Solution Explanation:

This problem requires finding the top three highest unique salaries for each department in an employee database. Two solutions are presented: one using pandas in Python and another using MySQL.

Solution 1: Using Pandas (Python)

This solution leverages the power of pandas for data manipulation.

Steps:

1. Identify Top 3 Salaries per Department:

   • employee.drop_duplicates(["salary", "departmentId"]): This removes duplicate salary entries within each department, ensuring only unique salaries are considered.
   • .groupby("departmentId")["salary"].nlargest(3): Groups the data by department ID and selects the three largest salaries within each group.
   • .groupby("departmentId").min(): This step is crucial. After getting the top 3, this finds the minimum of those top 3 salaries for each department. This gives us the cutoff salary; anyone above or equal to this cutoff is in the top 3.

2. Join with Department Names:

   • employee["Department"] = department.set_index("id")["name"][employee["departmentId"]].values: This efficiently joins the employee DataFrame with the department DataFrame to add the department name to each employee's record.

3. Filter for Top Earners:

   • employee["cutoff"] = salary_cutoff[employee["departmentId"]].values: Adds a new column 'cutoff' representing the minimum salary in the top 3 for each department.
   • employee[employee["salary"] >= employee["cutoff"]]: Filters the DataFrame to keep only employees whose salary is greater than or equal to the cutoff.

4. Rename and Select Columns:

   • .rename(columns={"name": "Employee", "salary": "Salary"}): Renames columns for better readability.
   • [["Department", "Employee", "Salary"]]: Selects the desired columns for the output.

Time Complexity: The pandas solution's time complexity is dominated by the groupby and nlargest operations. These are generally O(N log K), where N is the number of employees and K is the number of top salaries to select (in this case, K=3). The other operations are linear, so the overall complexity is approximately O(N log K). For a fixed K (like 3), this simplifies to O(N log 3) ≈ O(N).

Space Complexity: The space complexity is linear, O(N), as it creates several intermediate DataFrames proportional to the size of the input data.

Solution 2: Using MySQL

This solution uses window functions in MySQL to efficiently solve the problem.

Steps:

1. Rank Salaries Within Departments:

   • WITH T AS (...): A common table expression (CTE) called T is used.
   • DENSE_RANK() OVER (PARTITION BY departmentId ORDER BY salary DESC): This is the core of the solution. DENSE_RANK() assigns a rank to each employee's salary within their department, ordered descending by salary. DENSE_RANK() handles ties (same salaries) by assigning the same rank, unlike RANK() which would skip ranks.

2. Join and Filter:

   • SELECT d.name AS Department, t.name AS Employee, salary AS Salary: Selects the desired columns, aliasing them for clarity.
   • FROM T AS t JOIN Department AS d ON t.departmentId = d.id: Joins the CTE T with the Department table to get the department names.
   • WHERE rk <= 3: Filters the results to keep only employees with a rank of 3 or less (top 3).

Time Complexity: The MySQL solution's time complexity is dominated by the window function DENSE_RANK(). Window functions in MySQL generally have a complexity that is roughly linear to the number of rows, so the overall complexity is approximately O(N log N) in the worst case (though often optimized to be closer to linear in practice).

Space Complexity: The space complexity is dominated by the CTE T, which stores a copy of the Employee table with an added rank column. Thus, the space complexity is approximately O(N).

Summary Table:

| Solution | Language | Time Complexity | Space Complexity | |---------------|----------|-----------------|-----------------| | Solution 1 | Python | O(N) | O(N) | | Solution 2 | MySQL | O(N log N) | O(N) |

Note: The actual performance can vary depending on database optimizations and data characteristics. The pandas solution might be faster for smaller datasets, while the MySQL solution can be more efficient for very large datasets due to database optimizations.