MySQL solution-

select (salary * months)as earnings ,count(*) from employee group by 1 order by earnings desc limit 1;

MySQL solution

From my HackerRank solutions.

SELECT salary * months AS earnings, COUNT(*)
FROM Employee
GROUP BY earnings
ORDER BY earnings DESC
LIMIT 1;

Let me know if you have any questions.

MS SQL Solution

select max(months*salary),count(months*salary) from employee where employee_id in( select employee_id from employee where months*salary in (select max(months*salary) from employee));

Hello, There..

This is my first post on discussions.. Here is a solution for ORACLE:

SELECT * FROM (SELECT MAX(MONTHS*SALARY), COUNT(EMPLOYEE_ID) FROM EMPLOYEE GROUP BY MONTHS*SALARY ORDER BY MONTHS*SALARY DESC) WHERE ROWNUM = 1;

my sql querry:-

select max(salary * months), count(*) from employee 
where salary * months =  (select max(salary * months) from employee);