At Lululemon Athletica, SQL is used to analyze sales trends across different product lines and to manage the inventory management system based on real-time customer demand. That's why Lululemon frequently asks SQL query questions in interviews for Data Science, Data Engineering and Data Analytics jobs.
So, to help you study, here's 10 Lululemon Athletica SQL interview questions – can you solve them?
The business team at Lululemon wants to identify their VIP customers. A VIP customer is defined as a customer who has bought items worth over $5000 in total from the online shop. Write a SQL query to extract the user_id
s of the VIP customers and the total amount they have spent.
purchases
Example Input:purchase_id | user_id | purchase_date | product_id | amount_spent |
---|---|---|---|---|
101 | 12345 | 06/10/2022 | 50001 | 200 |
102 | 12345 | 07/10/2022 | 50001 | 300 |
103 | 67890 | 06/11/2022 | 69852 | 400 |
104 | 67890 | 07/20/2022 | 69852 | 5000 |
105 | 12345 | 07/25/2022 | 69852 | 4500 |
PostgreSQL Query:
SELECT user_id, SUM(amount_spent) as total_spent FROM purchases GROUP BY user_id HAVING SUM(amount_spent) > 5000;
This query groups all purchases by user_id
and calculates the sum of amount_spent
by each user. The HAVING clause filters out those users whose total spending is over $5000, which means these users are our VIP customers. The result will be a list of user_id
s and their corresponding total_spent
amounts. Examine the data to find the users that are most important to Lululemon's business.
To practice another SQL customer analytics question where you can solve it interactively and have your SQL solution instantly executed, try this Walmart SQL Interview Question:
Imagine you had a table of Lululemon employee salary data. Write a SQL query to find the 2nd highest salary at the company.
employees
Example Input:employee_id | salary |
---|---|
1 | 2500 |
2 | 800 |
3 | 1000 |
4 | 1200 |
second_highest_salary |
---|
1200 |
Test your SQL query for this interview question and run your code right in DataLemur's online SQL environment:
SELECT MAX(salary) AS second_highest_salary FROM employee WHERE salary < ( SELECT MAX(salary) FROM employee );
You can find a detailed solution with hints here: 2nd Highest Salary.
Also check out Lululemon's 2024 first quarter fiscal year results!
A join in SQL allows you to retrieve data from multiple tables and combine it into a single set of results.
To demonstrate the difference between left vs. right join, imagine you had two database tables, an advertising campaigns table which had information on Google Ads keywords and how much was bid for each keyword, and a sales table, which has data on how many products were sold and which Google Ads keyword drove that sale.
LEFT JOIN
: A LEFT JOIN retrieves all rows from the left table (in this case, the Advertising_Campaigns
table) and any matching rows from the right table (the Sales table). If there is no match in the right table, NULL
values will be returned for the right table's columns.
RIGHT JOIN
: A RIGHT JOIN retrieves all rows from the right table (in this case, the Sales table) and any matching rows from the left table (the Advertising_Campaigns
table). If there is no match in the left table, NULL
values will be returned for the left table's columns.
Lululemon has a database table 'sales' where every row represents a completed sale, and another table 'returns' where every row represents a returned item. Write a SQL query that provides the total sales and total returns for each product on a monthly basis.
sales
Example Input:sale_id | product_id | sale_date | quantity |
---|---|---|---|
7542 | 101 | 2022-07-01 | 5 |
8427 | 102 | 2022-07-05 | 7 |
6347 | 103 | 2022-07-10 | 3 |
9856 | 101 | 2022-07-20 | 2 |
6437 | 102 | 2022-08-01 | 9 |
7984 | 103 | 2022-08-15 | 6 |
9873 | 102 | 2022-08-22 | 4 |
6628 | 101 | 2022-08-29 | 8 |
returns
Example Input:return_id | product_id | return_date | quantity |
---|---|---|---|
2251 | 101 | 2022-07-15 | 1 |
3456 | 101 | 2022-08-10 | 2 |
4567 | 102 | 2022-08-12 | 3 |
2765 | 103 | 2022-09-02 | 5 |
SELECT EXTRACT(MONTH FROM s.sale_date) as month, s.product_id, SUM(s.quantity) as total_sales, COALESCE(SUM(r.quantity), 0) as total_returns FROM sales s LEFT JOIN returns r ON s.product_id = r.product_id AND EXTRACT(MONTH FROM s.sale_date) = EXTRACT(MONTH FROM r.return_date) GROUP BY month, s.product_id ORDER BY month, s.product_id;
In this query, we create a report that shows total sales and returns for each product on a monthly basis. We perform a left join on the 'sales' and 'returns' table, matching rows based on their product_id and the month of sale/return. We then sum the 'quantity' column from both 'sales' and 'returns' tables for each product and month. We use the COALESCE
function to handle null values in case a product has no returns in a given month.
To solve another window function question on DataLemur's free online SQL coding environment, try this Google SQL Interview Question:
The DISTINCT
keyword added to a SELECT
statement can be used to get records without duplicates.
For example, say you had a table of Lululemon customers:
name | city |
---|---|
Akash | SF |
Brittany | NYC |
Carlos | NYC |
Diego | Seattle |
Eva | SF |
Faye | Seattle |
Suppose you wanted to figure out which cities the customers lived in, but didn't want duplicate results, you could write a query like this:
SELECT DISTINCT city FROM lululemon_customers;
Your result would be:
city |
---|
SF |
NYC |
Seattle |
Lululemon wants to analyse the shopping habits of their customers. They are particularly interested in identifying the average purchase price per customer. Your task is to write a SQL query that retrieves each customer's identifier and the average price of their purchases.
purchases
Example Input:purchase_id | customer_id | purchase_date | product_id | price |
---|---|---|---|---|
101 | 300 | 01/01/2022 | 10001 | 120 |
102 | 300 | 02/01/2022 | 10002 | 80 |
103 | 400 | 01/01/2022 | 10003 | 140 |
104 | 400 | 03/01/2022 | 10002 | 80 |
105 | 500 | 01/01/2022 | 10004 | 160 |
customer_id | average_price |
---|---|
300 | 100 |
400 | 110 |
500 | 160 |
SELECT customer_id, AVG(price) AS average_price FROM purchases GROUP BY customer_id;
In the above PostgreSQL query, we use the AVG()
function to find the average price per customer. By grouping our purchases data by customer_id
, we can find the average amount spent on individual purchases for each customer.
To practice a very similar question try this interactive Amazon Highest-Grossing Items Question which is similar for calculating metrics based on customer spending or this Walmart Histogram of Users and Purchases Question which is similar for analyzing users' purchase habits.
UNION ALL
typically give the same results as a FULL OUTER JOIN
?No, in 99% of normal cases a UNION ALL
and FULL OUTER JOIN
do NOT produce the same result.
You can think of UNION ALL
more as set addition when it combines the results of two tables.
You can think of a FULL OUTER JOIN
more as set multiplication, producing all combos of table 1 with combos of table 2.
Given the "ads" table which includes clicks and impressions of digital ads for each product, and "conversions" table which includes the views and purchases of the product, calculate the click-through rate (CTR) and conversion rate (CR) for each product.
ads
Example Input:ads_id | product_id | date | impressions | clicks |
---|---|---|---|---|
1 | 200 | 01/05/2022 | 1000 | 50 |
2 | 200 | 01/06/2022 | 800 | 40 |
3 | 300 | 01/05/2022 | 1500 | 90 |
4 | 300 | 01/06/2022 | 1200 | 60 |
conversions
Example Input:conversion_id | product_id | date | views | purchases |
---|---|---|---|---|
1 | 200 | 01/05/2022 | 300 | 55 |
2 | 200 | 01/06/2022 | 250 | 50 |
3 | 300 | 01/05/2022 | 500 | 95 |
4 | 300 | 01/06/2022 | 450 | 75 |
SELECT a.product_id, SUM(a.clicks)::FLOAT / SUM(a.impressions)::FLOAT AS CTR, SUM(c.purchases)::FLOAT / SUM(c.views)::FLOAT AS CR FROM ads a JOIN conversions c ON a.product_id = c.product_id AND a.date = c.date GROUP BY a.product_id;
This query first joins the tables 'ads' and 'conversions' on 'product_id' and 'date' because we are calculating the CTR and CR for the same product and the same day. Then calculate the SUM of clicks and impressions for each product in 'ads' table to get the CTR, and calculate the SUM of purchases and views in 'conversions' table to get the CR. The double colons and 'FLOAT' are used for casting the sums to a float type to return a decimal number.
product_id | CTR | CR |
---|---|---|
200 | 0.054 | 0.21 |
300 | 0.062 | 0.17 |
Summary: Product 200 had higher conversion rate (21%) than product 300 (17%), but product 300 had a higher click-through rate (6.2%) than product 200 (5.4%).
To practice a related SQL interview question on DataLemur's free interactive SQL code editor, solve this SQL interview question asked by Facebook:
As a data analyst at Lululemon, we would like you to be able to analyze the average sale price by product category on an annual basis.
sales
Example Input:sale_id | sale_date | product_id | sale_price |
---|---|---|---|
00001 | 2022-01-05 | 101 | 120.00 |
00002 | 2022-03-12 | 101 | 115.00 |
00003 | 2022-04-25 | 102 | 65.00 |
00004 | 2022-03-03 | 103 | 80.00 |
00005 | 2022-07-11 | 103 | 75.00 |
products
Example Input:product_id | product_category |
---|---|
101 | Leggings |
102 | Tops |
103 | Shorts |
SELECT EXTRACT(YEAR FROM sale_date) AS year, p.product_category, AVG(s.sale_price) AS average_sale_price FROM sales s JOIN products p ON s.product_id = p.product_id GROUP BY year, p.product_category ORDER BY year, average_sale_price DESC;
This query first joins the sale and product tables on product_id. It then groups by year (extracted from sale_date) and product_category, calculating the average sale_price. The result will show the average sale price per product category per year, helping us understand which product category had the highest average sale price for each year.
One way to find duplicate records in a table is by using GROUP BY
, and then seeing which groups have more than one occurence:
SELECT x, y, COUNT(*) as occurrences FROM lululemon_table GROUP BY x, y HAVING occurrences> 1;
Another way is by using the EXISTS
operator:
SELECT * FROM lululemon_table t1 WHERE EXISTS ( SELECT 1 FROM lululemon_table t2 WHERE t1.column_name = t2.column_name AND t1.id <> t2.id );
The best way to prepare for a Lululemon SQL interview is to practice, practice, practice.
In addition to solving the earlier Lululemon SQL interview questions, you should also solve the 200+ tricky sql questions on DataLemur which come from companies like Netflix, Google, and Amazon.
Each problem on DataLemur has hints to guide you, step-by-step solutions and best of all, there's an online SQL coding environment so you can instantly run your query and have it checked.
To prep for the Lululemon SQL interview it is also wise to solve interview questions from other apparel companies like:
However, if your SQL skills are weak, don't worry about diving straight into solving questions – improve your SQL foundations with this SQL interview tutorial.
This tutorial covers SQL topics like Union vs. UNION ALL and Subqueries – both of these come up routinely in Lululemon interviews.
In addition to SQL query questions, the other topics tested in the Lululemon Data Science Interview are:
To prepare for Lululemon Data Science interviews read the book Ace the Data Science Interview because it's got: