Solution Explanation

This problem involves selecting employees (seniors and juniors) to hire within a budget of $70000, prioritizing seniors first and then juniors. The solution uses SQL and a clever approach to achieve this in an efficient manner.

Approach

The solution uses Common Table Expressions (CTEs) to break down the problem into smaller, manageable parts. The core idea is to cumulatively sum salaries for seniors and then juniors, keeping track of the running total to stay within budget.

1. Senior CTE (s):

This CTE selects employee_id and calculates a running sum of salaries for seniors using the SUM() OVER (ORDER BY salary) window function. This function calculates the cumulative sum of salaries as they are ordered by salary in ascending order. The cur column represents the running total of salaries for seniors hired so far.

2. Junior CTE (j):

This CTE selects employee_id for juniors and computes a running sum of their salaries. However, it's crucial to note the IFNULL() function. This function handles the case where no senior employees can be hired within the budget. It checks the maximum cumulative salary from the s CTE that's less than or equal to 70000. If no such senior exists (meaning MAX(cur) is NULL), it defaults to 0; otherwise, it uses the maximum cumulative senior salary. The running sum for juniors (cur) then begins from this value (either 0 or the maximum senior cumulative salary) and continues summing junior salaries.

3. Final SELECT Statement:

Finally, a UNION combines the results from the s and j CTEs. It selects employee_id from s where the cumulative senior salary (cur) is less than or equal to 70000 and then selects employee_id from j where the cumulative junior salary (cur) is also less than or equal to 70000. This ensures that only the employees whose inclusion does not exceed the budget are selected.

MySQL Code Explained

WITH
    s AS (
        SELECT
            employee_id,
            SUM(salary) OVER (ORDER BY salary) AS cur
        FROM Candidates
        WHERE experience = 'Senior'
    ),
    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'
    )
SELECT
    employee_id
FROM s
WHERE cur <= 70000
UNION
SELECT
    employee_id
FROM j
WHERE cur <= 70000;

The code directly implements the approach described above. Each part corresponds to the steps outlined in the approach section.

Time Complexity Analysis

The time complexity of this solution is dominated by the sorting within the SUM() OVER (ORDER BY salary) window functions. The ORDER BY clause requires a sort, which has a time complexity of O(N log N), where N is the number of rows in the Candidates table. The other operations (cumulative sum, selection) are linear (O(N)) in nature. Therefore, the overall time complexity is O(N log N).

Space Complexity Analysis

The space complexity is determined by the size of the CTEs. In the worst case, if all candidates are hired, the CTEs will store data proportional to the size of the input table. Hence, the space complexity is O(N).