Solution Explanation for LeetCode 1875: Group Employees of the Same Salary

This problem requires grouping employees with the same salary into teams, with each team having at least two members. Employees with unique salaries are excluded. Teams are assigned IDs based on salary rank (lowest salary gets team ID 1).

The solution uses SQL. The approach is broken down into several Common Table Expressions (CTEs):

1. S CTE: This selects distinct salaries that appear more than once (meaning at least two employees share that salary). The GROUP BY clause groups rows by salary, and the HAVING COUNT(1) > 1 clause filters out salaries with only one occurrence.

2. T CTE: This CTE ranks the salaries from S using ROW_NUMBER(). ROW_NUMBER() assigns a unique rank to each salary based on ascending order. This rank becomes the team_id.

3. Final SELECT Statement: This joins the original Employees table (e) with the T CTE (t) based on matching salaries. This adds the team_id to each employee's row. The result is then ordered by team_id (ascending) and employee_id (ascending) as specified.

Code (MySQL):

WITH
    S AS (
        SELECT salary
        FROM Employees
        GROUP BY salary
        HAVING COUNT(1) > 1
    ),
    T AS (
        SELECT salary, ROW_NUMBER() OVER (ORDER BY salary) AS team_id
        FROM S
    )
SELECT e.*, t.team_id
FROM
    Employees AS e
    JOIN T AS t ON e.salary = t.salary
ORDER BY 4, 1;

Time Complexity Analysis:

• The S CTE involves a GROUP BY operation, which has a time complexity of O(N log N) or O(N) depending on the database engine's optimization (N being the number of rows in the Employees table).
• The T CTE performs ROW_NUMBER() which is typically O(N) in most database systems.
• The final SELECT statement performs a JOIN operation, which is also O(N) on average (for properly indexed tables).
• The ORDER BY operation has a time complexity of O(N log N).

Therefore, the overall time complexity is dominated by the sorting in the ORDER BY clause, making it O(N log N).

Space Complexity Analysis:

The space complexity is determined by the size of the intermediate CTEs (S and T), which are both proportional to the number of unique salaries in the Employees table. In the worst case (all salaries are different), this would be O(N). However, in most practical scenarios where there's some salary repetition, it'll be considerably less than O(N). Thus, the space complexity is considered O(M) where M is the number of unique salaries present in the table (M <= N).