Data Analysts & Data Scientists at NETSCOUT write ad-hoc SQL queries daily as part of their job. They use SQL for analyzing network performance data and for managing large-scale customer databases even amidst the complexities of network management sector. For this reason NETSCOUT LOVES to ask folks interviewing at the company SQL interview questions.
So, to help you prepare, we've curated 9 NETSCOUT SQL interview questions – can you answer each one?
Sure, I understand you'd like a typical SQL interview question including the use of window functions, suited to NETSCOUT, a company which specializes in application and network performance management products. Let's say we have been working with two tables network_events
and devices
.
network_events
stores all events detected on the company's monitored networks, and devices
houses all devices in those networks. I'm considering a scenario where the interviewee is asked to identify the top 3 devices with the most network events each month.
Using the network_events
and devices
tables, can you write a SQL query to identify the top 3 devices that have triggered the most network events every month?
network_events
Example Input:event_id | device_id | timestamp_event |
---|---|---|
1 | 101 | 2022-01-03 10:00:00 |
2 | 202 | 2022-01-04 15:20:00 |
3 | 101 | 2022-01-04 18:00:00 |
4 | 303 | 2022-01-05 09:25:00 |
5 | 404 | 2022-01-06 10:30:00 |
devices
Example Input:device_id | device_name |
---|---|
101 | Router A |
202 | Switch B |
303 | Server C |
404 | Router D |
The following SQL query could solve this problem.
WITH monthly_rank as ( SELECT date_trunc('month', ne.timestamp_event) as month, d.device_name, COUNT(ne.event_id) as event_count, ROW_NUMBER() OVER (PARTITION BY date_trunc('month', ne.timestamp_event) ORDER BY COUNT(ne.event_id) DESC) as rank FROM network_events ne JOIN devices d ON ne.device_id = d.device_id GROUP BY month, device_name ) SELECT month, device_name, event_count FROM monthly_rank WHERE rank <= 3;
This query initially aggregates the event data by month and device, then assigns a rank to each device per month based on the number of events. Finally, it selects the top 3 devices from each month using the rank.
To solve a related window function SQL problem on DataLemur's free interactive coding environment, solve this Amazon SQL question asked in a BI Engineer interview:
Given a table of NETSCOUT employee salary information, write a SQL query to find the top 3 highest earning employees in each department.
employees
Example Input:employee_id | name | salary | department_id |
---|---|---|---|
1 | Emma Thompson | 3800 | 1 |
2 | Daniel Rodriguez | 2230 | 1 |
3 | Olivia Smith | 2000 | 1 |
4 | Noah Johnson | 6800 | 2 |
5 | Sophia Martinez | 1750 | 1 |
8 | William Davis | 6800 | 2 |
10 | James Anderson | 4000 | 1 |
department
Example Input:department_id | department_name |
---|---|
1 | Data Analytics |
2 | Data Science |
department_name | name | salary |
---|---|---|
Data Analytics | James Anderson | 4000 |
Data Analytics | Emma Thompson | 3800 |
Data Analytics | Daniel Rodriguez | 2230 |
Data Science | Noah Johnson | 6800 |
Data Science | William Davis | 6800 |
Check your SQL query for this problem and run your code right in DataLemur's online SQL environment:
We use the DENSE_RANK() window function to generate unique ranks for each employee's salary within their department, with higher salaries receiving lower ranks. Then, we wrap this up in a CTE and filter the employees with a ranking of 3 or lower.
WITH ranked_salary AS ( SELECT name, salary, department_id, DENSE_RANK() OVER ( PARTITION BY department_id ORDER BY salary DESC) AS ranking FROM employee ) SELECT d.department_name, rs.name, rs.salary FROM ranked_salary AS rs INNER JOIN department AS d ON rs.department_id = d.department_id WHERE rs.ranking <= 3 ORDER BY d.department_id, rs.salary DESC, rs.name ASC;
If the solution above is confusing, you can find a detailed solution here: Top 3 Department Salaries.
A non-correlated sub-query is not linked to the outer query and can be run on its own. It does not reference any columns from the outer query and is used to retrieve additional data for the outer query. On the other hand, a correlated sub-query is one that relies on the outer query and cannot be run independently. It uses the outer query to filter or transform data by referencing a column from it, while the outer query uses the results of the inner query.
Here is an example of a non-correlated sub-query:
SELECT t1.customer_id, t1.total_sales FROM netscout_sales t1 WHERE t1.total_sales > ( SELECT AVG(t2.total_sales) FROM netscout_sales t2 );
The sub-query in this case is non-correlated, as it does not reference any columns from the outer query.
Here is an example of a correlated sub-query:
SELECT t1.customer_id, t1.total_sales FROM netscout_sales t1 WHERE t1.total_sales > ( SELECT AVG(t2.total_sales) FROM netscout_sales t2 WHERE t2.customer_id = t1.customer_id );
This query selects the customer_id
and total sales of all NETSCOUT customers in the sales table whose total_sales
are greater than the average total_sales
of their own customer group. The sub-query in this case is correlated with the outer query, as it references the customer_id
column from the outer query (t1.customer_id
).
Non-correlated sub-queries are typically faster, as they only have to be executed once, wheareas correlated sub-queries are slower since they have to be re-evaluated for each row of the outer query.
As a data analyst at NETSCOUT, one of your tasks may involve segmenting and analyzing customer data for varied reasons such as targeting specific customer groups, identifying sales trends or tracking customer behavior.
Consider the fact you have a customers
database containing customer's personal information, the product they purchased and the purchase date. You have been tasked with writing an SQL query to find all clients from California (CA) who have made a purchase between Jan 1, 2022 and Dec 31, 2022 and have purchased product_id either 1001 or 2002.
customers
Example Input:customer_id | first_name | last_name | state | product_id | purchase_date |
---|---|---|---|---|---|
1289 | John | Doe | CA | 1001 | 04/03/2022 |
7925 | Alice | Fernandes | NY | 1001 | 06/16/2022 |
1023 | Mark | Watson | CA | 1045 | 10/21/2022 |
7492 | Linda | Newton | CA | 2002 | 07/18/2022 |
4691 | Donna | Macron | TX | 2002 | 03/16/2022 |
3528 | Paul | Singh | CA | 1001 | 01/06/2022 |
customer_id | first_name | last_name | state | product_id | purchase_date |
---|---|---|---|---|---|
1289 | John | Doe | CA | 1001 | 04/03/2022 |
7492 | Linda | Newton | CA | 2002 | 07/18/2022 |
3528 | Paul | Singh | CA | 1001 | 01/06/2022 |
SELECT customer_id, first_name, last_name, state, product_id, purchase_date FROM customers WHERE state = 'CA' AND purchase_date BETWEEN '01/01/2022' AND '12/31/2022' AND (product_id = 1001 OR product_id = 2002)
This query filters out customers from CA who made a purchase within 2022 and their product_id is either 1001 or 2002. The BETWEEN
command in SQL is used to filter the result within a specific range, and the AND
command combines multiple conditions.
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.
NETSCOUT is a company that provides application and network performance management products. They have both an online website for marketing and an e-commerce store where they sell their products. In a particular month, they want to analyze the click-through rate of their digital ads that leads a user to view their products and the conversion rate of these views to adding a product to the cart.
Here are the data tables:
web_clicks
Example Input:click_id | user_id | click_time | ad_id | product_id |
---|---|---|---|---|
1051 | 220 | 06/08/2022 10:10:00 | 40001 | 7001 |
1030 | 420 | 06/10/2022 10:21:00 | 30200 | 7002 |
1043 | 629 | 06/18/2022 16:30:00 | 40001 | 7001 |
1082 | 989 | 07/26/2022 10:10:00 | 30200 | 7002 |
1017 | 270 | 07/05/2022 14:00:00 | 30200 | 7001 |
product_views
Example Input:view_id | user_id | view_time | product_id |
---|---|---|---|
2201 | 220 | 06/08/2022 10:20:00 | 7001 |
2710 | 420 | 06/10/2022 10:30:00 | 7002 |
2153 | 629 | 06/18/2022 16:35:00 | 7001 |
2862 | 989 | 07/26/2022 10:20:00 | 7002 |
2027 | 270 | 07/05/2022 14:10:00 | 7001 |
cart_adds
Example Input:add_id | user_id | add_time | product_id |
---|---|---|---|
3311 | 220 | 06/08/2022 10:25:00 | 7001 |
3250 | 420 | 06/10/2022 10:35:00 | 7002 |
3413 | 629 | 06/18/2022 16:40:00 | 7001 |
3602 | 989 | 07/26/2022 10:25:00 | 7002 |
3227 | 270 | 07/05/2022 14:15:00 | 7001 |
WITH clicks_to_views AS ( SELECT wc.user_id, wc.click_time, wc.product_id as click_product_id, pv.view_time, pv.product_id as view_product_id FROM web_clicks wc JOIN product_views pv ON wc.user_id = pv.user_id AND wc.product_id = pv.product_id AND wc.click_time <= pv.view_time ), views_to_adds AS ( SELECT pv.user_id, pv.view_time, pv.product_id as view_product_id, ca.add_time, ca.product_id as add_product_id FROM product_views pv JOIN cart_adds ca ON pv.user_id = ca.user_id AND pv.product_id = ca.product_id AND pv.view_time <= ca.add_time ) SELECT count(ctv.user_id) AS total_clicks, count(distinct(ctv.user_id)) AS unique_clicks, count(vta.user_id) AS total_views, count(distinct(vta.user_id)) AS unique_views, count(vta.add_time) AS total_adds, count(distinct (vta.add_time)) AS unique_adds, (count(distinct(vta.user_id))::decimal / count(distinct(ctv.user_id))) * 100 AS click_to_view_ratio, (count(distinct (vta.add_time))::decimal / count(distinct(vta.user_id))) * 100 AS view_to_add_ratio FROM clicks_to_views ctv JOIN views_to_adds vta ON ctv.user_id = vta.user_id AND ctv.view_product_id = vta.view_product_id;
The query above connects the three actions - click, view, and add-to-cart - for the same user and the same product, and measures the conversion rate from click-to-view and view-to-add-to-cart. The time condition in the join ensures that the action sequence is respected. The ratio is calculated as a percentage (multiplied by 100).
To solve a similar problem about calculating rates, try this SQL interview question from TikTok on DataLemur's interactive coding environment:
A non-relational (NoSQL) database is any database that does not use the typical tabular format of rows and columns like in relational databases.
While knowing the four different types of NoSQL databases is beyond the scope of most Data Analyst & Data Science interviews, Data Engineers at NETSCOUT should vaguely refresh these concepts:
As a network performance company, Netscout is interested in understanding the overall reliability of their clients' networks. For a given month, can you calculate the average downtime each customer experienced? We will consider two tables: clients
and downtime_log
.
Assume clients
table contains the following fields: client_id
, client_name
, signup_date
.
The downtime_log
table contains the following fields: log_id
, client_id
, start_downtime
, end_downtime
.
Use the following sample data to solve this SQL problem:
clients
Example Input:
client_id | client_name | signup_date |
---|---|---|
101 | Client A | 01/01/2022 |
102 | Client B | 02/01/2022 |
103 | Client C | 03/01/2022 |
downtime_log
Example Input:
log_id | client_id | start_downtime | end_downtime |
---|---|---|---|
1 | 101 | 01/08/2022 10:00:00 | 01/08/2022 11:00:00 |
2 | 101 | 01/20/2022 12:00:00 | 01/20/2022 15:00:00 |
3 | 102 | 02/02/2022 18:00:00 | 02/02/2022 20:00:00 |
4 | 103 | 02/14/2022 08:00:00 | 02/14/2022 09:30:00 |
5 | 102 | 03/03/2022 10:00:00 | 03/03/2022 11:00:00 |
SELECT EXTRACT(MONTH FROM dl.start_downtime) AS month, c.client_name, AVG(EXTRACT(EPOCH FROM (dl.end_downtime - dl.start_downtime))/3600) AS avg_downtime_hours FROM clients c JOIN downtime_log dl ON c.client_id = dl.client_id GROUP BY month, c.client_name;
The above query extracts the month from the downtime start date and groups the results by it and the client name. For each group, it calculates the average downtime in hours. The downtime itself is calculated as the difference between the end_downtime
and start_downtime
timestamps and is converted to hours by dividing the result by 3600 (as the EXTRACT(EPOCH FROM timestamp)
function returns the timestamp in seconds).
Assume you are a data scientist for NETSCOUT and you've received a request from your marketing team. They are planning for an email campaign targetting customers in Texas, whose business name contain the word "Systems". Use the customer records database to find all customers from Texas whose names contain the word "Systems".
Given the table customers
:
customers
Example Input:customer_id | business_name | state |
---|---|---|
101 | Alpha Systems Incorporated | TX |
102 | Beta Network | NY |
103 | Charlie Systems LLC | TX |
104 | Delta Solutions | CA |
105 | Echo Systems | TX |
The expected output of your SQL query would be:
customer_id | business_name | state |
---|---|---|
101 | Alpha Systems Incorporated | TX |
103 | Charlie Systems LLC | TX |
105 | Echo Systems | TX |
To solve this problem, you would use the SQL LIKE
keyword, along with the wildcard character %
to match customers whose name contains the word "Systems" and are from Texas. Here's one example of how to accomplish this:
SELECT customer_id, business_name, state FROM customers WHERE business_name LIKE '%Systems%' AND state = 'TX';
This SQL statement will filter the customers
table for businesses in Texas ('TX') whose names contain the string 'Systems'. The '%' characters are wildcards that match any number of characters. The 'Systems' string can appear anywhere in the business name due to the wildcard characters before and after it.
The best way to prepare for a SQL interview, besides making sure you have strong SQL fundamentals, is to practice a ton of real SQL questions that were asked in recent job interviews.
In addition to solving the earlier NETSCOUT 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 problem on DataLemur has multiple hints, detailed solutions and crucially, there is an online SQL coding environment so you can instantly run your query and have it checked.
To prep for the NETSCOUT SQL interview you can also be helpful to solve interview questions from other tech companies like:
But if your SQL foundations are weak, don't worry about diving straight into solving questions – strengthen your SQL foundations with this free SQL tutorial.
This tutorial covers SQL topics like grouping by multiple columns and aggregate functions like SUM()/COUNT()/AVG() – both of which pop up often in NETSCOUT interviews.
Besides SQL interview questions, the other topics covered in the NETSCOUT Data Science Interview include:
I'm sorta biased, but I believe the best way to study for NETSCOUT Data Science interviews is to read the book I wrote: Ace the Data Science Interview.
The book covers 201 data interview questions sourced from Microsoft, Amazon & startups. It also has a crash course covering SQL, Product-Sense & ML. And finally it's vouched for by the data community, which is why it's earned 1000+ reviews on Amazon and a 4.5-star rating.