Multiple CTEs in one query in PostgreSQL - Time & Space Complexity

Time Complexity: Multiple CTEs in one query

O(n)

Understanding Time Complexity

When using multiple CTEs (Common Table Expressions) in one query, it's important to understand how the query's work grows as the data grows.

We want to know how the total steps needed change when the input data gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following query with multiple CTEs.

WITH cte1 AS (
  SELECT * FROM orders WHERE order_date > '2023-01-01'
),
cte2 AS (
  SELECT customer_id, COUNT(*) AS order_count FROM cte1 GROUP BY customer_id
),
cte3 AS (
  SELECT customer_id FROM cte2 WHERE order_count > 5
)
SELECT * FROM customers WHERE id IN (SELECT customer_id FROM cte3);
    

This query filters recent orders, counts orders per customer, then selects customers with more than 5 orders.

Identify Repeating Operations

Look for repeated work inside the query.

• Primary operation: Scanning and grouping rows in the orders table.
• How many times: Each CTE runs once in sequence, but the grouping scans all filtered orders.

How Execution Grows With Input

As the number of orders grows, the work to scan and group them grows too.

Input Size (n)	Approx. Operations
10	About 10 scans and groups
100	About 100 scans and groups
1000	About 1000 scans and groups

Pattern observation: The work grows roughly in direct proportion to the number of orders.

Final Time Complexity

Time Complexity: O(n)

This means the query's work grows linearly with the number of orders processed.

Common Mistake

[X] Wrong: "Using multiple CTEs means the query runs multiple times and multiplies the work."

[OK] Correct: Each CTE runs once in order, and the total work depends mostly on the largest data scanned, not on the number of CTEs.

Interview Connect

Understanding how multiple CTEs affect query time helps you write clear and efficient queries, a useful skill in real projects and interviews.

Self-Check

What if we added a join inside one of the CTEs? How would that change the time complexity?

Practice

1. What is the main purpose of using multiple CTEs (Common Table Expressions) in a single PostgreSQL query?

easy

A. To speed up the database server hardware

B. To create multiple databases at once

C. To permanently store data in the database

D. To break a complex query into smaller, readable parts

Solution

1. Step 1: Understand what CTEs do

   CTEs let you write parts of a query separately with names, making complex queries easier to read.

2. Step 2: Identify the purpose of multiple CTEs

   Using multiple CTEs helps organize and simplify complex logic by splitting it into smaller named queries.

3. Final Answer:

   To break a complex query into smaller, readable parts -> Option D

4. Quick Check:

   Multiple CTEs = Query readability [OK]

Hint: Think of CTEs as named steps in a recipe [OK]

Common Mistakes:

• Confusing CTEs with permanent tables
• Thinking CTEs improve hardware speed
• Assuming CTEs create databases

2. Which of the following is the correct syntax to define two CTEs named cte1 and cte2 in one PostgreSQL query?

easy

A. WITH cte1 AS (SELECT * FROM table1), cte2 AS (SELECT * FROM table2) SELECT * FROM cte1 JOIN cte2 ON cte1.id = cte2.id;

B. WITH cte1 (SELECT * FROM table1); WITH cte2 (SELECT * FROM table2); SELECT * FROM cte1, cte2;

C. WITH cte1 = SELECT * FROM table1, cte2 = SELECT * FROM table2 SELECT * FROM cte1 JOIN cte2;

D. WITH cte1 AS SELECT * FROM table1, cte2 AS SELECT * FROM table2 SELECT * FROM cte1 JOIN cte2;

Solution

1. Step 1: Recall correct CTE syntax

   CTEs start with WITH, each named, followed by AS and a query in parentheses, separated by commas.

2. Step 2: Check each option's syntax

   WITH cte1 AS (SELECT * FROM table1), cte2 AS (SELECT * FROM table2) SELECT * FROM cte1 JOIN cte2 ON cte1.id = cte2.id; correctly uses commas, AS, and parentheses. Others miss parentheses or use wrong separators.

3. Final Answer:

   WITH cte1 AS (SELECT * FROM table1), cte2 AS (SELECT * FROM table2) SELECT * FROM cte1 JOIN cte2 ON cte1.id = cte2.id; -> Option A

4. Quick Check:

   Correct CTE syntax = WITH cte1 AS (SELECT * FROM table1), cte2 AS (SELECT * FROM table2) SELECT * FROM cte1 JOIN cte2 ON cte1.id = cte2.id; [OK]

Hint: Remember: WITH cte_name AS (query), separate multiple CTEs by commas [OK]

Common Mistakes:

• Omitting parentheses around CTE queries
• Using semicolons between CTEs
• Using '=' instead of AS

3. Given the following query, what will be the output?

WITH cte1 AS (SELECT 1 AS num), cte2 AS (SELECT num * 2 AS double_num FROM cte1) SELECT * FROM cte2;

medium

A. [{num: 2}]

B. [{num: 1, double_num: 2}]

C. [{double_num: 2}]

D. Syntax error

Solution

1. Step 1: Analyze cte1

   cte1 returns one row with column num = 1.

2. Step 2: Analyze cte2 using cte1

   cte2 selects num * 2 as double_num, so double_num = 1 * 2 = 2.

3. Step 3: Final SELECT from cte2

   Only double_num column is selected, so output is [{double_num: 2}].

4. Final Answer:

   [{double_num: 2}] -> Option C

5. Quick Check:

   cte2 output columns = double_num [OK]

Hint: Check which columns each CTE outputs carefully [OK]

Common Mistakes:

• Assuming cte2 includes all columns from cte1
• Expecting num column in final output
• Thinking query has syntax errors

4. Identify the error in this query using multiple CTEs:

WITH cte1 AS (SELECT id, name FROM users), cte2 AS (SELECT id, age FROM users) SELECT cte1.id, cte1.name, cte2.age FROM cte1 JOIN cte2 ON cte1.id = cte2.user_id;

medium

A. CTEs must be defined with semicolons

B. Column cte2.user_id does not exist

C. Missing comma between CTEs

D. Cannot join two CTEs in one query

Solution

1. Step 1: Check CTE definitions

   Both cte1 and cte2 select id and other columns from users table correctly.

2. Step 2: Check JOIN condition

   Query joins on cte1.id = cte2.user_id, but cte2 has no user_id column, only id.

3. Step 3: Identify error

   Using cte2.user_id causes a column not found error.

4. Final Answer:

   Column cte2.user_id does not exist -> Option B

5. Quick Check:

   Check column names in JOIN = Column cte2.user_id does not exist [OK]

Hint: Verify column names in JOIN conditions carefully [OK]

Common Mistakes:

• Assuming user_id exists in cte2
• Thinking semicolons separate CTEs
• Believing CTEs cannot be joined

5. You want to calculate the average order amount per customer and then find customers whose average order is above 100. Which query correctly uses multiple CTEs to achieve this?

hard

A. WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg;

B. WITH avg_orders AS (SELECT customer_id, SUM(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg;

C. WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg;

D. WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id) SELECT * FROM avg_orders WHERE avg_amount > 100;

Solution

1. Step 1: Calculate average order per customer

   WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; correctly uses AVG(amount) with GROUP BY customer_id in avg_orders CTE.

2. Step 2: Filter customers with avg_amount > 100

   WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; uses a second CTE high_avg to select customers meeting the condition.

3. Step 3: Final SELECT returns customers with high average orders

   WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; selects all from high_avg, giving correct result.

4. Final Answer:

   WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; -> Option A

5. Quick Check:

   Correct aggregation and filtering = WITH avg_orders AS (SELECT customer_id, AVG(amount) AS avg_amount FROM orders GROUP BY customer_id), high_avg AS (SELECT customer_id FROM avg_orders WHERE avg_amount > 100) SELECT * FROM high_avg; [OK]

Hint: Use one CTE for aggregation, another for filtering [OK]

Common Mistakes:

• Using SUM instead of AVG for average calculation
• Missing GROUP BY clause
• Not filtering in a separate CTE