COUNT, SUM, AVG, MIN, MAX in PostgreSQL - Time & Space Complexity

Time Complexity: COUNT, SUM, AVG, MIN, MAX

O(n)

Understanding Time Complexity

When using aggregate functions like COUNT, SUM, AVG, MIN, and MAX in a database, it's important to understand how the time to get results changes as the data grows.

We want to know how the work done by these functions grows when the number of rows increases.

Scenario Under Consideration

Analyze the time complexity of the following SQL query using aggregate functions.

SELECT COUNT(*), SUM(price), AVG(price), MIN(price), MAX(price)
FROM sales;
    

This query calculates the total number of sales, the sum and average of prices, and the minimum and maximum price from the sales table.

Identify Repeating Operations

Look for repeated actions in the query execution.

• Primary operation: Scanning each row in the sales table once.
• How many times: Exactly once for all rows, since all aggregates can be computed in a single pass.

How Execution Grows With Input

As the number of rows increases, the database reads each row once to update the aggregate values.

Input Size (n)	Approx. Operations
10	10 reads and updates
100	100 reads and updates
1000	1000 reads and updates

Pattern observation: The work grows directly with the number of rows; doubling rows doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to compute these aggregates grows linearly with the number of rows in the table.

Common Mistake

[X] Wrong: "Aggregate functions like COUNT or SUM run instantly no matter how big the table is."

[OK] Correct: The database must look at each row to include it in the calculation, so more rows mean more work and more time.

Interview Connect

Understanding how aggregate functions scale helps you explain query performance clearly and shows you know how databases handle data as it grows.

Self-Check

"What if we added a WHERE clause that filters most rows before aggregation? How would the time complexity change?"

Practice

1. What does the SQL function COUNT(column_name) do in a PostgreSQL query?

easy

A. Finds the average of all values in the specified column.

B. Counts the number of non-null values in the specified column.

C. Calculates the sum of all values in the specified column.

D. Returns the maximum value in the specified column.

Solution

1. Step 1: Understand COUNT function behavior

   The COUNT function counts how many rows have a non-null value in the specified column.

2. Step 2: Differentiate from other aggregate functions

   SUM adds values, AVG finds average, MAX finds maximum, so they do not count rows.

3. Final Answer:

   Counts the number of non-null values in the specified column. -> Option B

4. Quick Check:

   COUNT = number of non-null rows [OK]

Hint: COUNT counts non-null rows, not sums or averages [OK]

Common Mistakes:

• Thinking COUNT sums values
• Assuming COUNT includes nulls
• Confusing COUNT with AVG or MAX

2. Which of the following is the correct syntax to find the average of the price column in a table named products?

easy

A. SELECT AVG price FROM products;

B. SELECT AVG[price] FROM products;

C. SELECT AVG(price) FROM products;

D. SELECT AVERAGE(price) FROM products;

Solution

1. Step 1: Recall correct function syntax

   Aggregate functions like AVG require parentheses around the column name: AVG(column_name).

2. Step 2: Check each option's syntax

   SELECT AVG(price) FROM products; uses correct syntax: AVG(price). Options A, B, and C have syntax errors or invalid function names.

3. Final Answer:

   SELECT AVG(price) FROM products; -> Option C

4. Quick Check:

   AVG(column) uses parentheses [OK]

Hint: Use parentheses around column in aggregate functions [OK]

Common Mistakes:

• Omitting parentheses in AVG
• Using wrong function name like AVERAGE
• Using brackets instead of parentheses

3. Given the table sales with column amount containing values (100, 200, NULL, 300), what is the result of this query?

SELECT COUNT(amount), SUM(amount), AVG(amount), MIN(amount), MAX(amount) FROM sales;

medium

A. (4, 600, 150, 100, 300)

B. (4, 600, 200, 100, 300)

C. (3, 600, 150, 100, 300)

D. (3, 600, 200, 100, 300)

Solution

1. Step 1: Count non-null values in amount

   Values are 100, 200, NULL, 300. COUNT(amount) counts non-null values = 3.

2. Step 2: Calculate SUM, AVG, MIN, MAX ignoring NULL

   SUM = 100+200+300 = 600; AVG = 600/3 = 200; MIN = 100; MAX = 300.

3. Final Answer:

   (3, 600, 200, 100, 300) -> Option D

4. Quick Check:

   COUNT=3, SUM=600, AVG=200 [OK]

Hint: Aggregate functions ignore NULL except COUNT(*) [OK]

Common Mistakes:

• Counting NULL as a value in COUNT(column)
• Dividing AVG by total rows including NULL
• Confusing COUNT(column) with COUNT(*)

4. Identify the error in this query:

SELECT COUNT(*), SUM(price), AVG(price) FROM products WHERE price > 'ten';

medium

A. Comparing price (numeric) with string 'ten' causes error.

B. Using COUNT(*) with WHERE clause is invalid.

C. SUM(price) cannot be used with AVG(price) in same query.

D. Missing GROUP BY clause for aggregate functions.

Solution

1. Step 1: Analyze WHERE clause condition

   The condition price > 'ten' compares a numeric column to a string, which causes a type error in PostgreSQL.

2. Step 2: Check other parts of the query

   COUNT(*) with WHERE is valid; SUM and AVG can be used together; GROUP BY not needed without grouping.

3. Final Answer:

   Comparing price (numeric) with string 'ten' causes error. -> Option A

4. Quick Check:

   Numeric compared to string causes error [OK]

Hint: Compare numeric columns only with numbers, not strings [OK]

Common Mistakes:

• Assuming COUNT(*) cannot be used with WHERE
• Thinking GROUP BY is always required with aggregates
• Mixing data types in WHERE conditions

5. You have a table orders with columns customer_id and total_amount. How do you write a query to find the average order amount for each customer, but only include customers with at least 5 orders?

hard

A. SELECT customer_id, AVG(total_amount) FROM orders GROUP BY customer_id HAVING COUNT(*) >= 5;

B. SELECT customer_id, AVG(total_amount) FROM orders WHERE COUNT(*) >= 5 GROUP BY customer_id;

C. SELECT customer_id, AVG(total_amount) FROM orders GROUP BY customer_id WHERE COUNT(*) >= 5;

D. SELECT customer_id, AVG(total_amount) FROM orders HAVING COUNT(*) >= 5 GROUP BY customer_id;

Solution

1. Step 1: Group orders by customer_id

   Use GROUP BY customer_id to aggregate orders per customer.

2. Step 2: Filter groups with HAVING clause

   HAVING COUNT(*) >= 5 filters customers with at least 5 orders.

3. Step 3: Calculate average order amount per customer

   Use AVG(total_amount) to find average order value per customer.

4. Final Answer:

   SELECT customer_id, AVG(total_amount) FROM orders GROUP BY customer_id HAVING COUNT(*) >= 5; -> Option A

5. Quick Check:

   GROUP BY + HAVING filters groups [OK]

Hint: Use HAVING to filter groups after GROUP BY [OK]

Common Mistakes:

• Using WHERE instead of HAVING for group filters
• Placing HAVING before GROUP BY
• Not grouping by customer_id