Find Highest Salary Employees by Department (Subquery & Join) in MySQL
Intuition\nTo solve the problem of retrieving the employee with the highest salary within each department, we need to:\n\n- Identify the maximum salary within e
# Intuition\nTo solve the problem of retrieving the employee with the highest salary within each department, we need to:\n\n- Identify the maximum salary within each department.\n- Filter employees to match this maximum salary.\n- Combine this information with department names to present the result.\n\n# Approach\n1. `Subquery for Maximum Salary`: Use a subquery to compute the maximum salary (MAX(Salary)) for each department by grouping the Employee table by departmentId.\n\n2. `Main Query Join`: Join the Employee and Department tables, filtering the results based on the subquery to ensure only employees with the maximum salary are selected.\n\n3. `Select Columns`: Ensure the final output includes the department name, employee name, and salary.\n\n# Complexity\n- Time complexity: $$O(nlogn)$$ \n`Subquery`: The subquery involves a GROUP BY operation which can be \n$$O(nlogn)$$ due to sorting.\n`Join and Filter`: The join and filtering operations are generally $$O(n)$$, assuming indexes are in place.\n\n- Space complexity: $$O(n)$$ space required for intermediate result storage and the final join operation grows linearly with the number of rows in the tables.\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT \n    Department.name AS Department, \n    Employee.name AS Employee, \n    Salary\nFROM \n    Employee\nJOIN \n    Department \n    ON Department.id = Employee.departmentId\nWHERE \n    (Employee.departmentId, Salary) IN (\n        SELECT \n            departmentId, \n            MAX(Salary) \n        FROM \n            Employee \n        GROUP BY \n            departmentId\n    );\n```
['Database', 'MySQL']