Show each employee alongside the lowest-paid employee name in their department.

Problem Statement

<p>Show each employee alongside the lowest-paid employee name in their department.</p>

Examples

Input: employees table: +----+---------+------------+--------+ | id | name | department | salary | +----+---------+------------+--------+ | 1 | Alice | IT | 90000 | | 2 | Bob | IT | 120000 | | 3 | Charlie | HR | 60000 | | 4 | Dana | HR | 75000 | +----+---------+------------+--------+

Output: +---------+------------+--------+---------------+ | name | department | salary | lowest_earner | +---------+------------+--------+---------------+ | Dana | HR | 75000 | Charlie | | Charlie | HR | 60000 | Charlie | | Bob | IT | 120000 | Alice | | Alice | IT | 90000 | Alice | +---------+------------+--------+---------------+

Explanation: The window function evaluates the calculation across the partition in the specified ordering.

Complexity

Time Complexity: -

Space Complexity: -

Hints

💡 Hint 1: Use a window function with OVER (...) to compute values across rows related to the current row without collapsing them like GROUP BY. 💡 Hint 2: Check if PARTITION BY is required to split windows by category or if only an ORDER BY inside OVER (...) is needed. 💡 Hint 3: Structure the query as: <window_func>() OVER (PARTITION BY ... ORDER BY ...) AS <alias>;

Editorial & Approach

Problem Overview & Intuition

To solve "LAST_VALUE Window Function", we query the relational database engine using declarative SQL. The goal is to show each employee alongside the lowest-paid employee name in their department. By formulating an optimal execution plan with appropriate projection and filtering, the database engine executes the query with minimal overhead.

Step-by-Step Approach

Analyze Schema: Identify the target tables, necessary foreign keys, and expected output columns.
Construct Filtering & Logic: Apply row projections to isolate the requested data.
Format & Order: Sort the resulting records according to specified order criteria.

Optimal Implementation (SQL)

SELECT name, department, salary, LAST_VALUE(name) OVER (PARTITION BY department ORDER BY salary DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS lowest_earner FROM employees;

Complexity Analysis

Time Complexity O(N log N) for sorting or partitioning rows.

Space Complexity O(N) for intermediate group hash tables or window buffers.

Key Considerations & Edge Cases

Empty Tables: The query executes safely returning zero rows without syntax error.
NULL Values: Columns containing NULL values are properly handled by standard ANSI SQL semantics.
Case Sensitivity: String comparisons and keywords adhere to PostgreSQL/standard SQL rules.

LAST_VALUE Window Function

Hard

Show each employee alongside the lowest-paid employee name in their department.

Example Scenarios

1Example 1

Input:

employees table

id	name	department	salary
1	Alice	IT	90000
2	Bob	IT	120000
3	Charlie	HR	60000
4	Dana	HR	75000

Output:

name	department	salary	lowest_earner
Dana	HR	75000	Charlie
Charlie	HR	60000	Charlie
Bob	IT	120000	Alice
Alice	IT	90000	Alice

Explanation:

The window function evaluates the calculation across the partition in the specified ordering.

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