I’ve interviewed a lot of data scientist candidates and have found there are a a lot of SQL interview questions for data science that eventually boil down to three generalized types of conceptual understandings.

If you’re an interviewing data scientist, these problems are a must know! They’re also great filter questions to test more than the basic one-hour primer study of a candidate that read the differences between an INNER and LEFT JOIN. Here’s an example question from each.

### 1. Getting the first or last value for each user in a `transactions` table.

``````
`transactions`
+---------------+---------+
| user_id       | int     |
| created_at    | datetime|
| product       | varchar |
+---------------+---------+
Question: Given the user transactions table above,
write a query to get the first purchase for each user.
``````

Why does this matter? How would you query for the first time a person commented on a post and read the post itself? How do we cohort users by start date? All of these analyses need this concept of querying based on first or last time and it definitely can be solved without using an expensive partition function.

Explanation:

``````
We want to take a table that looks like this:

user_id | created_at | product
--------+------------+--------
123    | 2019-01-01 | apple
456    | 2019-01-02 | banana
123    | 2019-01-05 | pear
456    | 2019-01-10 | apple
789    | 2019-01-11 | banana

and turn it into this

user_id | created_at | product
---------+------------+--------
123     | 2019-01-01 | apple
456     | 2019-01-02 | banana
789     | 2019-01-11 | banana

How do we get there?
``````

We can solve this problem by doing a multi-column join.

First, how do we figure out the first time each user purchased? This should be pretty simple and can be done by a simply GROUP BY aggregation and aggregating for the minimum datetime. Notice how the table has a created_at column. This is the column that determines which row is the first purchase for the specific user, so we can write a query with an aggregation to get the minimum datetime for every user.

``````
SELECT user_id, MIN(created_at) AS min_created_at
FROM transactions
GROUP BY 1
``````

Awesome. Now all we have to do is join this table back to the original on two columns: user_id and created_at. The self join will effectively filter for the first purchase. Then all we have to do is grab all of the columns on the left side table.

``````
SELECT t.user_id, t.created_at, t.product
FROM transactions AS t
INNER JOIN (
SELECT user_id, MIN(created_at) AS min_created_at
FROM transactions
GROUP BY 1
) AS t1
ON t.user_id = t1.user_id
t.created_at = t1.min_created_at
``````

### 2. Knowing the difference between a LEFT JOIN and INNER JOIN in practice.

``````
`users`
+---------+---------+
| id      | int     |
| name    | varchar |
| city_id | int     |<-+
+---------+---------+  |
|
|
`cities`               |
+---------+---------+  |
| id      | int     |<-+
| name    | varchar |
+---------+---------+
Question: Given the `users` and `cities` tables above,
write a query to return the list of cities without any users.
``````

Why does this matter? Anyone can memorize the definitions of an inner join and left join when asked during an interview. The Venn diagram provides an adequate explanation. But can the candidate actually implement the difference when in practice?

Explanation:

What is the actual difference between a LEFT JOIN and INNER JOIN?

``````
INNER JOIN: returns rows when there is a match in both tables.
LEFT JOIN: returns all rows from the left table, even if there are no matches in the right table.
``````

Okay, so we know that each user in the `users` table must live in a city given the `city_id` field. However the `cities` table doesn’t have a `user_id` field. In which if we run an INNER JOIN between these two tables joined by the `city_id` in each table, we’ll get all of the cities that have users and all of the cities without users will be filtered out.

``````
SELECT cities.name, users.id
FROM cities
ON users.city_id = cities.id
``````

But what if we run a LEFT JOIN between cities and users?

``````
cities.name  | users.id
_____________|__________
seattle      | 123
seattle      | 124
portland     | null
san diego    | 534
san diego    | 564
``````

Here we see that since we are keeping all of the values on the LEFT side of the table, since there’s no match on the city of Portland to any users that exist in the database, the city shows up as NULL. Therefore now all we have to do is run a WHERE filter to where any value in the users table is NULL.

``````
SELECT cities.name, users.id
FROM cities
ON users.city_id = cities.id
WHERE users.id IS NULL
``````

### 3. Aggregations with a conditional statement

``````
`transactions`
+---------------+---------+
| user_id       | int     |
| created_at    | datetime|
| product       | varchar |
+---------------+---------+
Question: Given the same user transactions table as before,
write a query to get the total purchases made in the morning
versus afternoon/evening (AM vs PM) by day.
``````

Why does this matter? If you can’t use conditional statements and/or aggregate with conditional statements, there’s no way to run any kind of analytics. How do you look at differences in populations based on new features or variables?

Explanation:

Notice whenever the question asks for a versus statement, we’re comparing two groups. Every time we have to compare two groups we must use a GROUP BY. It’s in the name. Heh.

In this case, we need to create a separate column to actually run our GROUP BY on, which in this case, is the difference between AM or PM in the `created_at` field. In that case, let’s create a condition in SQL to differentiate them.

``````
CASE WHEN
HOUR(created_at) > 11
THEN 'PM' ELSE 'AM' END AS time_of_day
``````

Pretty simple. We can cast the `created_at` column to the hour and set the new column value `time_of_day` as AM or PM based on this condition. Now we just have to run a GROUP BY on the original `created_at` field truncated to the day AND the new column we created that differentiates each row value. The last aggregation will then be the output variable we want which is total purchases by running the COUNT function.

``````
SELECT
DATE_TRUNC('day', created_at) AS date
, CASE WHEN
HOUR(created_at) > 11
THEN 'PM' ELSE 'AM' END AS time_of_day
, COUNT(*)
FROM transactions
GROUP BY 1,2
``````