At Squarespace, SQL is used all the damn time for querying and manipulating customer data for insight generation, and managing databases to ensure the smooth operation of the company's website hosting services. So, it shouldn't surprise you that Squarespace typically asks SQL coding questions during interviews for Data Analyst, Data Science, and BI jobs.
To help you practice for the Squarespace SQL interview, we've curated 8 Squarespace SQL interview questions – able to solve them?
A power user for Squarespace is defined as a user who has created a large number of published websites. Considering Squarespace's business model, the aim would be to identify users who are actively publishing websites more frequently as these are the valuable customers for the business.
Write a SQL query to identify these power users. More specifically, find the users who have published more than 50 websites in the last month.
We have two tables:
users
Example Input:user_id | join_date | country |
---|---|---|
123 | 2022-05-08 | USA |
265 | 2021-07-10 | Canada |
362 | 2020-12-18 | UK |
192 | 2022-03-26 | Australia |
981 | 2019-01-05 | USA |
websites
Example Input:website_id | published_date | user_id |
---|---|---|
6789 | 2022-06-10 | 123 |
5432 | 2022-05-30 | 123 |
9876 | 2022-06-18 | 265 |
4567 | 2022-06-31 | 265 |
1234 | 2022-06-15 | 981 |
4321 | 2022-06-20 | 981 |
SELECT u.user_id, u.join_date, u.country, COUNT(w.website_id) AS websites_published FROM users u JOIN websites w on u.user_id = w.user_id WHERE w.published_date >= DATE_SUB(CURRENT_DATE, INTERVAL 1 MONTH) GROUP BY u.user_id, u.join_date, u.country HAVING COUNT(w.website_id) > 50 ORDER BY websites_published DESC;
In this SQL query, we are joining the users with the websites based on the user_id. We filter out the websites that are published in the last 1 month. Then we group by user_id, join_date, and country to get the counts of how many websites each user has published.
We include only the rows where the number of published websites is more than 50. Finally, we provide the output sorted by the number of websites published by the user in descending order.
To practice a similar customer analytics SQL question where you can solve it right in the browser and have your SQL solution instantly executed, try this Walmart Labs SQL Interview Question:
Suppose you are given a dataset containing product reviews from users. Each record contains a review ID, the ID of the user who submitted the review, the date (with time) when the review was submitted, the ID of the product, and the number of stars given to the product.
Your task is to write a SQL query that calculates the average product rating for each product for each month. The ratings range from 1 to 5 stars.
product_reviews
Example Input:review_id | user_id | submit_date | product_id | stars |
---|---|---|---|---|
3174 | 174 | 06/15/2022 00:00:00 | 98765 | 5 |
5621 | 781 | 06/29/2022 00:00:00 | 65432 | 4 |
4910 | 391 | 07/03/2022 00:00:00 | 98765 | 3 |
8219 | 837 | 07/21/2022 00:00:00 | 65432 | 2 |
6710 | 543 | 08/19/2022 00:00:00 | 12345 | 4 |
month | product_id | avg_stars |
---|---|---|
6 | 98765 | 5.00 |
6 | 65432 | 4.00 |
7 | 98765 | 3.00 |
7 | 65432 | 2.00 |
8 | 12345 | 4.00 |
SELECT EXTRACT(MONTH FROM submit_date) as month, product_id, AVG(stars) as avg_stars FROM product_reviews GROUP BY month, product_id ORDER BY month, product_id;
In this query, we first extract the month from the submit_date
using the EXTRACT
function of PostgreSQL. We then calculate the average of the stars for each product for each month using the AVG
function and GROUP BY
both month
and product_id
. To make the results more readable, we sort the results by month
and product_id
using the ORDER BY
clause.
To solve another window function question on DataLemur's free online SQL coding environment, solve this Amazon SQL Interview Question:
Database normalization is the process of breaking down a table into smaller and more specific tables and defining relationships between them via foreign keys. This minimizes redundancy, and creates a database that's more flexible, scalable, and easier to maintain. It also helps to ensure the integrity of the data by minimizing the risk of data inconsistencies and anomalies.
At Squarespace, tracking and analyzing our website traffic is a crucial part of our business strategy. We're interested in understanding more about the users who visit our site. We have a visits
table that tracks user visits to our website. Each time a user visits the website, a new row is added to the visits
table. We would like to know how many unique users there were for each month.
The visits
table has the following schema:
visits
Example Input:visit_id | user_id | visit_date |
---|---|---|
1 | 100 | 2022/01/01 |
2 | 200 | 2022/01/02 |
3 | 100 | 2022/01/03 |
4 | 300 | 2022/01/04 |
5 | 200 | 2022/02/01 |
6 | 400 | 2022/02/02 |
7 | 500 | 2022/02/03 |
8 | 600 | 2022/02/04 |
9 | 700 | 2022/03/01 |
10 | 700 | 2022/03/02 |
11 | 800 | 2022/03/03 |
12 | 900 | 2022/03/04 |
Write a SQL query to get the month-wise count of unique users. The output should have two columns - month
and unique_users_count
. Order the results based on the month
.
SELECT DATE_TRUNC('month', visit_date) AS month, COUNT(DISTINCT user_id) AS unique_users_count FROM visits GROUP BY month ORDER BY month;
This query first truncates the visit_date
field to a 'month' level using the DATE_TRUNC
function. This gives us the first day of each month.
The query then groups the records based on these 'month' values and calculates the count of distinct user_ids in each group using the COUNT(DISTINCT user_id)
function.
Finally, it orders the result set based on the 'month' in ascending order.
UNION ALL
and a FULL OUTER JOIN
?No, in almost all cases, and for all practical purposes, UNION ALL
and FULL OUTER JOIN
do NOT produce the same result.
While both are similar, in that they combine two tables, you can think of joins as increasing the width of the resulting table (you'll have more columns in the result set for a left/inner/right join), whereas a union is used to combine rows which increases the height of the result set but keeps the column count the same.
Suppose Squarespace wants to analyze the effectiveness of its website layout by looking at clickthrough conversion rates. A clickthrough conversion occurs when a customer views a product and subsequently adds the product to their cart. Using the below tables where 'website_clicks' provides information on site visitors who viewed a product, and 'cart_adds' records when a product is added to a cart, write a SQL query to calculate the click-through conversion rate for each product.
website_clicks
Example Input:click_id | user_id | click_time | product_id |
---|---|---|---|
101 | 121 | 06/08/2022 12:00:00 | 001 |
202 | 215 | 06/08/2022 12:05:00 | 001 |
303 | 333 | 06/08/2022 12:10:00 | 002 |
404 | 444 | 06/08/2022 12:15:00 | 003 |
505 | 555 | 06/08/2022 12:20:00 | 003 |
cart_adds
Example Input:add_id | user_id | add_time | product_id |
---|---|---|---|
1111 | 121 | 06/08/2022 12:10:00 | 001 |
2222 | 333 | 06/08/2022 12:20:00 | 002 |
3333 | 444 | 06/08/2022 12:25:00 | 003 |
SELECT w.product_id, COUNT(DISTINCT w.user_id) AS views, COUNT(DISTINCT c.user_id) AS adds, (COUNT(DISTINCT c.user_id)::float / COUNT(DISTINCT w.user_id)) * 100 AS conversion_rate FROM website_clicks w LEFT JOIN cart_adds c ON w.product_id = c.product_id AND w.user_id = c.user_id GROUP BY w.product_id;
This solution starts by joining the 'website_clicks' and 'cart_adds' tables on both 'product_id' and 'user_id' to ensure we are accurately tracking each user’s journey. Next, we count the unique views and adds for each product, using product_id to group the data. Lastly, we calculate the conversion rate by dividing the count of unique adds by the count of unique views, converting to float to allow for decimal places, and then multiplying by 100 to return a percentage. Be sure to handle NULL results appropriately, as not all products viewed will be added to the cart.
To practice a similar problem on DataLemur's free online SQL code editor, solve this Meta SQL interview question:
A UNIQUE constraint ensures that all values in a column are different. It is often used in conjunction with other constraints, such as NOT NULL, to ensure that the data meets certain conditions.
For example, if you had Squarespace employee data stored in a database, here's some constraints you'd use:
CREATE TABLE squarespace_employees ( employee_id INTEGER PRIMARY KEY, first_name VARCHAR(255) NOT NULL, last_name VARCHAR(255) NOT NULL, email VARCHAR(255) NOT NULL UNIQUE, department VARCHAR(255) NOT NULL );
In the Squarespace employee example, the UNIQUE constraint is applied to the "email" field to ensure that each employee has a unique email address. This helps to ensure the integrity of the data in the database and prevents errors that could occur if two employees had the same email address.
A company, Squarespace, is interested to analyze the average purchase amount made by its customers in different regions. There are two tables available for this analysis:
customers
table, with the customer id, name, and their corresponding region.purchases
table, with customer id, date of purchase, and the purchase amount.Write a SQL query to join these two tables and find the average purchase amount by customers in each region.
customers
Table:customer_id | name | region |
---|---|---|
101 | Alice | North |
102 | Bob | South |
103 | Charlie | East |
104 | David | West |
105 | Eve | North |
purchases
Table:purchase_id | customer_id | date_of_purchase | amount |
---|---|---|---|
1001 | 101 | 06/01/2022 | 100.50 |
1002 | 101 | 06/15/2022 | 200.75 |
1003 | 102 | 06/10/2022 | 300.25 |
1004 | 103 | 06/18/2022 | 150.00 |
1005 | 104 | 07/05/2022 | 250.00 |
1006 | 105 | 07/12/2022 | 325.50 |
1007 | 105 | 07/20/2022 | 375.75 |
In PostgreSQL, the SQL query for this problem would be:
SELECT c.region, AVG(p.amount) AS avg_purchase_amount FROM customers c JOIN purchases p ON c.customer_id = p.customer_id GROUP BY c.region ORDER BY avg_purchase_amount DESC;
This SQL query joins the customers
table with the purchases
table on the customer_id
field. It then groups together records for each region and calculates the average purchase amount for each of these groups. Finally, it orders these by the average purchase amount in descending order, so regions with higher average purchase amounts will come first.
Because joins come up routinely during SQL interviews, practice this Spotify JOIN SQL question:
Assuming that you've already got basic SQL skills, the next best tip we have to prepare for the Squarespace SQL interview is to solve as many practice SQL interview questions as you can!
Besides solving the earlier Squarespace SQL interview questions, you should also solve the 200+ SQL questions from real Data Science & Analytics interviews which come from companies like Microsoft, Google, and Facebook.
Each DataLemur SQL question has multiple hints, full answers and crucially, there's an interactive SQL code editor so you can easily right in the browser your query and have it executed.
To prep for the Squarespace SQL interview it is also helpful to solve SQL problems from other tech companies like:
In case your SQL coding skills are weak, don't worry about jumping right into solving questions – go learn SQL with this SQL tutorial for Data Analytics.
This tutorial covers SQL topics like math functions and filtering strings based on patterns – both of these come up frequently during Squarespace SQL interviews.
Beyond writing SQL queries, the other types of problems to practice for the Squarespace Data Science Interview are:
To prepare for Squarespace Data Science interviews read the book Ace the Data Science Interview because it's got: