Advanced SQL Interview Questions for Experienced Professionals

Are you an experienced professional appearing for a senior-level SQL job interview? It will require a thorough understanding of advanced SQL queries, functions, and database design principles. Usually, scenario-based and technical SQL interview questions are asked to assess your proficiency and capability for problem-solving in a corporate environment. This blog will cover various SQL interview questions for experienced professionals, from basic concepts to complex scenarios. This will go a long way in establishing your confidence to excel in your next SQL interview and display your competence effectively.

SQL Interview Questions for Experienced Professionals [3 to 5 Years Experience]

Advanced SQL interview questions are designed to assess the depth of knowledge and expertise in handling complex database queries. These questions test your ability to optimize performance, work with advanced functions, and manage large datasets effectively. 

Here are some advanced SQL interview questions and answers for experienced professionals

Q1. What is a Common Table Expression (CTE), and how does it differ from a subquery?

Answer: A common table expression (CTE) is a transient result set established within a SQL query utilizing the WITH keyword. In contrast to subqueries, common table expressions (CTEs) enhance readability and may be referenced several times inside a query, simplifying intricate searches. Furthermore, CTEs provide recursive inquiries, which are challenging to implement with conventional subqueries.

Q2. What is the method to obtain the nth highest wage from an employee table?

Answer: To obtain the nth highest salary, use the DENSE_RANK() function. This function assigns ranks to salaries in descending order, allowing you to filter for the desired rank efficiently. Here’s an SQL code to obtain the nth highest wage from an employee table:

SELECT salary FROM (
SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rank
FROM employees
) AS ranked_salaries
WHERE rank = n;

Replace n with the desired rank.

Q3. Explain the distinction between INNER JOIN and LEFT JOIN.

Answer: INNER JOIN retrieves solely the rows that exhibit a match in both associated tables. Conversely, a LEFT JOIN retrieves all data from the left table and the corresponding rows from the right table. Without a match, NULL values are returned for columns from the right table.

Q4. Can you explain indexing and the various SQL index types to me?

Answer: Indexing is a technique used to enhance database performance by speeding up data retrieval processes. There are several types of indexes, including:

• Clustered Index: Sorts and stores rows based on key values.
• Non-clustered Index: Maintains a separate structure that references data rows.
• Unique Index: Ensures that all values in an indexed column are distinct.

Q5. Give an example of a window function in SQL.

Window functions perform calculations across a set of rows related to the current row without collapsing them into a single output row. For example:

SELECT employee_id, salary,
AVG(salary) OVER (PARTITION BY department) AS department_avg_salary
FROM employees;

This will give the result for the average salary in each of the departments.

Q6. Can you explain recursive CTEs and tell me when they might be useful?

Answer: Recursive CTEs use self-referencing queries to return data in a hierarchical framework. This makes them invaluable when dealing with data that resembles organizational systems or tree-like hierarchies. For instance: 

WITH RECURSIVE employee_hierarchy AS (
SELECT employee_id, manager_id FROM employees WHERE manager_id IS NULL
UNION ALL
SELECT e.employee_id, e.manager_id
FROM employees e
JOIN employee_hierarchy eh ON e.manager_id = eh.employee_id)
SELECT * FROM employee_hierarchy;

Q7. When running SQL queries, how can you find the slow spots and fix them?

Answer: To identify slow spots in SQL queries, you can analyze execution plans using either EXPLAIN or EXPLAIN ANALYSE. Check for problems like unnecessary joins, incomplete indexing, or full table scans. Then, you can optimize using CTEs or temporary tables, rewriting subqueries, and adding indexes.

Q8. What do the SQL commands DELETE, TRUNCATE, and DROP do?

Answer: These SQL commands can be used to remove rows from a table. Here’s an overview of each command: 

• DELETE: It removes specific rows based on conditions defined in a WHERE clause.
• TRUNCATE: This command deletes all rows from a table without logging individual row deletions. It cannot be used with a WHERE clause.
• DROP: The DROp command will completely remove a table structure along with all associated data and metadata.

Q9. How would you implement error handling in a stored procedure?

Answer: Error handling in SQL stored procedures can be done using TRY…CATCH blocks. This structure allows you to execute SQL statements while catching any exceptions that occur, enabling efficient error management and reporting. Here’s a code to implement error handling in a stored procedure:

BEGIN TRY
-- SQL statements
END TRY
BEGIN CATCH
SELECT ERROR_MESSAGE() AS ErrorMessage;
END CATCH;

Q10. Explain what ‘database normalization’ is and why it’s crucial.

Answer: Data normalization is the act of arranging data in a manner that eliminates duplication and improves data integrity. It entails breaking down big tables into smaller, related tables. Data normalization is crucial for minimizing data abnormalities and ensuring efficient queries and updates.

Pro Tip: Acing the SQL interview questions for experienced roles requires practicing SQL queries in various scenarios. To explore more interview questions and strengthen your interview preparation, check out our guide on SQL coding interview questions and answers. 

Q11. Give an example of how to use SQL’s PARTITION BY clause.

You can use PARTITION BY with window functions to create subsets of a result set that the window function can work on instead of the whole thing. Here’s a code to illustrate the PARTITION clause:

SELECT employee_id, salary, RANK() OVER (PARTITION BY department ORDER BY salary DESC) AS dept_rank
FROM employees;

Here, the rank is calculated for each department.

Q12. Why would you use a stored procedure in SQL?

Answer: Stored procedures have many advantages, including better performance (because of precompiled execution plans), more security (since SQL code may be abstracted), and more modularity. Also, they can optimize themselves apart from the application code and support complicated transactions.

Q13. Can you explain the GROUP BY clause and what it does?

Answer: The GROUP BY clause groups rows that share common values in specified columns so that aggregate functions like SUM or COUNT can be applied to each group. This functionality is essential for summarizing data effectively within SQL queries.

Q14. Please define a self-join.

Answer: A self-join in SQL is a type of join where a table is joined with itself. It is used to compare rows within the same table, often by using aliases to differentiate between the two instances of the table. This join is typically helpful when you need to find relationships within the same set of data, such as comparing employees to their managers in an employee table.

Q15. For each month, how can you determine which three products brought in the most money from a sales table?

Answer: To rank products based on the total revenue generated each month, you can use ROW_NUMBER() along with PARTITION BY to create monthly subsets of sales data. This approach will allow you to identify the top three products per month efficiently by ordering results according to revenue generated.

Advanced SQL Queries Interview Questions for Experienced Professionals [5 to 10 Years of Experience]

Advanced SQL query interview questions evaluate your ability to write complex queries and optimize database performance. These questions focus on your expertise in handling large datasets, advanced functions, and solving intricate problems using SQL. 

Here are some advanced SQL query interview questions for 5-year experienced candidates:

Q16. Formulate a query to identify duplicate email addresses within a customer table.

Answer: Identifying duplicate emails requires employing the GROUP BY clause on the email column and tallying the instances. Using HAVING COUNT(*) > 1 allows us to identify emails that occur multiple times. Here’s an example:

SELECT email, COUNT(*) AS duplicate_count
FROM customers
GROUP BY email
HAVING COUNT(*) > 1;

Q17. What is the method to get the cumulative sales total for each product by date?

Answer: To obtain a cumulative sum for each product, we employ the SUM() function as a window function, utilizing PARTITION BY product and ORDER BY sale date. This computes a cumulative aggregate of sales in chronological order. Here’s an example:

SELECT product_id, sale_date, sale_amount,
SUM(sale_amount) OVER (PARTITION BY product_id ORDER BY sale_date) AS cumulative_sales
FROM sales;

Q18. What is the method to enumerate departments with a workforce exceeding 10 employees?

Answer: Use GROUP BY on the department, we enumerate employees within each department and implement HAVING COUNT(*) > 10 to filter departments with over 10 employees. The code for this function is: 

SELECT department_id, COUNT(*) AS employee_count
FROM employees
GROUP BY department_id
HAVING COUNT(*) > 10;

Q19. Formulate a query to identify the penultimate order for each customer.

Answer: Utilizing ROW_NUMBER(), we may rank each customer’s orders by order date in descending order. Filtering for rank 2 yields the second most recent order. Here’s an example:

SELECT customer_id, order_id, order_date
FROM (
SELECT customer_id, order_id, order_date,
ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY order_date DESC) AS rank
FROM orders
) AS ranked_orders
WHERE rank = 2;

Pro Tip: Explore our advanced SQL course to practice intricate queries and enhance your comprehension of SQL operations. This course will help you acquire practical experience in database design and performance optimization, and most importantly, prepare you to answer advanced SQL interview questions for experienced professionals.

Q20. What method would you employ to identify all products that have never been sold?

Answer: To identify products lacking corresponding sales records, employ a LEFT JOIN from the products table to the sales table and filter for NULL values in the sales data. Here’s an example:

SELECT p.product_id, p.product_name
FROM products p
LEFT JOIN sales s ON p.product_id = s.product_id
WHERE s.product_id IS NULL;

Q21. Formulate a query to extract clients who have made purchases in successive months.

Answer: To identify clients with purchases in consecutive months, you can use the LAG() function to compare each order date with the previous order date for each customer. This method allows you to ascertain whether purchases occurred in successive months. Here’s an example:

SELECT customer_id, order_date
FROM (
SELECT customer_id, order_date,
LAG(order_date) OVER (PARTITION BY customer_id ORDER BY order_date) AS previous_order_date
FROM orders
) AS consecutive_orders
WHERE DATEDIFF(month, previous_order_date, order_date) = 1

Q22. How can one ascertain rows in which a column value has altered in comparison to the preceding row?

Answer: By utilizing the LAG() function, you can retrieve the value from the preceding row and compare it with the current row’s value. This approach enables you to identify any changes within a specified column across your dataset. Here’s an example:

SELECT id, column_name
FROM (
SELECT id, column_name,
LAG(column_name) OVER (ORDER BY id) AS previous_value
FROM table_name
) AS value_changes
WHERE column_name <> previous_value;

Q23. Formulate a query to retrieve the top N rows based on salary for each department.

Answer: To obtain the top N earners within each department, employ the ROW_NUMBER() function along with PARTITION BY to rank employees based on their salaries. You can then filter the results to include only those with ranks less than or equal to N. Here’s an example:

SELECT department_id, employee_id, salary FROM (
SELECT department_id, employee_id, salary,
ROW_NUMBER() OVER (PARTITION BY department_id ORDER BY salary DESC) AS rank
FROM employees
) AS ranked_employees
WHERE rank <=N;

Q24. What is the method for calculating the percentage of total sales for each product?

Answer: To determine the proportion of each product’s sales, calculate the overall sales, divide each product’s sales by this total, and then multiply by 100. Here is the code for this:

SELECT product_id, sale_amount,
(sale_amount / SUM(sale_amount) OVER ()) * 100 AS percentage_of_total
FROM sales;

Q25. Formulate a query to obtain the employees whose earnings exceed those of their manager.

Answer: By executing a self-join on the employees table, you can compare each employee’s salary with that of their manager. This method allows you to identify employees who earn more than their supervisors. Here’s an example:

SELECT e.employee_id, e.salary, m.employee_id AS manager_id, m.salary AS manager_salary
FROM employees e
JOIN employees m ON e.manager_id = m.employee_id
WHERE e.salary > m.salary;

Q26. What is the method for calculating the moving average of sales over three days?

Answer: To compute the moving sales average over three days, use the AVG() function as a window function. Define your window as ‘ROWS BETWEEN 2 PRECEDING AND CURRENT ROW’ to include sales from the current day and the two previous days in your average calculation. Here’s an example:

SELECT sale_date, sale_amount,
AVG(sale_amount) OVER (ORDER BY sale_date ROWS BETWEEN 2 PRECEDING AND CURRENT ROW)
AS moving_avg
FROM sales;

Q27. Formulate a question to ascertain the earliest and latest order dates for each customer.

Answer: Utilize the MIN() and MAX() functions, organized by customer ID, to retrieve the first and final order dates. The code used for this is:

SELECT customer_id, MIN(order_date) AS first_order, MAX(order_date) AS last_order
FROM orders
GROUP BY customer_id;

Q28. What method would you apply to identify employees with the highest salaries within each department?

Answer: To identify the person with the highest salary in each department, use the RANK() function to rank wages in descending order within each department and apply a filter for rank = 1. Here’s an example:

SELECT department_id, employee_id, salary FROM (
SELECT department_id, employee_id, salary,
RANK() OVER (PARTITION BY department_id ORDER BY salary DESC) AS rank
FROM employees
) AS ranked_salaries
WHERE rank = 1;

Q29. What methods can be employed to find products that rank in the top 5% of sales among all products?

Answer: Use the PERCENT_RANK() function to allocate a percentile rank to each product’s sales and filter for ranks that are more than or equal to 0.95, indicating the top 5%. Here’s an example:

SELECT product_id, total_sales
FROM (
SELECT product_id, total_sales,
PERCENT_RANK() OVER (ORDER BY total_sales DESC) AS sales_percentile
FROM product_sales
) AS ranked_products
WHERE sales_percentile >= 0.95;

SQL Scenario-Based Interview Questions for Experienced Professionals 

Scenario-based SQL interview questions assess your practical problem-solving skills in real-world database situations. These questions test your ability to write efficient queries, optimize performance, and handle complex data relationships within a database environment. 

Here are some SQL scenario-based interview questions for experienced professionals:

Q30. How would you find out how many employees are on the clock on a specific date if you had a table with their start and finish dates included?

Answer: To find out how many employees are currently on the clock on a specific date, you can query a table that contains their start and finish dates. The goal is to count employees who started work on or before the specified date and either have no end date or an end date that is after the specified date. To achieve this, you can use the following SQL query: 

SELECT COUNT(*) AS active_employees
FROM employees
WHERE start_date <= 'target_date' AND (end_date IS NULL OR end_date > 'target_date');

In this query, replace ‘target_date’ with the actual date you want to check. This will give you the total number of active employees on that specific date.

Q31. Is there a way to get the running balance for each client only by looking at their transactions?

Answer: Yes, you can achieve this by sorting transaction records by date and using the ‘SUM()’ function along with window functions to calculate a running balance for each client. This method allows you to see how balances change over time based on transaction history. The following is the code used for the function:

SELECT customer_id, transaction_date, amount,
SUM(amount) OVER (PARTITION BY customer_id ORDER BY transaction_date) AS running_balance FROM transactions;

Q32. Within a table that lists products and their daily sales, how can one locate those that have had no purchases in the past 30 days?

Answer: Use an LEFT JOIN to join the sales table, and then filter for NULL values during the last 30 days. The code given below is used to locate the function: 

SELECT p.product_id
FROM products p
LEFT JOIN sales s ON p.product_id = s.product_id AND s.sale_date >= DATE_SUB(CURDATE(), INTERVAL 30 DAY)
WHERE productids IS NULL;

Q33. What method would you employ to identify clients who made a minimum of one purchase per month during the past year?

Answer: You can use GROUP BY with COUNT(DISTINCT month) to find clients who have conducted transactions every month over the past year. The code below is used for this method is as follows:

SELECT customer_id
FROM transactions
WHERE transaction_date >= DATE_SUB(CURDATE(), INTERVAL 1 YEAR)
GROUP BY customer_id
HAVING COUNT(DISTINCT MONTH(transaction_date)) = 12;

Q34. How can one identify months in which overall sales exceeded those of the corresponding month in the prior year?

Answer: By using the LAG() function, you can compare the current month’s sales against those from the same month in the previous year. The code is as follows:

SELECT month, year, total_sales
FROM (
SELECT month, year, total_sales,
LAG(total_sales) OVER (PARTITION BY month ORDER BY year) AS last_year_sales
FROM monthly_sales
) AS sales_comparison
WHERE total_sales > last_year_sales;

Q35. What is the method to obtain records that are similar across all columns except for one particular column?

Answer: Utilise GROUP BY on all columns except the variable one and apply HAVING COUNT(*) > 1 to identify near-duplicates. The code to obtain the records is as follows:

SELECT column1, column2, ..., columnN
FROM table_name
GROUP BY column1, column2, ..., columnN
HAVING COUNT(*) > 1;

Q36. What is the method to determine the second-highest compensation for each department within an employee table?

Answer: You can achieve this by using ‘DENSE_RANK()’ with ‘PARTITION BY department’ and ‘ORDER BY salary’ in descending order, then filtering for ranks equal to 2. Here’s a code:

SELECT department_id, employee_id, salary
FROM (
SELECT department_id, employee_id, salary,
DENSE_RANK() OVER (PARTITION BY department_id ORDER BY salary DESC) AS rank
FROM employees
) AS ranked_salaries
WHERE rank = 2;

Q37. What is the method to identify all consumers who have never completed a transaction?

Answer: Use an LEFT JOIN between the customers and order tables. Thereafter, filter for NULL values in the orders table. The code for the method is as follows:

SELECT c.customer_id, c.customer_name
FROM customers c
LEFT JOIN orders o ON c.customer_id = o.customer_id
WHERE o.customer_id IS NULL;

Q38. How would you compute the month-over-month growth in sales using a table containing product sales data?

Answer: Use LAG() to obtain the sales from the preceding month, followed by the computation of the growth rate. The code for computing this is:

SELECT month, year, total_sales,
(total_sales - LAG(total_sales) OVER (ORDER BY year, month))/LAG(total_sales) OVER (ORDER BY year, month) * 100 AS month_over_month_growth
FROM monthly_sales;

Q39. What is the method to identify employees who have been supervised by multiple managers?

Answer: Utilise GROUP BY on employee ID and apply HAVING COUNT(DISTINCT manager_id) > 1. This query reveals employees who report to more than one manager throughout their tenure. The code below can be used for this function:

SELECT employee_id
FROM employee_manager
GROUP BY employee_id
HAVING COUNT(DISTINCT manager_id) > 1;

Q40. How can one discover orders where the order quantity exceeds the typical order amount for a specific customer?

Answer: Utilize a WINDOW function to compute the average order amount for each customer, thereafter filtering for orders that exceed this average. Here’s a code:

SELECT order_id, customer_id, order_amount
FROM (
SELECT order_id, customer_id, order_amount,
AVG(order_amount) OVER (PARTITION BY customer_id) AS avg_order_amount
FROM orders
) AS customer_orders
WHERE order_amount > avg_order_amount;

Q41. What is the method to obtain the fifth highest wage from an employee table?

Answer: To obtain the fifth highest salary, we can utilize the DENSE_RANK() function, which allocates a unique rank to each different salary in descending order. By enclosing this in a subquery, we filter for a rank of five.

SELECT salary FROM (
SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rank
FROM employees
) AS ranked_salaries
WHERE rank = 5;

This query will provide the fifth-highest distinct salary, regardless of duplication.

Q42. What method would you utilize to identify all employees whose remuneration exceeds the average compensation within their respective departments?

Answer: To identify employees earning above the departmental average, utilize the AVG() method as a window function. This enables the calculation of the departmental average for each employee without the need for grouping, facilitating straightforward filtering. Here’s a code:

SELECT employee_id, department_id, salary
FROM (
SELECT employee_id, department_id, salary,
AVG(salary) OVER (PARTITION BY department_id) AS avg_salary
FROM employees
) AS dept_salaries
WHERE salary> avg_salary;

Conclusion 

Preparing for advanced SQL interview questions for experienced professionals requires a thorough understanding of SQL concepts and functions. Also, understand more advanced concepts like queries, and database optimization strategies. By understanding these key concepts and practicing scenario-based SQL interview questions, you can effectively demonstrate your experience. This will help showcase your problem-solving skills and improve your chances of success in the interview.

If you are looking to interview at Amazon and want to prepare thoroughly, check out our guide on Amazon SQL interview questions and answers. 

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