Solution Explanation

This problem requires finding all employees who report, directly or indirectly, to the head of the company (employee ID 1). The solution uses SQL joins to traverse the hierarchical manager-employee relationship. The constraint that the indirect reporting chain doesn't exceed three levels simplifies the query.

Approach

The provided SQL solution cleverly uses multiple joins to achieve this. Let's break down the query:

1. SELECT e1.employee_id: This selects the employee ID from the first table (e1) which will eventually represent our result set.

2. FROM Employees AS e1 JOIN Employees AS e2 ON e1.manager_id = e2.employee_id: This joins the Employees table to itself (aliased as e1 and e2). e1 represents an employee, and e2 represents their direct manager. The join condition ensures we only consider pairs where e1's manager ID matches e2's employee ID.

3. JOIN Employees AS e3 ON e2.manager_id = e3.employee_id: This adds another self-join. e3 represents the manager of e2 (i.e., the manager's manager). This extends the chain to two levels of indirect reporting.

4. WHERE e1.employee_id != 1 AND e3.manager_id = 1: This crucial filter ensures we only include employees who are not the CEO (e1.employee_id != 1) and whose ultimate manager (e3) is the CEO (e3.manager_id = 1). This condition effectively filters out the CEO and any employees who are not in the CEO's reporting chain (even indirectly).

The query effectively traverses the management hierarchy in a depth of up to three levels (CEO -> Manager -> Sub-Manager -> Employee). If an employee's chain leads to the CEO within those three levels, the query will include the employee's employee_id in the result.

Time Complexity Analysis

The time complexity of the SQL query is primarily determined by the join operations. In the worst case, the number of comparisons needed for the joins is proportional to the product of the number of rows in the tables being joined. Assuming N is the number of rows in the Employees table, the join operations have a complexity of approximately O(N^3) in the worst case (due to three self joins). However, because of indexing and query optimization techniques used by database engines (especially if employee_id and manager_id are indexed), the actual performance will generally be far better than O(N^3). In practice, it will likely closer to O(N log N) or even better depending on the database system and its indexing.

Space Complexity Analysis

The space complexity is determined by the temporary space needed during the join operations and the space needed to store the resulting table. The space complexity is linearly proportional to the number of rows in the Employees table, hence O(N) in the worst case. However, much of this is already in memory as the tables must be loaded. The additional memory for joins and sorting is usually small compared to the input table.

Code in Other Languages

While the problem is inherently SQL-based, it would be inefficient (and not recommended) to solve this in languages like Python or Java without using a database. These languages would require custom implementations of graph traversal algorithms which is far less efficient. The optimal approach directly leverages the database's capabilities for handling relational data.