SQL (Intermediate)

Weather Observation Station 5

Question Link

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.

Sample Input

For example, CITY has four entries: DEF, ABC, PQRS and WXY.

Sample Output

ABC 3
PQRS 4

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

(SELECT
    CITY, LENGTH(CITY)
FROM STATION s
ORDER BY LENGTH(CITY), CITY
LIMIT 1)
UNION
(SELECT
    CITY, LENGTH(CITY)
FROM STATION s
ORDER BY LENGTH(CITY) DESC, CITY ASC
LIMIT 1)

Weather Observation Station 20

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 decimal places.

Input Format

The STATION table is described as follows:

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

SELECT ROUND(LAT_N,4)
FROM (
  SELECT 
    LAT_N, 
    ROW_NUMBER() OVER (ORDER BY LAT_N) AS row_num,
    COUNT(*) OVER () AS total_rows
  FROM STATION
) AS subquery
WHERE row_num IN (FLOOR((total_rows + 1) / 2), CEIL((total_rows + 1) / 2));

New Companies

Question Link

Each of the companies follows this hierarchy:

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

Sample Input

Company Table:

Lead_Manager Table:

Senior_Manager Table:

Manager Table:

Employee Table:

Sample Output

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

Solution:

SELECT
    e.company_code,
    c.founder,
    COUNT(DISTINCT e.lead_manager_code) AS ld_mngr_cnt,
    COUNT(DISTINCT e.senior_manager_code) AS snr_mngr_cnt,
    COUNT(DISTINCT e.manager_code) AS mngr_cnt,
    COUNT(DISTINCT e.employee_code) AS emp_cnt
FROM employee e
JOIN company c 
ON e.company_code = c.company_code
GROUP BY e.company_code, c.founder
ORDER BY e.company_code

Link to Full Output