Correlated subquery execution model in SQL - Time & Space Complexity

Time Complexity: Correlated subquery execution model

O(n * m)

Understanding Time Complexity

When using correlated subqueries, the database runs a smaller query for each row of the main query.

We want to understand how this repeated work grows as the data gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following SQL query with a correlated subquery.

SELECT e.employee_id, e.name
FROM employees e
WHERE e.salary > (
  SELECT AVG(salary)
  FROM employees
  WHERE department_id = e.department_id
);
    

This query finds employees whose salary is above the average salary in their own department.

Identify Repeating Operations

Look for repeated work inside the query.

• Primary operation: The subquery runs once for each employee row.
• How many times: As many times as there are employees (n times).

How Execution Grows With Input

Each employee triggers a subquery that scans employees in their department.

Input Size (n)	Approx. Operations
10	About 10 times scanning small groups
100	About 100 times scanning groups of employees
1000	About 1000 times scanning groups, more work overall

Pattern observation: The total work grows roughly with the square of the number of employees if departments are evenly sized.

Final Time Complexity

Time Complexity: O(n * m)

This means the query work grows with the number of employees times the average size of their departments.

Common Mistake

[X] Wrong: "The subquery runs just once, so the query is fast regardless of data size."

[OK] Correct: The subquery depends on each employee's department, so it runs repeatedly, increasing total work as data grows.

Interview Connect

Understanding how correlated subqueries work helps you explain query performance clearly and shows you can think about how databases handle repeated work.

Self-Check

"What if the subquery was uncorrelated and ran only once? How would the time complexity change?"

Practice

1. What is a correlated subquery in SQL?

easy

A. A subquery that runs independently of the outer query

B. A subquery that uses values from the outer query to filter results

C. A query that joins two tables without conditions

D. A query that only returns aggregate values

Solution

1. Step 1: Understand subquery types

   A correlated subquery depends on the outer query's current row to run.

2. Step 2: Identify correlation

   It uses columns from the outer query inside the subquery's WHERE clause.

3. Final Answer:

   A subquery that uses values from the outer query to filter results -> Option B

4. Quick Check:

   Correlated subquery = uses outer query values [OK]

Hint: Correlated subqueries reference outer query columns [OK]

Common Mistakes:

• Thinking subquery runs once independently
• Confusing with JOIN operations
• Assuming it returns only aggregates

2. Which of the following is the correct syntax for a correlated subquery?

easy

A. SELECT e.name FROM employees e WHERE e.salary > (SELECT AVG(salary) FROM employees WHERE department = e.department)

B. SELECT e.name FROM employees e WHERE salary > 50000

C. SELECT e.name FROM employees e WHERE e.salary > (SELECT AVG(salary) FROM employees)

D. SELECT e.name FROM employees e JOIN departments d ON e.department = d.id

Solution

1. Step 1: Identify correlation in subquery

   SELECT e.name FROM employees e WHERE e.salary > (SELECT AVG(salary) FROM employees WHERE department = e.department) uses 'e.department' inside the subquery, linking it to the outer query.

2. Step 2: Check other options

   SELECT e.name FROM employees e WHERE e.salary > (SELECT AVG(salary) FROM employees) has no reference to outer query in subquery; the option with 'salary > 50000' lacks a subquery and the JOIN option is not a subquery.

3. Final Answer:

   SELECT e.name FROM employees e WHERE e.salary > (SELECT AVG(salary) FROM employees WHERE department = e.department) -> Option A

4. Quick Check:

   Correlation needs outer query column inside subquery [OK]

Hint: Look for outer query columns inside subquery WHERE clause [OK]

Common Mistakes:

• Missing outer query reference inside subquery
• Confusing JOIN with subquery
• Using subquery without correlation

3. Given the tables employees(id, name, department, salary) and the query:

SELECT e1.name FROM employees e1 WHERE e1.salary > (SELECT AVG(e2.salary) FROM employees e2 WHERE e2.department = e1.department);

What does this query return?

medium

A. Employees whose salary is above the average salary of their own department

B. Employees whose salary is above the average salary of all employees

C. Employees with salary above 50000

D. All employees regardless of salary

Solution

1. Step 1: Understand the subquery correlation

   The subquery calculates average salary for the department of the current employee (e1.department).

2. Step 2: Compare salaries

   The outer query selects employees whose salary is greater than that department average.

3. Final Answer:

   Employees whose salary is above the average salary of their own department -> Option A

4. Quick Check:

   Salary > department average = Employees whose salary is above the average salary of their own department [OK]

Hint: Check which outer column is used inside subquery condition [OK]

Common Mistakes:

• Assuming average is for all employees
• Ignoring the correlation condition
• Confusing with simple WHERE salary > value

4. Identify the error in the following correlated subquery:

SELECT c.customer_id FROM customers c WHERE c.orders_count > (SELECT AVG(o.orders_count) FROM orders o WHERE o.customer_id = c.customer_id);

medium

A. The subquery will return multiple rows causing an error

B. The subquery uses a wrong table alias 'o' which is not defined

C. The subquery compares orders_count incorrectly; should use SUM instead of AVG

D. The subquery references the outer query correctly; no error

Solution

1. Step 1: Analyze correlation

   The subquery correctly uses 'c.customer_id' from the outer query in its WHERE clause.

2. Step 2: Check aggregation and output

   AVG(o.orders_count) is an aggregate that returns a single scalar value, even for customers with multiple orders.

3. Final Answer:

   The subquery references the outer query correctly; no error -> Option D

4. Quick Check:

   Subquery must return single value for comparison [OK]

Hint: Ensure subquery returns one value for comparison [OK]

Common Mistakes:

• Thinking AVG returns multiple rows without GROUP BY
• Assuming alias 'o' is undefined
• Believing SUM is needed instead of AVG

5. You want to find all products whose price is higher than the average price of products in the same category. Which query correctly uses a correlated subquery to achieve this?

hard

A. SELECT p.product_name FROM products p JOIN categories c ON p.category = c.id WHERE p.price > c.avg_price

B. SELECT p.product_name FROM products p WHERE p.price > (SELECT AVG(price) FROM products)

C. SELECT p.product_name FROM products p WHERE p.price > (SELECT AVG(price) FROM products WHERE category = p.category)

D. SELECT product_name FROM products WHERE price > ALL (SELECT price FROM products)

Solution

1. Step 1: Identify correlation condition

   SELECT p.product_name FROM products p WHERE p.price > (SELECT AVG(price) FROM products WHERE category = p.category) uses 'p.category' inside the subquery to calculate average price per category, correlating outer and inner queries.

2. Step 2: Verify other options

   SELECT p.product_name FROM products p WHERE p.price > (SELECT AVG(price) FROM products) compares to overall average, not per category; the option using JOIN with categories uses a JOIN but no subquery; the option using > ALL compares to all individual prices, not average.

3. Final Answer:

   SELECT p.product_name FROM products p WHERE p.price > (SELECT AVG(price) FROM products WHERE category = p.category) -> Option C

4. Quick Check:

   Correlated subquery filters by category [OK]

Hint: Use outer query column inside subquery WHERE for correlation [OK]

Common Mistakes:

• Using overall average instead of per category
• Confusing JOIN with subquery
• Using ALL instead of AVG in subquery