Solution Explanation for Calculating Salaries with Taxes

This problem involves calculating employee salaries after applying taxes based on the maximum salary within each company. The tax rate varies depending on the maximum salary: 0% if less than $1000, 24% if between $1000 and $10000 (inclusive), and 49% if greater than $10000. The final salary needs to be rounded to the nearest integer.

The most efficient way to solve this is using SQL. The solution leverages subqueries and conditional logic to achieve the desired outcome.

Approach

1. Find Maximum Salaries: A subquery is used to find the maximum salary for each company. This is done by grouping the Salaries table by company_id and using the MAX() aggregate function to find the maximum salary (top) within each group.

2. Join with Main Table: The result of the subquery (maximum salaries for each company) is joined with the original Salaries table using a JOIN clause based on company_id. This links each employee's record with their company's maximum salary.

3. Apply Taxes Based on Maximum Salary: A CASE statement applies the appropriate tax rate based on the top (maximum salary) from the joined subquery. The calculated salary after tax is then rounded using the ROUND() function.

4. Return Result: The query returns the company_id, employee_id, employee_name, and the calculated salary (after tax) for each employee.

MySQL Code Explained

SELECT
    s.company_id,
    employee_id,
    employee_name,
    ROUND(
        CASE
            WHEN top < 1000 THEN salary  -- 0% tax
            WHEN top >= 1000 AND top <= 10000 THEN salary * 0.76 -- 24% tax
            ELSE salary * 0.51  -- 49% tax
        END
    ) AS salary
FROM
    Salaries AS s
    JOIN (
        SELECT company_id, MAX(salary) AS top
        FROM Salaries
        GROUP BY company_id
    ) AS t
        ON s.company_id = t.company_id;

• SELECT ... FROM Salaries AS s: This selects the relevant columns from the Salaries table, aliased as s.
• JOIN (...) AS t ON s.company_id = t.company_id: This joins Salaries with the subquery result (t), matching companies based on company_id.
• SELECT company_id, MAX(salary) AS top FROM Salaries GROUP BY company_id: This subquery finds the maximum salary (top) for each company.
• ROUND(CASE ... END) AS salary: This CASE statement determines the tax rate and applies it, then rounds the result to the nearest integer.

Time Complexity Analysis

The time complexity is dominated by the GROUP BY operation in the subquery, which has a time complexity of O(N log N) or O(N) depending on the database's specific implementation (N being the number of rows in the Salaries table). The join operation is also O(N) in the average case. Therefore, the overall time complexity of the query is O(N log N) in the worst case or O(N) on average. The space complexity is O(M), where M is the number of distinct company IDs, as that's the space needed to store the intermediate results of the subquery.

No other language-specific solutions are needed as the problem is inherently solved efficiently using SQL. Other approaches would be significantly less efficient.