Count how many employees earn above 80000 and how many earn 80000 or below.

Problem Statement

<p>Count how many <code>employees</code> earn above 80000 and how many earn 80000 or below.</p>

Examples

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

Output: +--------------+--------+ | high_earners | others | +--------------+--------+ | 2 | 2 | +--------------+--------+

Explanation: The records are grouped by category and the aggregate calculation is applied to produce the summary result.

Complexity

Time Complexity: -

Space Complexity: -

Hints

💡 Hint 1: Use a conditional CASE WHEN ... THEN ... ELSE ... END expression to transform values or implement if-else logic. 💡 Hint 2: Each condition is evaluated in order; provide a sensible ELSE fallback if unhandled cases exist. 💡 Hint 3: Place the CASE expression in the SELECT list with an alias: SELECT CASE WHEN condition THEN val ELSE default END AS column_name FROM ...;

Editorial & Approach

Problem Overview & Intuition

To solve "Conditional Counting (COUNT with CASE)", we query the relational database engine using declarative SQL. The goal is to count how many employees earn above 80000 and how many earn 80000 or below. 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: Ensure columns match the expected project schema in order.

Optimal Implementation (SQL)

SELECT COUNT(CASE WHEN salary > 80000 THEN 1 END) AS high_earners, COUNT(CASE WHEN salary <= 80000 THEN 1 END) AS others FROM employees;

Complexity Analysis

Time Complexity O(N)

Space Complexity O(1)

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.

Conditional Counting (COUNT with CASE)

Medium

Count how many employees earn above 80000 and how many earn 80000 or below.

Example Scenarios

1Example 1

Input:

employees table

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

Output:

high_earners	others
2	2

Explanation:

The records are grouped by category and the aggregate calculation is applied to produce the summary result.

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