Amazon SQL Interview Questions

In an Amazon SQL interview, candidates can expect a range of questions that test their SQL knowledge and problem-solving skills. These questions often involve topics such as data manipulation, query optimization, and database design. To succeed, candidates should be well-versed in SQL fundamentals and prepared to solve complex scenarios with confidence.

In this article, We will see some most asked SQL questions in the Amazon interview questions along with detailed explanations to help candidates prepare effectively.

Top 15 Amazon SQL Interview Questions

Let’s Learn some important and most-asked Amazon SQL interview problems with the examples and output shown below:

1. SQL Query to Calculate Average Review Ratings

Suppose you need to analyze customer feedback for products over time and you have a below table Reviews containing review_id, product_id, user_id, rating and review_date. Your task is to calculate the average review ratings for each product every month.

review_id product_id user_id rating review_date
1 101 201 4 2024-01-10
2 102 202 5 2024-01-15
3 101 203 3 2024-02-10
4 103 201 4 2024-02-20
5 101 204 5 2024-02-25
SELECT
product_id,
EXTRACT(YEAR FROM review_date) AS year,
EXTRACT(MONTH FROM review_date) AS month,
AVG(rating) AS avg_rating
FROM
Reviews
GROUP BY
product_id,
EXTRACT(YEAR FROM review_date),
EXTRACT(MONTH FROM review_date)
ORDER BY
product_id,
year,
month;

Output:

Explanation:

In the above output, It shows the average review ratings for each product in each month of the year 2024. For example, product 101 received an average rating of 4 in both January and February 2024 while product 102 received a rating of 5 in January 2024.

This analysis provides insights into how customer’s perceptions of products change over time and helps in product development and marketing strategies.

2. What are the Strategies for Optimizing Slow SQL Queries?

Optimizing a slow SQL query in a large Amazon database involves several strategies tailored to handle the scale and complexity of such databases:

  1. Indexing: Identify and create indexes on columns used in the query’s WHERE, JOIN, and ORDER BY clauses to speed up data retrieval.
  2. Query Optimization: Rewrite the query to use efficient join methods, reduce the number of calculations, and minimize the amount of data being processed.
  3. Partitioning: If applicable, partition large tables based on a key (e.g., date range) to distribute data and queries across multiple servers, reducing query times.
  4. Caching: Implement caching mechanisms such as Amazon ElastiCache to store frequently accessed data and reduce the load on the database.
  5. Database Sharding: If your database supports it, consider sharding to horizontally partition data across multiple servers, reducing the data size each server needs to handle.
  6. Schema Optimization: Redesign the database schema to eliminate unnecessary joins, normalize or denormalize tables, and optimize data types and sizes.

Implementing these strategies requires careful planning and testing to ensure they effectively improve the performance of your SQL queries in a large Amazon database.

3. What are the Different Types of Joins?

Below are the different types of SQL Joins are defined below:

  1. INNER JOIN: Returns records that have matching values in both tables.
  2. LEFT JOIN (or LEFT OUTER JOIN): Returns all records from the left table and the matched records from the right table. If there is no match, the result is NULL on the right side.
  3. RIGHT JOIN (or RIGHT OUTER JOIN): Returns all records from the right table and the matched records from the left table. If there is no match, the result is NULL on the left side.
  4. FULL JOIN (or FULL OUTER JOIN): Returns records when there is a match in either the left or right table. It combines the result of both LEFT JOIN and RIGHT JOIN.
  5. CROSS JOIN: Returns the Cartesian product of the two tables i.e. all possible combinations of rows.
  6. SELF JOIN: Joins a table to itself typically used to compare rows within the same table.

These joins help in combining data from multiple tables based on a related column between them.

4. Determining Top-Selling Products Based on Analyzing Sales Data for Profitable Insights?

Consider a scenario where you are analyzing sales data for a retail company and you are task to identify the top-selling products to understand their performance. You have a table named orders containing order_id, product_id and order_amount. By calculating the total revenue generated by each product and sorting the products based on their revenue your goal is to determine the most profitable products in your inventory.

To understand we will consider the below Orders table:

order_id product_id order_amount
1 101 50.00
2 102 75.00
3 101 100.00
4 103 120.00
5 101 80.00
6 102 90.00
7 101 110.00
8 103 130.00
9 102 85.00
10 101 95.00
11 102 70.00
12 103 140.00

Query:

SELECT
product_id,
SUM(order_amount) AS total_revenue
FROM
orders
GROUP BY
product_id
ORDER BY
total_revenue DESC;

Output:

Explanation:

This query calculates the total revenue generated by each product by summing up the `order_amount` for each `product_id` in the `orders` table. It then groups the results by `product_id` and orders them in descending order of total revenue. The output shows the `product_id` and the corresponding total revenue for each product, helping to identify the top-grossing products.

5. Write a Query to Identify High-Spending Customers on Analyzing Purchases Exceeding $100.

Consider a retail business scenario where you need to identify customers who have made purchases exceeding $100. You have two tables called customers (containing customer_id and name) and orders (containing order_id, customer_id, and order_amount). The task is to join these tables and calculate the total purchase amount for each customer and select customers whose total purchase amount exceeds $100.

To understand we will consider the below customers and orders table:

Customers Table:

customer_id name
1 Alice
2 Bob
3 Charlie
4 David

Orders Table:

order_id customer_id order_amount
1 1 50.00
2 2 75.00
3 3 100.00
4 1 120.00
5 2 80.00
6 4 150.00

Query:

SELECT c.customer_id, c.name
FROM customers c
JOIN orders o ON c.customer_id = o.customer_id
GROUP BY c.customer_id, c.name
HAVING SUM(o.order_amount) > 100;

Output:

Explanation:

This query retrieves the customer_id and name of customers who have made purchases exceeding $100. It joins the customers and orders tables on customer_id, groups the results by customer_id and name and calculates the total purchase amount for each customer using SUM(o.order_amount). The HAVING clause filters out groups where the total purchase amount is not greater than $100.

6. How to use SQL to generate a histogram showing the count of comments made by each user?

Consider a social media platform where your task is to analyze user engagement by counting the number of comments made by each user. The below usersusers table while the below comments table. You need to join these tables then calculate the count of comments for each user, and present the results in descending order of comment count to identify the most active users.

To understand we will consider the below Users and Comments table:

Users Table:

user_id name
1 Alice
2 Bob
3 Charlie
4 David

Comments Table:

comment_id user_id comment_text
1 1 Great product!
2 2 I really like this.
3 1 Could be better.
4 3 Not satisfied.
5 1 Excellent service.
6 2 Could use some improvements.
7 4 Fast delivery.
8 3 Highly recommend.
9 1 Will buy again.
10 2 Good value for money.

Query:

SELECT u.user_id, u.name, COUNT(c.comment_id) AS comment_count
FROM users u
LEFT JOIN comments c ON u.user_id = c.user_id
GROUP BY u.user_id, u.name
ORDER BY comment_count DESC;

Output:

Output Explanation: This query calculates the count of comments made by each user by joining the users and `comments tables on user_id. It then groups the results by user_id and name showing the comment count for each user and ordering the results in descending order of comment count to highlight the most active users.

7. How to Construct an SQL Query to Retrieve a List of Distinct Pairs of Locations?

In a travel planning scenario, your task is to generate a list of all possible unique location pairs for creating travel routes. The below locations table contains the names of various destinations. You need to join the table with itself to pair each location with every other location, ensuring that each pair is listed only once and avoiding duplicates.

location_name
New York
Los Angeles
Chicago
Houston
Phoenix

Query:

SELECT DISTINCT
l1.location_name AS location1,
l2.location_name AS location2
FROM
locations l1
JOIN locations l2 ON l1.location_name < l2.location_name;

Output:

Explanation:

This query generates all unique pairs of locations by joining the locations table with itself ensuring that each pair is listed only once. It uses the condition l1.location_name < l2.location_name to avoid duplicate pairs and provide a comprehensive list of unique location combinations.

8. How would you formulate an SQL query to determine the daily Aggregate count of new users?

Consider a membership-based platform, your task is to track daily user registrations. The below `users` table contains user_id and registration_date. Your goal is to create a report showing the daily count of new users (new_users) and the cumulative count of users (cumulative_count) over time which is ordered by registration date.

user_id registration_date
1 2024-01-01
2 2024-01-01
3 2024-01-02
4 2024-01-02
5 2024-01-03
6 2024-01-03
7 2024-01-03
8 2024-01-04
9 2024-01-04
10 2024-01-04

Query:

SELECT
registration_date,
COUNT(user_id) AS new_users,
SUM(COUNT(user_id)) OVER (ORDER BY registration_date) AS cumulative_count
FROM
users
GROUP BY
registration_date
ORDER BY
registration_date;

Output:

Explanation:

This query calculates the daily count of new users (`new_users`) by grouping registrations by date. The `SUM()` window function is used to calculate the cumulative count of users (`cumulative_count`) over time, ordered by registration date, providing a daily overview of new user registrations and the total user base growth.

9. How to Construct an SQL Query to Analyze the everyday Dispersion of Conversations Initiated by Users in year 2020?

In a social networking platform, your task is to analyze user engagement by tracking the daily count of unique users initiating conversations in 2020. The below conversation table contains `user_id` and conversation_date and your goal is to generate a report showing the daily distribution of conversations initiated by users in the specified year.

conversation_id user_id conversation_date
1 1 2020-01-01
2 2 2020-01-01
3 1 2020-01-02
4 3 2020-01-02
5 1 2020-01-03
6 2 2020-01-03
7 4 2020-01-03
8 1 2020-01-04
9 3 2020-01-04
10 4 2020-01-04

Query:

SELECT
DATE_FORMAT(conversation_date, '%Y-%m-%d') AS date,
COUNT(DISTINCT user_id) AS conversations_initiated
FROM
conversations
WHERE
YEAR(conversation_date) = 2020
GROUP BY
date
ORDER BY
date;

Output:

Explanation:

This query calculates the daily count of unique users who initiated conversations in 2020. It formats the date (`conversation_date`) to show only the year, month, and day, then groups the results by date to show the daily distribution of conversations initiated by users in 2020.

10. SQL Query to Identify the Second-Highest Salary of Employees in the Engineering Department

In a company’s HR department, your task is to find the second-highest salary in the engineering department. The `employees` table while the `departments` table. Your goal is to generate a report showing the second-highest salary for the engineering department.

departments:

department_id department_name
1 Engineering
2 Sales
3 Marketing

employees:

employee_id department_id salary
1 1 60000.00
2 1 75000.00
3 1 80000.00
4 2 50000.00
5 2 55000.00
6 3 45000.00

Query:

SELECT
department_name,
MAX(salary) AS second_highest_salary
FROM
(
SELECT
d.department_name,
e.salary,
RANK() OVER (PARTITION BY d.department_name ORDER BY e.salary DESC) AS salary_rank
FROM
employees e
JOIN departments d ON e.department_id = d.department_id
WHERE
d.department_name = 'Engineering'
) ranked_salaries
WHERE
salary_rank = 2
GROUP BY
department_name;

Output:

Explanation:

This query uses a subquery to rank salaries within the engineering department (`department_name = ‘Engineering’`) in descending order. It then selects the maximum salary that has a rank of 2, which corresponds to the second-highest salary in the engineering department, and groups the result by department name.

11. SQL Query to Determine which Manager Supervises the Biggest team

Consider a company’s HR department and your task to identifying the manager who oversees the largest team. The below employee’s table. Your goal is to generate a report showing the manager ID and the size of the team they oversee.

employee_id manager_id department_id
1 3 1
2 3 1
3 NULL 1
4 3 1
5 4 2
6 4 2
7 4 2
8 4 2
9 5 2
10 5 2

Query:

SELECT
manager_id,
COUNT(employee_id) AS team_size
FROM
employees
GROUP BY
manager_id
ORDER BY
team_size DESC
LIMIT 1;

Output:

Explanation:

This query calculates the size of each team managed by a manager by counting the number of employees (`employee_id`) for each `manager_id`. It then orders the results by team size in descending order and limits the output to only the first row, effectively identifying the manager who oversees the largest team.

12. SQL Query to List Shipments that were Delivered within the Customer’s Membership Period

In a logistics and membership management system, you need to identify which shipments were delivered within the active membership period of each customer. You have a table shipments_customers that contains details of shipments and the membership periods of customers.

shipment_id customer_id delivery_date membership_start_date membership_end_date
1 1 2024-01-05 2024-01-01 2024-12-31
2 2 2024-02-10 2024-02-01 2024-02-28
3 1 2024-03-15 2024-01-01 2024-12-31
4 3 2024-04-20 2024-03-01 2024-05-31
5 2 2024-05-25 2024-02-01 2024-02-28
6 1 2024-06-30 2024-01-01 2024-12-31

Query:

SELECT
shipment_id,
customer_id,
delivery_date
FROM
shipments_customers
WHERE
delivery_date BETWEEN membership_start_date AND membership_end_date;

Output:

Explanation:

This query retrieves shipments where the delivery date falls within the customer’s membership period. The WHERE clause checks if delivery_date is between membership_start_date and membership_end_date.

13. Customers Who Purchased Both A and B Without Buying C

Given a list of customers and their purchases, find the customers who bought both products A and B but did not buy product C. To understand we will consider the below customers and purchases table:

customers table:

customer_id name
1 John Doe
2 Jane Smith
3 Alice Johnson
4 Bob Brown
5 Charlie Black

purchases table:

purchase_id customer_id product_id
1 1 A
2 1 B
3 2 A
4 2 C
5 3 B
6 3 A
7 4 A
8 4 B
9 5 C

Query:

SELECT c.customer_id, c.name
FROM customers c
WHERE c.customer_id IN (
SELECT customer_id
FROM purchases
WHERE product_id = 'A'
AND customer_id IN (SELECT customer_id FROM purchases WHERE product_id = 'B')
AND customer_id NOT IN (SELECT customer_id FROM purchases WHERE product_id = 'C')
);

Output:

Explanation:

This output indicates that John Doe, Alice Johnson, and Bob Brown purchased both products A and B but did not purchase product C.

14. Generating a Report of Product Names, Sale Years, and Prices per Sale ID from the Sales Table Using SQL

In a situation where a business needs to report on sales data, it is essential to extract key details such as the product name, the year the sale was made, and the price for each sale. This query helps in generating such a report from the below sales table.

sale_id product_name sale_date price
101 Laptop 2024-01-15 1200.00
102 Smartphone 2024-02-20 800.00
103 Tablet 2024-03-10 600.00
104 Laptop 2024-04-05 1100.00
105 Smartwatch 2024-05-25 300.00

Query:

SELECT
sale_id,
product_name,
EXTRACT(YEAR FROM sale_date) AS year,
price
FROM
sales;

Output:

Explanation:

This output shows the sale ID, product name, year of the sale, and the price for each sale in the sales table, which helps in generating a detailed sales report for the business.

15. Calculating the Total Quantity Sold for Each Product ID Using SQL

In a scenario where a business needs to analyze the total quantity sold for each product, this query calculates the total quantity sold for every unique product ID in the below sales table.

sale_id product_id quantity
1 101 5
2 102 3
3 101 2
4 103 4
5 102 6
6 101 3
7 104 5
8 102 2
9 103 3

Query:

SELECT
product_id,
SUM(quantity) AS total_quantity_sold
FROM
sales
GROUP BY
product_id;

Output:

Explanation:

This output shows the total quantity sold for each product ID in the sales table, which can be useful for inventory management and sales analysis.

Conclusion

Overall, preparing for an Amazon SQL interview requires a solid understanding of SQL fundamentals and the ability to apply them to real-world scenarios. By practicing with the questions provided in this article and understanding the underlying concepts, candidates can enhance their chances of success in the interview process. It’s essential to approach each question methodically, and demonstrate problem-solving skills. With thorough preparation and a strategic mindset candidates can confidently navigate Amazon SQL interviews and showcase their expertise in SQL.