MySQL Solution for Department Highest Salary Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to find employees who have the highest salary in each department. To achiev
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to find employees who have the highest salary in each department. To achieve this, we need to compare the salary of each employee within their respective department and select the ones with the maximum salary.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach is to join the Employee and Department tables based on the departmentId foreign key. By performing this join, we can access the department name for each employee. Then, we use a subquery to calculate the maximum salary for each department. Finally, we filter the results by selecting only the employees whose department and salary match the maximum values.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this solution depends on the size of the Employee table. Let\'s denote m as the number of rows in the Employee table. The subquery that calculates the maximum salary for each department has a complexity of O(m). The join operation and filtering have a complexity of O(m) as well. Therefore, the overall time complexity is O(m).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is determined by the number of rows returned in the result table. If there are k departments and each department has one employee with the highest salary, the result will have k rows. Therefore, the space complexity is O(k).\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT d.name AS Department, e.name AS Employee, e.Salary\nFROM Employee e\nINNER JOIN Department d ON e.departmentId = d.id\nWHERE (e.departmentId, e.salary) IN (\n    SELECT departmentId, MAX(salary)\n    FROM Employee\n    GROUP BY departmentId\n);\n\n```
['MySQL']