Solution Explanation:

This problem requires finding the most experienced employees for each project. The solution uses a combination of SQL joins and window functions for efficient retrieval.

Approach:

1. Join Tables: We begin by joining the Project and Employee tables using the employee_id as the common key. This combines project information with employee experience data in a single dataset. An INNER JOIN is used to only include projects with associated employees.

2. Window Function (RANK()): A window function, specifically RANK(), is employed. RANK() assigns a rank within each project_id group based on experience_years in descending order. Employees with the same experience within a project receive the same rank.

3. Filtering for Top Rank: After ranking, we filter the results, selecting only rows where the rank (rk) is 1. This ensures we only retrieve the employees with the highest experience years for each project. If multiple employees share the highest experience, they will all have a rank of 1 and be included.

Time Complexity Analysis:

The time complexity is dominated by the JOIN operation and the RANK() window function. In general, JOIN operations can have a time complexity of O(N*M) in the worst case (N and M being the number of rows in the respective tables), though optimized database implementations often achieve better performance. The RANK() window function, while operating on a sorted set, has a time complexity proportional to the number of rows in the joined table after grouping. Therefore, the overall time complexity is approximately O(N log N) where N is the total number of rows in the joined table, assuming efficient indexing and sorting are used by the database.

Space Complexity Analysis:

The space complexity depends on the size of the intermediate results. The JOIN operation creates a temporary table holding the combined data. The RANK() function also requires temporary space for its calculations. The space complexity is therefore proportional to the size of the joined table, which is approximately O(N) where N is the number of rows in the joined table.

Code in Different Languages (MySQL):

The provided solution is already in MySQL. Other SQL dialects (like PostgreSQL, SQL Server, Oracle) may have slightly different syntax for window functions, but the core logic remains the same.

WITH
    T AS (
        SELECT
            *,
            RANK() OVER (
                PARTITION BY project_id
                ORDER BY experience_years DESC
            ) AS rk
        FROM
            Project
            JOIN Employee USING (employee_id)
    )
SELECT project_id, employee_id
FROM T
WHERE rk = 1;

Note: To run this code, you would need to have the Project and Employee tables created in your MySQL database with the specified schema. The USING (employee_id) clause provides a shorthand for joining on the common column name. If column names differ, you should use the ON keyword with explicit column comparisons instead.