type
Name	String
Occupation	String

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
Ashely	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<=2<=3<=3), and then alphabetically by profession (doctor<=singer, and actor<=professor).

-- Requirement 1
SELECT 
    CONCAT(Name, '(', LEFT(Occupation, 1), ')') AS NameWithOccupation
FROM 
    OCCUPATIONS
ORDER BY 
    Name;

-- Requirement 2
SELECT 
    CONCAT('There are a total of ', COUNT(*)) AS occupation_count, 
    CONCAT(LOWER(Occupation), 's.') AS occupation
FROM 
    OCCUPATIONS
GROUP BY 
    Occupation
ORDER BY 
    occupation_count, 
    occupation;

African Cities

📆 07/29/2024

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.

SELECT CITY.NAME
FROM CITY
LEFT JOIN COUNTRY
ON CITY.COUNTRYCODE = COUNTRY.CODE
WHERE COUNTRY.CONTINENT = 'Africa';

Weather Observation Station 3

📆 07/29/2024 🚀

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.

SELECT DISTINCT(CITY)
FROM STATION
WHERE ID%2 = 0;

Weather Observation Station 20

📆 07/29/2024 🚀

A median is defined as a number separating the higher half of a data set from the lower half. Query the median of the Northern Latitudes (LAT_N) from STATION and round your answer to 4 decimal places.

SELECT ROUND(M.median, 4)
FROM (
    SELECT LAT_N AS median
    FROM (
        SELECT LAT_N, 
               @row_num := @row_num + 1 AS row_num, 
               @total_rows := @row_num
        FROM STATION, (SELECT @row_num := 0) r
        ORDER BY LAT_N
    ) AS ordered
    WHERE row_num IN (FLOOR((@total_rows + 1) / 2), CEIL((@total_rows + 1) / 2))
) AS M;

📆 07/29/2024

Query the Western Longitude (LONG_W)where the smallest Northern Latitude (LAT_N) in STATION is greater than 38.7780. Round your answer to 4 decimal places.

SELECT ROUND(LONG_W, 4)
FROM STATION
WHERE LAT_N = (
    SELECT MIN(LAT_N)
    FROM STATION
    WHERE LAT_N > 38.7780
);

📆 07/29/2024

Query the smallest Northern Latitude (LAT_N) from STATION that is greater than 38.7780. Round your answer to 4 decimal places.

SELECT ROUND(MIN(LAT_N), 4)
FROM STATION
WHERE LAT_N > 38.7780;

📆 07/29/2024

Query the Western Longitude (LONG_W) for the largest Northern Latitude (LAT_N) in STATION that is less than 137.2345. Round your answer to 4 decimal places.

SELECT ROUND(LONG_W, 4)
FROM STATION
WHERE LAT_N = (
    SELECT MAX(LAT_N)
    FROM STATION
    WHERE LAT_N < 137.2345
);

📆 07/29/2024

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.

SELECT SUM(CITY.POPULATION)
FROM CITY
LEFT JOIN COUNTRY
ON CITY.COUNTRYCODE = COUNTRY.CODE
WHERE COUNTRY.CONTINENT = 'Asia';

📆 07/28/2024

Query the sum of Northern Latitudes (LAT_N) from STATION having values greater than 38.7880 and less than 137.2345. Truncate your answer to 4 decimal places.

SELECT TRUNCATE(SUM(LAT_N), 4)
FROM STATION
WHERE LAT_N > 38.7880
AND LAT_N < 137.2345;

📆 07/28/2024

Query the greatest value of the Northern Latitudes (LAT_N) from STATION that is less than 137.2345. Truncate your answer to 4 decimal places.

SELECT 
    TRUNCATE(MAX(LAT_N), 4)
FROM 
    STATION
WHERE 
    LAT_N < 137.2345;

📆 07/24/2024

Query the following two values from the STATION table:

The sum of all values in LAT_N rounded to a scale of 2 decimal places.
The sum of all values in LONG_W rounded to a scale of 2 decimal places.

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

Output Format

Your results must be in the form:

lat lon

where lat is the sum of all values in LAT_N and lon is the sum of all values in LONG_W. Both results must be rounded to a scale of 2 decimal places.

SELECT 
    ROUND(SUM(LAT_N), 2) AS lat, 
    ROUND(SUM(LONG_W), 2) AS lon
FROM STATION;

📆 07/06/2024

We define an employee's total earnings to be their monthly salary x months worked, and the maximum total earnings to be the maximum total earnings for any employee in the Employee table. Write a query to find the maximum total earnings for all employees as well as the total number of employees who have maximum total earnings. Then print these values as 2 space-separated integers.

SELECT
    MAX(earnings) AS max_earnings,
    COUNT(*) AS num_employees
FROM (
    SELECT
        salary * months AS earnings
    FROM
        Employee
) AS TotalEarnings
WHERE
    earnings = (
        SELECT
            MAX(salary * months)
        FROM
            Employee
    );

📆 07/06/2024

Samantha was tasked with calculating the average monthly salaries for all employees in the EMPLOYEES table, but did not realize her keyboard's 0 key was broken until after completing the calculation. She wants your help finding the difference between her miscalculation (using salaries with any zeros removed), and the actual average salary.

Write a query calculating the amount of error (i.e.: average monthly salaries), and round it up to the next integer.

WITH ActualAverage AS (
    SELECT AVG(salary) AS actual_avg_salary
    FROM EMPLOYEES
),
MiscalculatedAverage AS (
    SELECT AVG(CAST(REPLACE(CAST(salary AS CHAR), '0', '') AS DECIMAL)) AS miscalculated_avg_salary
    FROM EMPLOYEES
)
SELECT CEIL(ABS(ActualAverage.actual_avg_salary - MiscalculatedAverage.miscalculated_avg_salary)) AS error
FROM ActualAverage, MiscalculatedAverage;

WITH, CAST, REPLACE, CEIL

📆 07/04/2024

Query the difference between the maximum and minimum populations in CITY.

SELECT MAX(POPULATION) - MIN(POPULATION)
FROM CITY;

MAX, MIN

📆 07/03/2024

Query the sum of the populations for all Japanese cities in CITY. The COUNTRYCODE for Japan is JPN.

SELECT (SUM(POPULATION))
FROM CITY
WHERE COUNTRYCODE = 'JPN';

SUM

📆 07/01/2024

Query the average population for all cities in CITY, rounded down to the nearest integer.

SELECT ROUND(AVG(POPULATION), 0)
FROM CITY;

ROUND

📆 06/30/2024

Query the average population of all cities in CITY where District is California.

SELECT AVG(POPULATION)
FROM CITY
WHERE DISTRICT = 'California;

📆 06/29/2024

Query the total population of all cities in CITY where District is California.

SELECT SUM(POPULATION)
FROM CITY
WHERE District = 'California';

📆 06/22/2024

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 TriangleType
FROM TRIANGLES;

CASE, WHEN, END

peppapig13132/hackerrank-mysql.md

Average Population of Each Continent

Input Format

Weather Observation Station 18

Input Format

New Companies

Input Format

Sample Input

Sample Output

Explanation

Binary Tree Nodes

Sample Input

Sample Output

Explanation

Occupations

Input Format

Sample Input

Sample Output

Explanation

The PADS

Input Format

Sample Input

Sample Output

Explanation

African Cities

Weather Observation Station 3

Weather Observation Station 20