CareerCup

select Name from EMPLOYEE where Dept=(select Dept from (select sum(Dept),Dept from EMPLOYEE group by Dept order by 1 desc limit 1));

SELECT e.name FROM Employee e
INNER JOIN
(
SELECT depId, count(*) FROM Employee GROUP BY depId ORDER BY 2 DESC LIMIT 1
) t ON t.depId = e.depId

LOL.

Hash[Dept_Name] = Num_Employees
Hash[Employee_Name]=Dept_Name

Now, search for the Dept_Name with highest Employee Number. Then, return all those employees in the second hash table that work in that department. This is an O(n) approach.
We can maintain a Max_Heap where keys are the Num_Employees and values are Dept_Name and then return all employees that work in the Dept corresponding to the root of the Max_Heap.