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Most Candidates Fail These SQL Ideas in Information Interviews


Most Candidates Fail These SQL Concepts in Data Interviews
Picture by writer | Canva

 

An interviewer’s job is to search out essentially the most appropriate candidates for the marketed place. In doing so, they are going to gladly arrange SQL interview inquiries to see if they will catch you off guard. There are a number of SQL ideas at which candidates usually fail.

Hopefully, you’ll be a kind of who keep away from that future, as I’ll clarify these ideas intimately under, full with examples of tips on how to clear up sure issues accurately.

 
Most Candidates Fail These SQL Concepts in Data Interviews
 

1. Window Features

 
Why It’s Arduous: Candidates memorize what every window perform does however don’t actually perceive how window frames, partitions, or ordering really work.

Widespread Errors: A standard mistake just isn’t specifying ORDER BY in rating window features or worth window features, akin to LEAD() or LAG(), and anticipating the question to work or for the outcome to be deterministic.

Instance: In this instance, you could discover customers who made a second buy inside 7 days of any earlier buy.

You would possibly write this question.

WITH ordered_tx AS (
  SELECT user_id,
         created_at::date AS tx_date,
         LAG(created_at::DATE) OVER (PARTITION BY user_id) AS prev_tx_date
  FROM amazon_transactions
)

SELECT DISTINCT user_id
FROM ordered_tx
WHERE prev_tx_date IS NOT NULL AND tx_date - prev_tx_date <= 7;

 

At first look, every thing might sound proper. The code even outputs one thing which may seem like an accurate reply.

 
Window Functions
 

To start with, we’re fortunate that the code works in any respect! This occurs just because I’m writing it in PostgreSQL. In another SQL flavors, you’d get an error since ORDER BY is obligatory in rating and analytical window features.

Second, the output is incorrect; I highlighted some rows that shouldn’t be there. Why do they seem, then?

They seem as a result of we didn’t specify an ORDER BY clause within the LAG() window perform. With out it, the row order is bigoted. So, we’re evaluating the present transaction to some random earlier row for that person, not the one which occurred instantly earlier than it in time.

This isn’t what the query asks. We have to examine every transaction to the earlier one by date. In different phrases, we have to specify this explicitly within the ORDER BY clause throughout the LAG() perform.

WITH ordered_tx AS (
  SELECT user_id,
         created_at::date AS tx_date,
         LAG(created_at::DATE) OVER (PARTITION BY user_id ORDER BY created_at) AS prev_tx_date
  FROM amazon_transactions
)

SELECT DISTINCT user_id
FROM ordered_tx
WHERE prev_tx_date IS NOT NULL AND tx_date - prev_tx_date <= 7;

 

2. Filtering With Aggregates (Particularly HAVING vs. WHERE)

 
Why It’s Arduous: Folks usually don’t perceive the execution order in SQL, which is: FROM -> WHERE -> GROUP BY -> HAVING -> SELECT -> ORDER BY. This order implies that WHERE filters rows earlier than aggregation, and HAVING filters after. That additionally, logically, means which you can’t use mixture features within the WHERE clause.

Widespread Mistake: Making an attempt to make use of mixture features in WHERE in a grouped question and getting an error.

Instance: This interview query asks you to search out the overall income made by every vineyard. Solely wineries the place 90 is the bottom variety of factors for any of their varieties ought to be thought-about.

Many will see this as a simple query and rapidly write this question.

SELECT vineyard,
       selection,
       SUM(worth) AS total_revenue
FROM winemag_p1
WHERE MIN(factors) >= 90
GROUP BY vineyard, selection
ORDER BY vineyard, total_revenue DESC;

 

Nonetheless, that code will throw an error stating that mixture features aren’t allowed within the WHERE clause. This beautiful a lot explains every thing. The answer? Transfer the filtering situation from WHERE to HAVING.

SELECT vineyard,
       selection,
       SUM(worth) AS total_revenue
FROM winemag_p1
GROUP BY vineyard, selection
HAVING MIN(factors) >= 90
ORDER BY vineyard, total_revenue DESC;

 

3. Self-Joins for Time-Based mostly or Occasion-Based mostly Comparisons

 
Why It’s Arduous: The concept of becoming a member of a desk with itself is sort of unintuitive, so candidates usually neglect it’s an possibility.

Widespread Mistake: Utilizing subqueries and complicating the question when becoming a member of a desk with itself could be easier and quicker, particularly when filtering by dates or occasions.

Instance: Right here’s a query asking you to point out the change of each foreign money’s alternate fee between 1 January 2020 and 1 July 2020.

You’ll be able to clear up this by writing an outer correlated subquery that fetches the July 1 alternate charges, then subtracts the January 1 alternate charges, which come from the inside subquery.

SELECT jan_rates.source_currency,
  (SELECT exchange_rate 
   FROM sf_exchange_rate 
   WHERE source_currency = jan_rates.source_currency AND date="2020-07-01") - jan_rates.exchange_rate AS distinction
FROM (SELECT source_currency, exchange_rate
      FROM sf_exchange_rate
      WHERE date="2020-01-01"
) AS jan_rates;

 

This returns an accurate output, however such an answer is unnecessarily sophisticated. A a lot easier resolution, with fewer strains of code, includes self-joining a desk with itself after which making use of two date filtering circumstances within the WHERE clause.

SELECT jan.source_currency,
       jul.exchange_rate - jan.exchange_rate AS distinction
FROM sf_exchange_rate jan
JOIN sf_exchange_rate jul ON jan.source_currency = jul.source_currency
WHERE jan.date="2020-01-01" AND jul.date="2020-07-01";

 

4. Subqueries vs. Widespread Desk Expressions (CTEs)

 
Why It’s Arduous: Folks usually get caught on subqueries as a result of they study them earlier than Widespread Desk Expressions (CTEs) and proceed utilizing them for any question with layered logic. Nonetheless, subqueries can get messy in a short time.

Widespread Mistake: Utilizing deeply nested SELECT statements when CTEs could be a lot easier.

Instance: Within the interview query from Google and Netflix, you could discover the highest actors primarily based on their common film score throughout the style wherein they seem most often.

The answer utilizing CTEs is as follows.

WITH genre_stats AS
  (SELECT actor_name,
          style,
          COUNT(*) AS movie_count,
          AVG(movie_rating) AS avg_rating
   FROM top_actors_rating
   GROUP BY actor_name,
            style),
            
max_genre_count AS
  (SELECT actor_name,
          MAX(movie_count) AS max_count
   FROM genre_stats
   GROUP BY actor_name),
     
top_genres AS
  (SELECT gs.*
   FROM genre_stats gs
   JOIN max_genre_count mgc ON gs.actor_name = mgc.actor_name
   AND gs.movie_count = mgc.max_count),
     
top_genre_avg AS
  (SELECT actor_name,
          MAX(avg_rating) AS max_avg_rating
   FROM top_genres
   GROUP BY actor_name),
   
filtered_top_genres AS
  (SELECT tg.*
   FROM top_genres tg
   JOIN top_genre_avg tga ON tg.actor_name = tga.actor_name
   AND tg.avg_rating = tga.max_avg_rating),
     ranked_actors AS
  (SELECT *,
          DENSE_RANK() OVER (
                             ORDER BY avg_rating DESC) AS rank
   FROM filtered_top_genres),
   
final_selection AS
  (SELECT MAX(rank) AS max_rank
   FROM ranked_actors
   WHERE rank <= 3)
   
SELECT actor_name,
       style,
       avg_rating
FROM ranked_actors
WHERE rank <=
    (SELECT max_rank
     FROM final_selection);

 

It’s comparatively sophisticated, but it surely nonetheless consists of six clear CTEs, with the code’s readability enhanced by clear aliases.

Curious what the identical resolution would seem like utilizing solely subqueries? Right here it’s.

SELECT ra.actor_name,
       ra.style,
       ra.avg_rating
FROM (
    SELECT *,
           DENSE_RANK() OVER (ORDER BY avg_rating DESC) AS rank
    FROM (
        SELECT tg.*
        FROM (
            SELECT gs.*
            FROM (
                SELECT actor_name,
                       style,
                       COUNT(*) AS movie_count,
                       AVG(movie_rating) AS avg_rating
                FROM top_actors_rating
                GROUP BY actor_name, style
            ) AS gs
            JOIN (
                SELECT actor_name,
                       MAX(movie_count) AS max_count
                FROM (
                    SELECT actor_name,
                           style,
                           COUNT(*) AS movie_count,
                           AVG(movie_rating) AS avg_rating
                    FROM top_actors_rating
                    GROUP BY actor_name, style
                ) AS genre_stats
                GROUP BY actor_name
            ) AS mgc
            ON gs.actor_name = mgc.actor_name AND gs.movie_count = mgc.max_count
        ) AS tg
        JOIN (
            SELECT actor_name,
                   MAX(avg_rating) AS max_avg_rating
            FROM (
                SELECT gs.*
                FROM (
                    SELECT actor_name,
                           style,
                           COUNT(*) AS movie_count,
                           AVG(movie_rating) AS avg_rating
                    FROM top_actors_rating
                    GROUP BY actor_name, style
                ) AS gs
                JOIN (
                    SELECT actor_name,
                           MAX(movie_count) AS max_count
                    FROM (
                        SELECT actor_name,
                               style,
                               COUNT(*) AS movie_count,
                               AVG(movie_rating) AS avg_rating
                        FROM top_actors_rating
                        GROUP BY actor_name, style
                    ) AS genre_stats
                    GROUP BY actor_name
                ) AS mgc
                ON gs.actor_name = mgc.actor_name AND gs.movie_count = mgc.max_count
            ) AS top_genres
            GROUP BY actor_name
        ) AS tga
        ON tg.actor_name = tga.actor_name AND tg.avg_rating = tga.max_avg_rating
    ) AS filtered_top_genres
) AS ra
WHERE ra.rank <= (
    SELECT MAX(rank)
    FROM (
        SELECT *,
               DENSE_RANK() OVER (ORDER BY avg_rating DESC) AS rank
        FROM (
            SELECT tg.*
            FROM (
                SELECT gs.*
                FROM (
                    SELECT actor_name,
                           style,
                           COUNT(*) AS movie_count,
                           AVG(movie_rating) AS avg_rating
                    FROM top_actors_rating
                    GROUP BY actor_name, style
                ) AS gs
                JOIN (
                    SELECT actor_name,
                           MAX(movie_count) AS max_count
                    FROM (
                        SELECT actor_name,
                               style,
                               COUNT(*) AS movie_count,
                               AVG(movie_rating) AS avg_rating
                        FROM top_actors_rating
                        GROUP BY actor_name, style
                    ) AS genre_stats
                    GROUP BY actor_name
                ) AS mgc
                ON gs.actor_name = mgc.actor_name AND gs.movie_count = mgc.max_count
            ) AS tg
            JOIN (
                SELECT actor_name,
                       MAX(avg_rating) AS max_avg_rating
                FROM (
                    SELECT gs.*
                    FROM (
                        SELECT actor_name,
                               style,
                               COUNT(*) AS movie_count,
                               AVG(movie_rating) AS avg_rating
                        FROM top_actors_rating
                        GROUP BY actor_name, style
                    ) AS gs
                    JOIN (
                        SELECT actor_name,
                               MAX(movie_count) AS max_count
                        FROM (
                            SELECT actor_name,
                                   style,
                                   COUNT(*) AS movie_count,
                                   AVG(movie_rating) AS avg_rating
                            FROM top_actors_rating
                            GROUP BY actor_name, style
                        ) AS genre_stats
                        GROUP BY actor_name
                    ) AS mgc
                    ON gs.actor_name = mgc.actor_name AND gs.movie_count = mgc.max_count
                ) AS top_genres
                GROUP BY actor_name
            ) AS tga
            ON tg.actor_name = tga.actor_name AND tg.avg_rating = tga.max_avg_rating
        ) AS filtered_top_genres
    ) AS ranked_actors
    WHERE rank <= 3
);

 

There may be redundant logic repeated throughout subqueries. What number of subqueries is that? I don’t know. The code is unattainable to keep up. Although I simply wrote it, I’d nonetheless want half a day to know it if I needed to vary one thing tomorrow. Moreover, the fully meaningless subquery aliases don’t assist.

 

5. Dealing with NULLs in Logic

 
Why It’s Arduous: Candidates usually assume that NULL is the same as one thing. It’s not. NULL isn’t equal to something — not even itself. Logic involving NULLs behaves in a different way from logic involving precise values.

Widespread Mistake: Utilizing = NULL as an alternative of IS NULL in filtering or lacking output rows as a result of NULLs break the situation logic.

Instance: There’s an interview query by IBM that asks you to calculate the overall variety of interactions and the overall variety of contents created for every buyer.

It doesn’t sound too tough, so that you would possibly write this resolution with two CTEs, the place one CTE counts the variety of interactions per buyer, whereas the opposite counts the variety of content material objects created by a buyer. Within the ultimate SELECT, you FULL OUTER JOIN the 2 CTEs, and you’ve got the answer. Proper?

WITH interactions_summary AS
  (SELECT customer_id,
          COUNT(*) AS total_interactions
   FROM customer_interactions
   GROUP BY customer_id),
   
content_summary AS
  (SELECT customer_id,
          COUNT(*) AS total_content_items
   FROM user_content
   GROUP BY customer_id)
   
SELECT i.customer_id,
  i.total_interactions,
  c.total_content_items
FROM interactions_summary AS i
FULL OUTER JOIN content_summary AS c ON i.customer_id = c.customer_id
ORDER BY customer_id;

 

Nearly proper. Right here’s the output. (By the way in which, you see double citation marks (“”) as an alternative of NULL. That’s how the StrataScratch UI shows it, however belief me, the engine nonetheless treats them for what they’re: NULL values).

 
5. Handling NULLs in Logic
 

The highlighted rows comprise NULLs. This makes the output incorrect. A NULL worth is neither the client ID nor the variety of interactions and contents, which the query explicitly asks you to point out.

What we’re lacking within the above resolution is COALESCE() to deal with NULLs within the ultimate SELECT. Now, all the shoppers with out interactions will get their IDs from the content_summary CTE. Additionally, for patrons that don’t have interactions, or content material, or each, we’ll now change NULL with 0, which is a sound quantity.

WITH interactions_summary AS
  (SELECT customer_id,
          COUNT(*) AS total_interactions
   FROM customer_interactions
   GROUP BY customer_id),
   
content_summary AS
  (SELECT customer_id,
          COUNT(*) AS total_content_items
   FROM user_content
   GROUP BY customer_id)
   
SELECT COALESCE(i.customer_id, c.customer_id) AS customer_id,
       COALESCE(i.total_interactions, 0) AS total_interactions,
       COALESCE(c.total_content_items, 0) AS total_content_items
FROM interactions_summary AS i
FULL OUTER JOIN content_summary AS c ON i.customer_id = c.customer_id
ORDER BY customer_id;

 

6. Group-Based mostly Deduplication

 
Why It’s Arduous: Group-based deduplication means you’re choosing one row per group, e.g., “most up-to-date”, “highest rating”, and many others. At first, it sounds such as you solely want to choose one row per person. However you may’t use GROUP BY except you mixture. Then again, you usually want a full row, not a single worth that aggregation and GROUP BY return.

Widespread Mistake: Utilizing GROUP BY + LIMIT 1 (or DISTINCT ON, which is PostgreSQL-specific) as an alternative of ROW_NUMBER() or RANK(), the latter if you would like ties included.

Instance: This query asks you to establish the best-selling merchandise for every month, and there’s no must separate months by yr. One of the best-selling merchandise is calculated as unitprice * amount.

The naive method could be this. First, extract the sale month from invoicedate, choose description, and discover the overall gross sales by summing unitprice * amount. Then, to get the overall gross sales by month and product description, we merely GROUP BY these two columns. Lastly, we solely want to make use of ORDER BY to kind the output from the very best to the worst-selling product and use LIMIT 1 to output solely the primary row, i.e., the best-selling merchandise.

SELECT DATE_PART('MONTH', invoicedate) AS sale_month,
       description,
       SUM(unitprice * amount) AS total_paid
FROM online_retail
GROUP BY sale_month, description
ORDER BY total_paid DESC
LIMIT 1;

 

As I stated, that is naive; the output considerably resembles what we’d like, however we’d like this for each month, not only one.

 
Group-Based Deduplication
 

One of many right approaches is to make use of the RANK() window perform. With this method, we observe an identical methodology to the earlier code. The distinction is that the question now turns into a subquery within the FROM clause. As well as, we use RANK() to partition the information by month after which rank the rows inside every partition (i.e., for every month individually) from the best-selling to the worst-selling merchandise.

Then, in the primary question, we merely choose the required columns and output solely rows the place the rank is 1 utilizing the WHERE clause.

SELECT month,
       description,
       total_paid
FROM
  (SELECT DATE_PART('month', invoicedate) AS month,
          description,
          SUM(unitprice * amount) AS total_paid,
          RANK() OVER (PARTITION BY DATE_PART('month', invoicedate) ORDER BY SUM(unitprice * amount) DESC) AS rnk
   FROM online_retail
   GROUP BY month, description) AS tmp
WHERE rnk = 1;

 

 

Conclusion

 
The six ideas we’ve lined generally seem in SQL coding interview questions. Take note of them, apply interview questions that contain these ideas, study the proper approaches, and also you’ll considerably enhance your probabilities in your interviews.
 
 

Nate Rosidi is an information scientist and in product technique. He is additionally an adjunct professor educating analytics, and is the founding father of StrataScratch, a platform serving to information scientists put together for his or her interviews with actual interview questions from prime firms. Nate writes on the newest tendencies within the profession market, provides interview recommendation, shares information science initiatives, and covers every thing SQL.



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