SQL is a crucial skill for data analysts and data engineers, and it is often tested during interviews at top tech companies like Square. If you’re preparing for a Square SQL interview, you’ll want to be well-versed in SQL concepts and be ready to tackle a range of problems. In this article, we’ll cover some of the top Square SQL interview questions you might encounter, along with explanations and tips for success.
Understanding the Interview Process
Before we dive into the questions, let’s briefly discuss the interview process at Square. For data analyst roles, the first round typically involves a SQL pair programming exercise. You’ll be presented with two tables of data and asked to answer 4-5 questions within an hour using a collaborative IDE like CoderPad.
For data engineer roles, the interview process may include a phone screen, technical rounds focusing on SQL and coding challenges, and onsite interviews assessing your technical, behavioral, and soft skills.
Top Square SQL Interview Questions
Here are some of the top Square SQL interview questions you might encounter:
Question 1: Find the Total Number of Unique Transactions, Senders, and Recipients with Successful Transactions
SELECT COUNT(DISTINCT payment_id) AS unique_transactions, COUNT(DISTINCT sender_id) AS unique_senders, COUNT(DISTINCT recipient_id) AS unique_recipientsFROM paymentsWHERE payment_state = 'SUCCESS';
This query retrieves the total number of unique transactions, senders, and recipients with successful transactions from the payments
table. It uses the COUNT(DISTINCT)
function to count the unique values in the payment_id
, sender_id
, and recipient_id
columns, filtering only the rows where payment_state
is ‘SUCCESS’.
Question 2: Find the Failure Rate of Transactions
SELECT ROUND( 100.0 * SUM(CASE WHEN payment_state = 'FAILURE' THEN 1 ELSE 0 END) / COUNT(*), 2 ) AS failure_rateFROM payments;
This query calculates the failure rate of transactions from the payments
table. It uses a CASE
statement to count the number of failed transactions, divides it by the total number of transactions, and multiplies by 100 to get the percentage. The ROUND
function is used to round the result to two decimal places.
Question 3: Find the Number of Customers Who Signed Up in January 2020 and Had a Combined Sending and Receiving Volume Greater Than $100 in Their First 30 Days
SELECT COUNT(DISTINCT customer_id)FROM ( SELECT customer_id, SUM( CASE WHEN sender_id = customer_id THEN amount_cents WHEN recipient_id = customer_id THEN -amount_cents ELSE 0 END ) AS net_amount_cents FROM customers c JOIN payments p ON c.customer_id IN (p.sender_id, p.recipient_id) WHERE c.created_at BETWEEN '2020-01-01' AND '2020-01-31' AND p.created_at BETWEEN c.created_at AND c.created_at + INTERVAL '30 days' AND p.payment_state = 'SUCCESS' GROUP BY customer_id HAVING ABS(net_amount_cents) > 10000) t;
This query finds the number of customers who signed up in January 2020 and had a combined sending and receiving volume greater than $100 (10,000 cents) in their first 30 days after sign-up. Here’s a breakdown of what’s happening:
- The subquery calculates the net amount (sending minus receiving) for each customer using a
CASE
statement and theSUM
function. - The
JOIN
clause combines thecustomers
andpayments
tables to include only transactions involving the customers who signed up in January 2020. - The
WHERE
clause filters for customers created in January 2020 and successful transactions within the first 30 days of their account creation. - The
GROUP BY
andHAVING
clauses filter for customers with a net transaction volume greater than $100 (10,000 cents). - The outer query counts the distinct
customer_id
values from the subquery result.
Question 4: Choose the Top 3 Departments with at Least 10 Employees and Arrange Them by the Percentage of Employees Making Over $100K
SELECT d.department_name, ROUND( 100.0 * SUM(CASE WHEN e.salary > 100000 THEN 1 ELSE 0 END) / COUNT(*), 2 ) AS pct_over_100kFROM departments dJOIN employees e ON d.department_id = e.department_idGROUP BY d.department_idHAVING COUNT(*) >= 10ORDER BY pct_over_100k DESCLIMIT 3;
This query finds the top 3 departments with at least 10 employees, ordered by the percentage of employees making over $100,000. Here’s how it works:
- The
JOIN
clause combines thedepartments
andemployees
tables to include employee salary information. - The
GROUP BY
clause groups the rows bydepartment_id
. - The
HAVING
clause filters for departments with at least 10 employees. - The
SELECT
statement calculates the percentage of employees making over $100,000 using aCASE
statement and theSUM
andCOUNT
functions. - The
ORDER BY
clause sorts the results by the percentage of employees making over $100,000 in descending order. - The
LIMIT
clause restricts the output to the top 3 departments.
Question 5: Return the Two Students with the Closest Test Scores and Handle Ties
SELECT s1.name AS student1, s2.name AS student2, ABS(s1.score - s2.score) AS score_diffFROM students s1CROSS JOIN students s2WHERE s1.id < s2.idORDER BY score_diff ASC, s1.name ASC, s2.name ASCLIMIT 1;
This query finds the two students with the closest test scores, handling ties by selecting the combination with the higher student names. Here’s how it works:
- The
CROSS JOIN
clause creates a Cartesian product of thestudents
table with itself, generating all possible pairs of students. - The
WHERE
clause filters out pairs where the first student’s ID is greater than or equal to the second student’s ID, avoiding duplicates. - The
ABS
function calculates the absolute difference between the two students’ scores. - The
ORDER BY
clause sorts the results by the score difference in ascending order, and in case of ties, by the student names in ascending order. - The
LIMIT
clause restricts the output to the first row, which contains the students with the minimum score difference and the highest names in case of ties.
Question 6: Check if Users’ Subscription Date Ranges Overlap
SELECT u1.user_id AS user1, u2.user_id AS user2, CASE WHEN (u1.start_date <= u2.end_date AND u1.end_date >= u2.start_date) OR (u2.start_date <= u1.end_date AND u2.end_date >= u1.start_date) THEN 'True' ELSE 'False' END AS overlapFROM subscriptions u1CROSS JOIN subscriptions u2WHERE u1.user_id < u2.user_id;
This query checks if any users’ subscription date ranges overlap and returns a ‘True’ or ‘False’ value for each pair of users. Here’s how it works:
- The
CROSS JOIN
clause creates a Cartesian product of thesubscriptions
table with itself, generating all possible pairs of subscriptions. - The
WHERE
clause filters out pairs where the first user’s ID is greater than or equal to the second user’s ID, avoiding duplicates. - The
CASE
statement checks if either user’s subscription date range overlaps with the other user’s range. If there’s an overlap, it returns ‘True’; otherwise, it returns ‘False’. - The
SELECT
statement retrieves the user IDs and the overlap indicator.
Conclusion
SQL is a crucial skill for data analysts and data engineers, and being able to solve complex SQL problems is essential for succeeding in interviews at top tech companies like Square. By practicing these types of SQL interview questions and understanding the underlying concepts, you’ll be better prepared to showcase your skills and stand out during the interview process.
Remember, it’s always a good idea to practice with real-world datasets and to familiarize yourself with the specific database management system (e.g., PostgreSQL) used by the company you’re interviewing with.
TOP 23 SQL INTERVIEW QUESTIONS & ANSWERS! (SQL Interview Tips + How to PASS an SQL interview!)
FAQ
What are the different types of SQL problems?