[MySQL/Hackerrank] Higher Than 75 Marks, Employee Names, Employee Salaries

Higher Than 75 Marks

Query the Name of any student in STUDENTS who scored higher than  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.

Sample Input

Sample Output

Ashley

Julia

Belvet

Explanation

Only Ashley, Julia, and Belvet have Marks > . If you look at the last three characters of each of their names, there are no duplicates and 'ley' < 'lia' < 'vet'.

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.

Sample Input

Sample Output

Angela

Bonnie

Frank

Joe

Kimberly

Lisa

Michael

Patrick

Rose

Todd

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 $2000 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.

Sample Input

Sample Output

Angela

Michael

Todd

Joe

SELECT NAME
FROM Employee 
WHERE salary > 2000
    AND months < 10
ORDER BY employee_id;