Solution Explanation for LeetCode 2004: The Number of Seniors and Juniors to Join the Company

This problem involves querying a database table (Candidates) to determine the optimal number of senior and junior employees to hire, given a budget constraint. The hiring process prioritizes seniors first, then uses the remaining budget for juniors.

The solution utilizes window functions in SQL to efficiently solve this. Let's break down the MySQL solution:

MySQL Solution:

The solution uses two Common Table Expressions (CTEs): s for seniors and j for juniors.

1. s CTE (Senors):

WITH
    s AS (
        SELECT
            employee_id,
            SUM(salary) OVER (ORDER BY salary) AS cur
        FROM Candidates
        WHERE experience = 'Senior'
    )

• SELECT employee_id, ...: Selects the employee ID.
• SUM(salary) OVER (ORDER BY salary) AS cur: This is the crucial part. It uses a window function SUM() OVER (ORDER BY salary) to cumulatively sum the salaries of seniors in ascending order of salary. cur represents the running total of salaries for seniors hired so far.

2. j CTE (Juniors):

j AS (
    SELECT
        employee_id,
        IFNULL(
            SELECT
                MAX(cur)
            FROM s
            WHERE cur <= 70000,
            0
        ) + SUM(salary) OVER (ORDER BY salary) AS cur
    FROM Candidates
    WHERE experience = 'Junior'
)

• IFNULL((SELECT MAX(cur) FROM s WHERE cur <= 70000), 0): This subquery finds the maximum cumulative salary of seniors hired within the budget ($70000). IFNULL handles the case where no seniors can be hired (it returns 0). This represents the amount of the budget already spent on senior hires.
• + SUM(salary) OVER (ORDER BY salary) AS cur: Similar to the s CTE, this cumulatively sums junior salaries, but it adds the amount spent on seniors to the running total. This ensures that the junior hiring starts with the remaining budget after senior hiring.

3. Final SELECT Statement:

SELECT
    'Senior' AS experience,
    COUNT(employee_id) AS accepted_candidates
FROM s
WHERE cur <= 70000
UNION ALL
SELECT
    'Junior' AS experience,
    COUNT(employee_id) AS accepted_candidates
FROM j
WHERE cur <= 70000;

• This part combines the results from both CTEs.
• WHERE cur <= 70000: Filters out candidates whose cumulative salary exceeds the budget.
• UNION ALL: Combines the results for seniors and juniors into a single result set.

Time Complexity Analysis:

The time complexity is dominated by the sorting and window function calculations within the CTEs. The ORDER BY clause in the window functions implies a sort, which has a time complexity of O(N log N), where N is the number of rows in the Candidates table. The window function itself is then applied in linear time, O(N). Therefore, the overall time complexity is O(N log N).

Space Complexity Analysis:

The space complexity is primarily determined by the size of the CTEs, which are proportional to the number of rows in the Candidates table. Thus, the space complexity is O(N).

This solution elegantly uses SQL's capabilities to efficiently solve the problem without needing complex procedural logic. The use of window functions makes it concise and readable.