FIRST_VALUE and LAST_VALUE in PostgreSQL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

Output of FIRST_VALUE with window frame

Given the table sales with columns region, month, and revenue, what is the output of the following query?

SELECT region, month, revenue,
       FIRST_VALUE(revenue) OVER (PARTITION BY region ORDER BY month ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS first_rev
FROM sales
ORDER BY region, month;

PostgreSQL

CREATE TABLE sales (region TEXT, month INT, revenue INT);
INSERT INTO sales VALUES
('East', 1, 100),
('East', 2, 150),
('East', 3, 120),
('West', 1, 200),
('West', 2, 180);

SELECT region, month, revenue,
       FIRST_VALUE(revenue) OVER (PARTITION BY region ORDER BY month ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS first_rev
FROM sales
ORDER BY region, month;

A [{"region": "East", "month": 1, "revenue": 100, "first_rev": 100}, {"region": "East", "month": 2, "revenue": 150, "first_rev": 100}, {"region": "East", "month": 3, "revenue": 120, "first_rev": 100}, {"region": "West", "month": 1, "revenue": 200, "first_rev": 200}, {"region": "West", "month": 2, "revenue": 180, "first_rev": 200}]

B [{"region": "East", "month": 1, "revenue": 100, "first_rev": 100}, {"region": "East", "month": 2, "revenue": 150, "first_rev": 150}, {"region": "East", "month": 3, "revenue": 120, "first_rev": 120}, {"region": "West", "month": 1, "revenue": 200, "first_rev": 200}, {"region": "West", "month": 2, "revenue": 180, "first_rev": 180}]

C [{"region": "East", "month": 1, "revenue": 100, "first_rev": 100}, {"region": "East", "month": 2, "revenue": 150, "first_rev": 150}, {"region": "East", "month": 3, "revenue": 120, "first_rev": 150}, {"region": "West", "month": 1, "revenue": 200, "first_rev": 200}, {"region": "West", "month": 2, "revenue": 180, "first_rev": 200}]

D [{"region": "East", "month": 1, "revenue": 100, "first_rev": 120}, {"region": "East", "month": 2, "revenue": 150, "first_rev": 120}, {"region": "East", "month": 3, "revenue": 120, "first_rev": 120}, {"region": "West", "month": 1, "revenue": 200, "first_rev": 180}, {"region": "West", "month": 2, "revenue": 180, "first_rev": 180}]

💡 Hint

Remember that FIRST_VALUE returns the first value in the window frame defined by the ORDER BY and ROWS clause.

The window frame is from the start (UNBOUNDED PRECEDING) to the current row, so FIRST_VALUE always returns the revenue of the first month in each region partition.

❓ query_result

intermediate

Output of LAST_VALUE with default frame

Consider the table orders with columns customer, order_date, and amount. What is the output of this query?

SELECT customer, order_date, amount,
       LAST_VALUE(amount) OVER (PARTITION BY customer ORDER BY order_date) AS last_amt
FROM orders
ORDER BY customer, order_date;

PostgreSQL

CREATE TABLE orders (customer TEXT, order_date DATE, amount INT);
INSERT INTO orders VALUES
('Alice', '2024-01-01', 50),
('Alice', '2024-01-05', 70),
('Bob', '2024-01-02', 30),
('Bob', '2024-01-03', 40);

SELECT customer, order_date, amount,
       LAST_VALUE(amount) OVER (PARTITION BY customer ORDER BY order_date) AS last_amt
FROM orders
ORDER BY customer, order_date;

A [{"customer": "Alice", "order_date": "2024-01-01", "amount": 50, "last_amt": 50}, {"customer": "Alice", "order_date": "2024-01-05", "amount": 70, "last_amt": 50}, {"customer": "Bob", "order_date": "2024-01-02", "amount": 30, "last_amt": 30}, {"customer": "Bob", "order_date": "2024-01-03", "amount": 40, "last_amt": 30}]

B [{"customer": "Alice", "order_date": "2024-01-01", "amount": 50, "last_amt": 70}, {"customer": "Alice", "order_date": "2024-01-05", "amount": 70, "last_amt": 70}, {"customer": "Bob", "order_date": "2024-01-02", "amount": 30, "last_amt": 40}, {"customer": "Bob", "order_date": "2024-01-03", "amount": 40, "last_amt": 40}]

C [{"customer": "Alice", "order_date": "2024-01-01", "amount": 50, "last_amt": null}, {"customer": "Alice", "order_date": "2024-01-05", "amount": 70, "last_amt": null}, {"customer": "Bob", "order_date": "2024-01-02", "amount": 30, "last_amt": null}, {"customer": "Bob", "order_date": "2024-01-03", "amount": 40, "last_amt": null}]

D [{"customer": "Alice", "order_date": "2024-01-01", "amount": 50, "last_amt": 50}, {"customer": "Alice", "order_date": "2024-01-05", "amount": 70, "last_amt": 70}, {"customer": "Bob", "order_date": "2024-01-02", "amount": 30, "last_amt": 30}, {"customer": "Bob", "order_date": "2024-01-03", "amount": 40, "last_amt": 40}]

💡 Hint

LAST_VALUE by default uses the window frame from the first row to the current row, so it returns the last value in that frame, which is the current row's value.

Without specifying ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING, LAST_VALUE returns the value of the current row, not the last in the partition.

📝 Syntax

advanced

Identify the syntax error in FIRST_VALUE usage

Which option contains a syntax error when using FIRST_VALUE in PostgreSQL?

A SELECT FIRST_VALUE(salary) OVER (PARTITION BY department ORDER BY salary ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) FROM employees;

B SELECT FIRST_VALUE(salary) FROM employees ORDER BY department;

C SELECT FIRST_VALUE(salary) OVER (PARTITION BY department ORDER BY salary) FROM employees;

D SELECT FIRST_VALUE(salary) OVER (ORDER BY department) FROM employees;

💡 Hint

Remember that FIRST_VALUE is a window function and requires an OVER clause.

Option B lacks the OVER clause, so it is a syntax error. FIRST_VALUE must be used with OVER to define the window.

❓ optimization

advanced

Optimizing LAST_VALUE for full partition

You want to get the last sale amount per customer over all their orders. Which query correctly uses LAST_VALUE to always return the last amount in the entire partition regardless of row position?

A SELECT customer, order_date, amount, LAST_VALUE(amount) OVER (PARTITION BY customer ORDER BY order_date ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING) AS last_amt FROM sales;

B SELECT customer, order_date, amount, LAST_VALUE(amount) OVER (PARTITION BY customer ORDER BY order_date) AS last_amt FROM sales;

C SELECT customer, order_date, amount, LAST_VALUE(amount) OVER (PARTITION BY customer ORDER BY order_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS last_amt FROM sales;

D SELECT customer, order_date, amount, LAST_VALUE(amount) OVER (PARTITION BY customer ORDER BY order_date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS last_amt FROM sales;

💡 Hint

To get the last value in the entire partition, the window frame must cover all rows.

Option C sets the frame from the first to the last row in the partition, so LAST_VALUE returns the last amount overall.

🧠 Conceptual

expert

Understanding window frame impact on FIRST_VALUE and LAST_VALUE

Which statement best explains why the output of FIRST_VALUE and LAST_VALUE can differ depending on the window frame specified?

A Because the window frame defines the subset of rows considered for each row, changing which values are first or last in that frame.

B Because FIRST_VALUE and LAST_VALUE always return the first and last values of the entire table regardless of window frame.

C Because FIRST_VALUE ignores the ORDER BY clause while LAST_VALUE uses it to find the last value.

D Because window frames only affect aggregate functions, not FIRST_VALUE or LAST_VALUE.

💡 Hint

Think about how the window frame limits the rows the function looks at for each row.

The window frame restricts the rows considered for each row, so FIRST_VALUE and LAST_VALUE return values based on that subset, not the entire partition.

Practice

1. What does the FIRST_VALUE function do in PostgreSQL?

easy

A. Returns the first value in a sorted group of rows.

B. Returns the last value in a sorted group of rows.

C. Returns the average value of a column.

D. Returns the total count of rows.

Solution

1. Step 1: Understand the purpose of FIRST_VALUE

   FIRST_VALUE returns the first value in a sorted window or group of rows.

2. Step 2: Compare with other options

   Options B, C, and D describe different functions or aggregates, not FIRST_VALUE.

3. Final Answer:

   Returns the first value in a sorted group of rows. -> Option A

4. Quick Check:

   FIRST_VALUE = first value in sorted rows [OK]

Hint: FIRST_VALUE always picks the first row's value in order [OK]

Common Mistakes:

• Confusing FIRST_VALUE with LAST_VALUE
• Thinking it returns an aggregate like sum or count
• Ignoring the sorting order in the window

2. Which of the following is the correct syntax to get the last value of a column score ordered by date using LAST_VALUE in PostgreSQL?

easy

A. SELECT LAST_VALUE(score) FROM results ORDER BY date;

B. SELECT LAST_VALUE(score) OVER (ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM results;

C. SELECT LAST_VALUE(score) OVER (PARTITION BY date) FROM results;

D. SELECT LAST_VALUE(score) OVER (ORDER BY date) FROM results;

Solution

1. Step 1: Understand LAST_VALUE default frame

   LAST_VALUE by default looks from current row to end, so it may not return the last value overall without framing.

2. Step 2: Use explicit window frame to get true last value

   Adding ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING ensures the entire partition is considered.

3. Final Answer:

   SELECT LAST_VALUE(score) OVER (ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM results; -> Option B

4. Quick Check:

   LAST_VALUE needs full frame to get last value [OK]

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

Common Mistakes:

• Omitting window frame causing wrong last value
• Using PARTITION BY incorrectly
• Using LAST_VALUE without OVER() clause

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

SELECT region, month, revenue, FIRST_VALUE(revenue) OVER (PARTITION BY region ORDER BY month) AS first_rev FROM sales ORDER BY region, month;

medium

A. The total revenue summed for each region.

B. The revenue of the last month for each region repeated on every row of that region.

C. The revenue of the first month for each region repeated on every row of that region.

D. The revenue of the current month only.

Solution

1. Step 1: Analyze FIRST_VALUE with PARTITION and ORDER BY

   FIRST_VALUE(revenue) OVER (PARTITION BY region ORDER BY month) returns the first revenue value per region ordered by month.

2. Step 2: Understand output per row

   For each row in a region, the first month's revenue is repeated as first_rev.

3. Final Answer:

   The revenue of the first month for each region repeated on every row of that region. -> Option C

4. Quick Check:

   FIRST_VALUE repeats first row's value per partition [OK]

Hint: FIRST_VALUE repeats first sorted value per group [OK]

Common Mistakes:

• Thinking it returns last value instead
• Assuming it sums or aggregates
• Ignoring PARTITION BY effect

4. You wrote this query but it returns unexpected results:

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

What is the likely cause?

medium

A. LAST_VALUE cannot be used without PARTITION BY clause.

B. ORDER BY employee_id is invalid in window functions.

C. salary column must be aggregated with SUM before LAST_VALUE.

D. LAST_VALUE needs a window frame to include all rows; default frame causes wrong output.

Solution

1. Step 1: Recall LAST_VALUE default frame behavior

   By default, LAST_VALUE looks from current row to end, so it may not return the last row's value overall.

2. Step 2: Identify missing window frame

   Without specifying ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING, the last value is not the last row's value.

3. Final Answer:

   LAST_VALUE needs a window frame to include all rows; default frame causes wrong output. -> Option D

4. Quick Check:

   LAST_VALUE requires full frame for correct last row [OK]

Hint: Add full frame to LAST_VALUE to fix wrong results [OK]

Common Mistakes:

• Assuming ORDER BY alone fixes LAST_VALUE
• Thinking PARTITION BY is mandatory
• Trying to aggregate salary before LAST_VALUE

5. You want to find the first and last sale amount for each customer ordered by sale date. Which query correctly returns both values per customer in PostgreSQL?

hard

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

B. SELECT customer_id, sale_date, sale_amount, FIRST_VALUE(sale_amount), LAST_VALUE(sale_amount) FROM sales GROUP BY customer_id;

C. SELECT customer_id, sale_date, sale_amount, FIRST_VALUE(sale_amount) OVER (ORDER BY sale_date), LAST_VALUE(sale_amount) OVER (ORDER BY sale_date) FROM sales;

D. SELECT customer_id, sale_date, sale_amount, MIN(sale_amount) OVER (PARTITION BY customer_id), MAX(sale_amount) OVER (PARTITION BY customer_id) FROM sales;

Solution

1. Step 1: Use FIRST_VALUE with PARTITION and ORDER BY

   FIRST_VALUE(sale_amount) OVER (PARTITION BY customer_id ORDER BY sale_date) gets first sale per customer.

2. Step 2: Use LAST_VALUE with full window frame

   LAST_VALUE needs ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING to get last sale per customer.

3. Step 3: Check other options

   SELECT customer_id, sale_date, sale_amount, FIRST_VALUE(sale_amount), LAST_VALUE(sale_amount) FROM sales GROUP BY customer_id; lacks OVER() and ORDER BY, C lacks PARTITION BY, D uses MIN/MAX which is different.

4. Final Answer:

   SELECT customer_id, sale_date, sale_amount, FIRST_VALUE(sale_amount) OVER (PARTITION BY customer_id ORDER BY sale_date), LAST_VALUE(sale_amount) OVER (PARTITION BY customer_id ORDER BY sale_date ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) FROM sales; -> Option A

5. Quick Check:

   Combine FIRST_VALUE and LAST_VALUE with proper framing [OK]

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

Common Mistakes:

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