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This approach involves using SQL queries to retrieve the required data. We'll utilize the GROUP BY clause to segregate data by department, and the MAX function to determine the highest salary within each department. By joining the Employee table and the Department table, we can extract both the department name and the pertinent employee information.
Time Complexity: O(N) where N is the number of employees, assuming indexing on join fields.
Space Complexity: O(N) for storing the intermediate join results and final results.
1SELECT dept.name AS Department, e.name AS Employee, e.salary AS Salary FROM Employee e JOIN Department dept ON e.departmentId = dept.id WHERE (e.departmentId, e.salary) IN ( SELECT departmentId, MAX(salary) FROM Employee GROUP BY departmentId )Here, we first perform a JOIN operation between the Employee and Department tables based on departmentId. The subquery inside filters out the department-wise maximum salary using GROUP BY and MAX functions. The result is a list of employees with the highest salaries in each department, which is then retrieved using the outer query.
This method leverages nested SQL queries to first compute the maximum salary in each department and then joins these results back to the Employee table. The primary objective is to match employees with these max salaries and retrieve both their names and department names.
Time Complexity: O(N log N) due to subquery operations and sorting for determining max.
Space Complexity: O(N) for storing intermediate JOIN and final results.
1WITH MaxSalaries AS ( SELECT departmentId, MAX(salary) AS max_salary FROM Employee GROUP BY departmentId This approach involves using an SQL subquery in conjunction with the JOIN operation. We first determine the maximum salary for each department by grouping the Employee table by departmentId. Then, we use this result to join back to the Employee table and select the employee records that match these maximum salaries.
Time Complexity: O(N * M), where N is the number of employees and M is the number of departments. The subquery is executed for each employee record, comparing salaries within the department.
Space Complexity: O(N), for storing the result set.
1SELECT d.name AS Department, e.name AS Employee, e.salary AS Salary
2This approach utilizes SQL window functions to eliminate the need for subqueries. By using the RANK() function over each department's salary list, you can directly identify which employees have the maximum salary in their respective departments.
Time Complexity: O(N log N), where N is the number of employees. The RANK function operates in a similar way to a sort operation within each partition of the dataset.
Space Complexity: O(N), for storing ranking results and generating output.
1SELECT Department, Employee, Salary
2FROM (
3 SELECT d.name We utilize a Common Table Expression (CTE) to store maximum salaries per department (MaxSalaries) and join this temporary result back with the Employee table to filter out those employees who have these maximum salaries. Another join fetches department names from the Department table. This ensures that the results list the highest-paid employees by department.
This query works in two main steps:
The query involves the following major elements: