FIRST_VALUE and LAST_VALUE in SQL - Time & Space Complexity

Time Complexity: FIRST_VALUE and LAST_VALUE

O(n)

Understanding Time Complexity

We want to understand how the time it takes to run queries using FIRST_VALUE and LAST_VALUE changes as the data grows.

How does the size of the data affect the work done by these functions?

Scenario Under Consideration

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

SELECT employee_id, department_id, salary,
       FIRST_VALUE(salary) OVER (PARTITION BY department_id ORDER BY salary DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS highest_salary,
       LAST_VALUE(salary) OVER (PARTITION BY department_id ORDER BY salary DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS lowest_salary
FROM employees;
    

This query finds the highest and lowest salary in each department for every employee row.

Identify Repeating Operations

Look for repeated work inside the query.

• Primary operation: Scanning each row and computing window functions per partition.
• How many times: Once per row, but window functions look at all rows in the same department.

How Execution Grows With Input

As the number of employees grows, the work to find first and last values per department grows too.

Input Size (n)	Approx. Operations
10	About 10 rows scanned, small partitions checked
100	About 100 rows scanned, partitions larger, more rows checked per partition
1000	About 1000 rows scanned, partitions much larger, more comparisons per row

Pattern observation: The work grows roughly in proportion to the number of rows because each row's window looks at its partition.

Final Time Complexity

Time Complexity: O(n)

This means the time to run the query grows roughly in direct proportion to the number of rows.

Common Mistake

[X] Wrong: "FIRST_VALUE and LAST_VALUE only look at one row, so they run instantly regardless of data size."

[OK] Correct: These functions scan all rows in each partition to find the first or last value, so bigger partitions mean more work.

Interview Connect

Understanding how window functions scale helps you explain query performance clearly and shows you know how databases handle grouped data.

Self-Check

What if we removed the PARTITION BY clause? How would the time complexity change?

Practice

1. What does the SQL function FIRST_VALUE(column_name) OVER (ORDER BY column_name) return?

easy

A. The average value of the column

B. The first value in the ordered set of rows for that column

C. The last value in the ordered set of rows for that column

D. The total count of rows in the table

Solution

1. Step 1: Understand FIRST_VALUE function

   It returns the first value in the ordered set of rows defined by the OVER clause.

2. Step 2: Analyze the ORDER BY clause

   The ORDER BY inside OVER() defines the order to pick the first value from.

3. Final Answer:

   The first value in the ordered set of rows for that column -> Option B

4. Quick Check:

   FIRST_VALUE returns first row value [OK]

Hint: FIRST_VALUE picks the first row after ordering [OK]

Common Mistakes:

• Confusing FIRST_VALUE with LAST_VALUE
• Thinking it returns an aggregate like average
• Ignoring the ORDER BY inside OVER()

2. Which of the following is the correct syntax to get the last value of salary ordered by hire_date in SQL?

easy

A. SELECT LAST_VALUE(salary) OVER (ORDER BY hire_date) FROM employees;

B. SELECT LAST_VALUE(salary) FROM employees ORDER BY hire_date;

C. SELECT LAST_VALUE(salary) OVER (PARTITION BY hire_date) FROM employees;

D. SELECT LAST_VALUE(salary) OVER (ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM employees;

Solution

1. Step 1: Understand LAST_VALUE default frame

   LAST_VALUE by default looks only at rows up to current row, so it may not return the true last value.

2. Step 2: Use frame clause to include all rows

   Adding ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ensures LAST_VALUE looks at the entire partition.

3. Final Answer:

   SELECT LAST_VALUE(salary) OVER (ORDER BY hire_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM employees; -> Option D

4. Quick Check:

   LAST_VALUE needs full frame to get last row [OK]

Hint: Add full frame clause to LAST_VALUE for correct last row [OK]

Common Mistakes:

• Omitting frame clause with LAST_VALUE
• Using PARTITION BY incorrectly
• Trying to use LAST_VALUE without OVER()

3. Given the table sales with columns region and amount, what will this query return?

SELECT region, amount, FIRST_VALUE(amount) OVER (PARTITION BY region ORDER BY amount DESC) AS max_amount FROM sales;

medium

A. The first amount in the whole table regardless of region

B. The lowest amount per region repeated on each row of that region

C. The highest amount per region repeated on each row of that region

D. Syntax error due to missing frame clause

Solution

1. Step 1: Analyze PARTITION BY and ORDER BY

   The window partitions rows by region and orders amounts descending within each region.

2. Step 2: Understand FIRST_VALUE with ordering

   FIRST_VALUE picks the first row's amount in each partition ordered by amount DESC, which is the max amount.

3. Final Answer:

   The highest amount per region repeated on each row of that region -> Option C

4. Quick Check:

   FIRST_VALUE with DESC order gives max per partition [OK]

Hint: Use ORDER BY DESC with FIRST_VALUE for max per group [OK]

Common Mistakes:

• Assuming FIRST_VALUE always returns lowest value
• Confusing PARTITION BY with filtering
• Expecting syntax error without frame clause here

4. Identify the error in this query:

SELECT employee_id, LAST_VALUE(salary) OVER (ORDER BY employee_id) AS last_sal FROM employees;

medium

A. LAST_VALUE needs a frame clause to get correct last value

B. Missing PARTITION BY clause causes error

C. ORDER BY cannot be used inside OVER()

D. salary column must be aggregated with GROUP BY

Solution

1. Step 1: Check LAST_VALUE default behavior

   By default, LAST_VALUE looks only up to current row, so it may not return the true last value.

2. Step 2: Add frame clause to fix

   Adding ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING makes LAST_VALUE consider all rows.

3. Final Answer:

   LAST_VALUE needs a frame clause to get correct last value -> Option A

4. Quick Check:

   LAST_VALUE without frame returns wrong last row [OK]

Hint: Add full frame to LAST_VALUE for correct last row [OK]

Common Mistakes:

• Thinking PARTITION BY is mandatory
• Believing ORDER BY is invalid inside OVER()
• Confusing LAST_VALUE with aggregate functions

5. You want to find the first and last sale amounts per customer ordered by sale date from the sales table. Which query correctly returns both values on each row?

hard

A. SELECT customer_id, sale_date, FIRST_VALUE(amount) OVER (PARTITION BY customer_id ORDER BY sale_date) AS first_sale, LAST_VALUE(amount) OVER (PARTITION BY customer_id ORDER BY sale_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS last_sale FROM sales;

B. SELECT customer_id, sale_date, FIRST_VALUE(amount) OVER (ORDER BY sale_date) AS first_sale, LAST_VALUE(amount) OVER (ORDER BY sale_date) AS last_sale FROM sales;

C. SELECT customer_id, sale_date, FIRST_VALUE(amount) AS first_sale, LAST_VALUE(amount) AS last_sale FROM sales GROUP BY customer_id;

D. SELECT customer_id, sale_date, MIN(amount) OVER (PARTITION BY customer_id) AS first_sale, MAX(amount) OVER (PARTITION BY customer_id) AS last_sale FROM sales;

Solution

1. Step 1: Use PARTITION BY customer_id

   This groups rows by customer to get first and last sales per customer.

2. Step 2: Use ORDER BY sale_date for chronological order

   Ordering by sale_date ensures first and last are by date.

3. Step 3: Add frame clause for LAST_VALUE

   LAST_VALUE needs ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING to get true last value.

4. Final Answer:

   Query in option A correctly returns first and last sale amounts per customer -> Option A

5. Quick Check:

   FIRST_VALUE and LAST_VALUE with proper frame and partition [OK]

Hint: Use full frame for LAST_VALUE with PARTITION and ORDER BY [OK]

Common Mistakes:

• Omitting frame clause for LAST_VALUE
• Using aggregate functions instead of window functions
• Not partitioning by customer_id