LEAD function for next row access in SQL - Time & Space Complexity

Time Complexity: LEAD function for next row access

O(n)

Understanding Time Complexity

We want to understand how the time needed to run a query using the LEAD function changes as the data grows.

Specifically, how does accessing the next row's value affect performance?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

SELECT employee_id, salary,
       LEAD(salary) OVER (ORDER BY employee_id) AS next_salary
FROM employees;
    

This query lists each employee's salary and the salary of the next employee ordered by their ID.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The database scans each row once and looks ahead to the next row's salary.
• How many times: Once for every row in the employees table.

How Execution Grows With Input

As the number of employees grows, the database does a similar amount of work for each employee.

Input Size (n)	Approx. Operations
10	About 10 lookups
100	About 100 lookups
1000	About 1000 lookups

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

Final Time Complexity

Time Complexity: O(n)

This means the time to run the query grows in a straight line as the number of rows increases.

Common Mistake

[X] Wrong: "LEAD causes the query to do extra loops inside loops, making it much slower than scanning once."

[OK] Correct: The LEAD function just looks ahead one row during a single pass, so it does not multiply the work by the number of rows.

Interview Connect

Understanding how window functions like LEAD work helps you explain query performance clearly and shows you know how databases handle row-by-row operations efficiently.

Self-Check

"What if we used LEAD with a PARTITION BY department? How would the time complexity change?"

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