LEAD function for next row access in SQL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

Output of LEAD function with default offset

Given the table Sales with columns id and amount, what is the output of this query?

SELECT id, amount, LEAD(amount) OVER (ORDER BY id) AS next_amount FROM Sales ORDER BY id;

SQL

CREATE TABLE Sales (id INT, amount INT);
INSERT INTO Sales VALUES (1, 100), (2, 200), (3, 300);

A [{"id":1,"amount":100,"next_amount":200},{"id":2,"amount":200,"next_amount":300},{"id":3,"amount":300,"next_amount":null}]

B [{"id":1,"amount":100,"next_amount":100},{"id":2,"amount":200,"next_amount":200},{"id":3,"amount":300,"next_amount":300}]

C [{"id":1,"amount":100,"next_amount":null},{"id":2,"amount":200,"next_amount":null},{"id":3,"amount":300,"next_amount":null}]

D [{"id":1,"amount":100,"next_amount":300},{"id":2,"amount":200,"next_amount":null},{"id":3,"amount":300,"next_amount":null}]

💡 Hint

LEAD returns the value of the next row in the order specified.

The LEAD function with default offset 1 returns the amount from the next row ordered by id. For the last row, there is no next row, so it returns null.

❓ query_result

intermediate

LEAD function with offset 2

What is the output of this query on the same Sales table?

SELECT id, amount, LEAD(amount, 2) OVER (ORDER BY id) AS next_2_amount FROM Sales ORDER BY id;

SQL

CREATE TABLE Sales (id INT, amount INT);
INSERT INTO Sales VALUES (1, 100), (2, 200), (3, 300), (4, 400);

A [{"id":1,"amount":100,"next_2_amount":400},{"id":2,"amount":200,"next_2_amount":null},{"id":3,"amount":300,"next_2_amount":null},{"id":4,"amount":400,"next_2_amount":null}]

B [{"id":1,"amount":100,"next_2_amount":200},{"id":2,"amount":200,"next_2_amount":300},{"id":3,"amount":300,"next_2_amount":400},{"id":4,"amount":400,"next_2_amount":null}]

C [{"id":1,"amount":100,"next_2_amount":null},{"id":2,"amount":200,"next_2_amount":null},{"id":3,"amount":300,"next_2_amount":null},{"id":4,"amount":400,"next_2_amount":null}]

D [{"id":1,"amount":100,"next_2_amount":300},{"id":2,"amount":200,"next_2_amount":400},{"id":3,"amount":300,"next_2_amount":null},{"id":4,"amount":400,"next_2_amount":null}]

💡 Hint

Offset 2 means skipping one row and looking two rows ahead.

LEAD(amount, 2) returns the amount value two rows after the current row ordered by id. Rows without a second next row return null.

📝 Syntax

advanced

Identify the syntax error in LEAD usage

Which option contains a syntax error when using the LEAD function?

A SELECT id, LEAD(amount, 1, 0) OVER (ORDER BY id) FROM Sales;

B SELECT id, LEAD(amount, 2) OVER (ORDER BY id) FROM Sales;

C SELECT id, LEAD(amount ORDER BY id) FROM Sales;

D SELECT id, LEAD(amount) OVER (ORDER BY id ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) FROM Sales;

💡 Hint

Check the placement of parentheses and commas in LEAD function.

Option C is missing a comma between the first argument and the ORDER BY clause inside LEAD, which is invalid syntax. LEAD requires the ORDER BY clause in the OVER() clause, not inside the function arguments.

❓ optimization

advanced

Optimizing LEAD usage with partitioning

Given a Sales table with columns region, id, and amount, which query correctly uses LEAD to get the next amount within each region?

A SELECT region, id, amount, LEAD(amount) OVER (PARTITION BY region ORDER BY id) AS next_amount FROM Sales;

B SELECT region, id, amount, LEAD(amount) OVER (ORDER BY id) AS next_amount FROM Sales;

C SELECT region, id, amount, LEAD(amount, 1) OVER (PARTITION BY id ORDER BY region) AS next_amount FROM Sales;

D SELECT region, id, amount, LEAD(amount) FROM Sales GROUP BY region, id, amount;

💡 Hint

Partitioning groups rows before applying LEAD.

Option A partitions rows by region, so LEAD returns the next amount within the same region ordered by id. Option A ignores regions, C partitions incorrectly, and D misuses aggregation.

🧠 Conceptual

expert

Understanding LEAD with default value

What will be the output of the next_amount column for the last row in this query?

SELECT id, amount, LEAD(amount, 1, 999) OVER (ORDER BY id) AS next_amount FROM Sales ORDER BY id;

Assuming the last row has id = 4 and amount = 400.

A NULL

B 999

C 400

D 0

💡 Hint

The third argument in LEAD is the default value if no next row exists.

LEAD(amount, 1, 999) returns the amount from the next row ordered by id. For the last row, since no next row exists, it returns the default value 999 instead of NULL.

Practice

1. What does the SQL LEAD function do in a query?

easy

A. Returns the value from the previous row in the result set

B. Returns the value from the next row in the result set

C. Calculates the sum of values in a column

D. Filters rows based on a condition

Solution

1. Step 1: Understand LEAD function purpose

   The LEAD function is used to access data from the next row relative to the current row without changing the current row position.

2. Step 2: Compare options with LEAD behavior

   Returns the value from the next row in the result set correctly describes this behavior. Options B, C, and D describe other functions or operations.

3. Final Answer:

   Returns the value from the next row in the result set -> Option B

4. Quick Check:

   LEAD = next row value [OK]

Hint: LEAD looks forward one row by default [OK]

Common Mistakes:

• Confusing LEAD with LAG (previous row)
• Thinking LEAD filters rows
• Assuming LEAD sums values

2. Which of the following is the correct syntax to use the LEAD function to get the next row's salary ordered by employee_id?

easy

A. LEAD(salary) OVER (ORDER BY employee_id)

B. LEAD(salary ORDER BY employee_id)

C. LEAD(salary) ORDER BY employee_id

D. LEAD(salary) BY employee_id

Solution

1. Step 1: Recall LEAD function syntax

   The LEAD function requires an OVER clause with an ORDER BY to define the row order for accessing the next row.

2. Step 2: Match syntax with options

   LEAD(salary) OVER (ORDER BY employee_id) correctly uses LEAD(salary) OVER (ORDER BY employee_id). Options A, B, and D have incorrect syntax.

3. Final Answer:

   LEAD(salary) OVER (ORDER BY employee_id) -> Option A

4. Quick Check:

   LEAD + OVER + ORDER BY = correct syntax [OK]

Hint: LEAD needs OVER(ORDER BY ...) syntax [OK]

Common Mistakes:

• Omitting OVER clause
• Placing ORDER BY inside LEAD parentheses
• Using BY instead of ORDER BY

3. Given the table Employees with columns id and salary sorted by id, what will be the output of this query?

SELECT id, salary, LEAD(salary) OVER (ORDER BY id) AS next_salary FROM Employees;

Assuming data:
id | salary
1 | 5000
2 | 6000
3 | 5500

medium

A. [{id:1, salary:5000, next_salary:6000}, {id:2, salary:6000, next_salary:null}, {id:3, salary:5500, next_salary:5500}]

B. [{id:1, salary:5000, next_salary:5500}, {id:2, salary:6000, next_salary:5000}, {id:3, salary:5500, next_salary:6000}]

C. [{id:1, salary:5000, next_salary:null}, {id:2, salary:6000, next_salary:null}, {id:3, salary:5500, next_salary:null}]

D. [{id:1, salary:5000, next_salary:6000}, {id:2, salary:6000, next_salary:5500}, {id:3, salary:5500, next_salary:null}]

Solution

1. Step 1: Understand LEAD output for each row

   LEAD(salary) returns the salary from the next row ordered by id. For id=1, next_salary is salary of id=2 (6000). For id=2, next_salary is salary of id=3 (5500). For id=3, no next row exists, so next_salary is NULL.

2. Step 2: Match output with options

   [{id:1, salary:5000, next_salary:6000}, {id:2, salary:6000, next_salary:5500}, {id:3, salary:5500, next_salary:null}] matches this output exactly. Other options have incorrect next_salary values.

3. Final Answer:

   [{id:1, salary:5000, next_salary:6000}, {id:2, salary:6000, next_salary:5500}, {id:3, salary:5500, next_salary:null}] -> Option D

4. Quick Check:

   LEAD returns next row's salary or NULL [OK]

Hint: Next row salary follows order, last row gets NULL [OK]

Common Mistakes:

• Mixing up next and previous rows
• Expecting last row to have a value
• Ignoring ORDER BY in LEAD

4. Identify the error in this SQL query using LEAD:

SELECT id, salary, LEAD(salary, 2) ORDER BY id FROM Employees;

medium

A. salary column does not exist

B. LEAD cannot take a second argument

C. Missing OVER clause with ORDER BY inside LEAD function

D. ORDER BY cannot be used in SELECT clause

Solution

1. Step 1: Check LEAD function syntax

   LEAD requires an OVER clause with ORDER BY to specify row order. The query misses the OVER(...) part.

2. Step 2: Analyze the query error

   ORDER BY is incorrectly placed outside OVER clause and LEAD function lacks OVER clause, causing syntax error.

3. Final Answer:

   Missing OVER clause with ORDER BY inside LEAD function -> Option C

4. Quick Check:

   LEAD needs OVER(ORDER BY ...) [OK]

Hint: LEAD always needs OVER with ORDER BY [OK]

Common Mistakes:

• Omitting OVER clause
• Placing ORDER BY outside OVER
• Assuming LEAD can't take offset argument

5. You want to find each employee's salary and the salary two rows ahead ordered by hire_date. If no such row exists, show 0 instead of NULL. Which query correctly uses LEAD to achieve this?

hard

A. SELECT id, salary, LEAD(salary, 2, 0) OVER (ORDER BY hire_date) AS salary_after_two FROM Employees;

B. SELECT id, salary, LEAD(salary, 0, 2) OVER (ORDER BY hire_date) AS salary_after_two FROM Employees;

C. SELECT id, salary, LEAD(salary, 2) OVER (ORDER BY hire_date) AS salary_after_two FROM Employees WHERE salary_after_two IS NOT NULL;

D. SELECT id, salary, LEAD(salary, 2, NULL) OVER (ORDER BY hire_date) AS salary_after_two FROM Employees;

Solution

1. Step 1: Understand LEAD parameters

   LEAD(column, offset, default) lets you specify how many rows ahead to look and what to return if no row exists.

2. Step 2: Match query to requirements

   SELECT id, salary, LEAD(salary, 2, 0) OVER (ORDER BY hire_date) AS salary_after_two FROM Employees; uses offset 2 and default 0, ordered by hire_date, matching the requirement. SELECT id, salary, LEAD(salary, 0, 2) OVER (ORDER BY hire_date) AS salary_after_two FROM Employees; has wrong offset and default order. SELECT id, salary, LEAD(salary, 2) OVER (ORDER BY hire_date) AS salary_after_two FROM Employees WHERE salary_after_two IS NOT NULL; filters out NULLs instead of replacing. SELECT id, salary, LEAD(salary, 2, NULL) OVER (ORDER BY hire_date) AS salary_after_two FROM Employees; returns NULL instead of 0.

3. Final Answer:

   SELECT id, salary, LEAD(salary, 2, 0) OVER (ORDER BY hire_date) AS salary_after_two FROM Employees; -> Option A

4. Quick Check:

   LEAD with offset 2 and default 0 [OK]

Hint: Use LEAD(column, offset, default) with OVER(ORDER BY ...) [OK]

Common Mistakes:

• Swapping offset and default arguments
• Filtering NULL instead of setting default
• Omitting OVER clause