SQL — 107 (Basics) Misc

Introduction

This series is the revision of SQL concepts. We are doing the basic questions, then intermediate, and so on.

Question 1

[Ref: Hackerrank]

Generate the following two result sets:

1. 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).
2. 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:

Occupation will only contain one of the following values: Doctor, Professor, Singer or Actor.

Sample Input

An OCCUPATIONS table that contains the following records:

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:

SELECT CONCAT(Name,'(',SUBSTRING(Occupation,1,1),')') FROM OCCUPATIONS ORDER BY Name;
SELECT CONCAT('There are a total of ',COUNT(Occupation), ' ', LOWER(Occupation),'s.') FROM OCCUPATIONS GROUP BY Occupation ORDER BY count(OCCUPATION) ASC;

Question 2

[Ref: Hackerrank]

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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 P1 and P2 and round it to a scale of 4 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.

Solution:

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

Question 3

[Ref: Hackerrank]

Consider P1(a,b) and P2(c,d) to be two points on a 2D plane where (a,b) are the respective minimum and maximum values of Northern Latitude (LAT_N) and are (c,d) the respective minimum and maximum values of Western Longitude (LONG_W) in STATION.

Query the Euclidean Distance between points P1 and P2 and format your answer to display 4 decimal digits.

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 ROUND(SQRT(POWER(MIN(LAT_N) -  MAX(LAT_N),2) +  POWER(MIN(LONG_W) - MAX(LONG_W),2)),4)  FROM STATION

Previous Posts

1. SQL 101
2. SQL 102 (Basics)
3. SQL — 103 (Basics)
4. SQL — 104 (Basics) Aggregation
5. SQL — 105 (Basics) Aggregation
6. SQL — 106 (Basics) Joins