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I started studying SQL from a very famous site - HackerRank. Here I will try to provide multiple approaches & solutions to the same problem. It will help you learn and understand SQL in a better way.

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SQL Problem Statement:

We define an employee's total earnings to be their monthly salary * 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.

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

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:
`69952 1`

Explanation:

The table and earnings data is depicted in the following diagram:

The maximum earnings value is 69952. The only employee with earnings = 69952 is Kimberly, so we print the maximum earnings value (69952) and a count of the number of employees who have earned \$69952 (which is 1) as two space-separated values.

### Solution-1: Using GROUP BY, ORDER BY & LIMIT (MySQL Query):

```SELECT (SALARY * MONTHS) as MAX_SAL, COUNT(*)
FROM EMPLOYEE
GROUP BY MAX_SAL
ORDER BY MAX_SAL DESC
LIMIT 1;```

NOTE:
1. '*' operator is used for multiplying Salary & Months.
2. Using ORDER BY MAX_SAL DESC, salaries are sorted in descending order of max_salary.
3. LIMIT is used to limit the number of rows in the output. LIMIT 1 will return only the first row of the output. (Which will be the maximum salary due to order by desc clause.)

### Solution-2: Using GROUP BY & SUB-QUERY (MySQL Query):

```SELECT SALARY*MONTHS, COUNT(*)
FROM EMPLOYEE
WHERE (SALARY*MONTHS) = (SELECT MAX(SALARY*MONTHS) FROM EMPLOYEE)
GROUP BY SALARY*MONTHS;```

### Solution-3: Using MAX & SUB-QUERY (MySQL Query):

```SELECT MAX(SALARY*MONTHS), COUNT(*)
FROM EMPLOYEE
WHERE (SALARY*MONTHS) = (SELECT MAX(SALARY*MONTHS) FROM EMPLOYEE);```