Interview Questions

Sai vuppalapati M. - "At a high level, the core challenge here revolves around building an effective recommendation algorithm for news. News is an inherently diverse category, spanning various topics and catering to a wide array of user types and personas, such as adults, business professionals, general readers, or specific cohorts with unique interests. Consequently, developing a single, one-size-fits-all recommendation algorithm is not feasible. To enhance the personalization of the news recommendation algorithm,"See full answer

Zain R. - "I've worked on projects not quite like this, but very similar, in the past - I'll borrow from that to answer this: The Broader Context this problem doesn't specify the type of data we're working with, or how it's being ingested to align with my personal background, I'll assume a picture that lends this problem well to being a computer vision (abbreviated "CV") related question: let's say we have a conveyor belt in a waste facility, which sequentially carries a stream of waste w"See full answer

Riku H. - "Great question - this is similiar to the "should uber go into rides for kids?" type of question so it's mostly strategy but also part design the way the question is phrased. Purely design would be " Tell me how you would build uber for kids" Assumption here is Uber in India functions the same as anywhere else in the world, is that correct? Yes, but in India they also take cash payments OK, what about safety regulations like car seats? (needs clarificaitons) What about driver licensing in Ind"See full answer

Salome L. - "SELECT MIN(id) AS id, TRIM(LOWER(email)) AS cleaned_email FROM users GROUP BY cleaned_email ORDER BY id `"See full answer

Akshay D. - "SELECT u.user_id, u.user_name, u.email, ROUND(AVG(CASE WHEN b.status = 'Unmatched' THEN 1.0 ELSE 0 END), 2) AS avgunmatchedbookings FROM users u LEFT JOIN bookings b ON u.userid = b.userid GROUP BY u.user_id, u.user_name, u.email; `"See full answer

Alexey T. - "-- The text of the task is a bit confusing. If the status is repeated several -- times, then in the end you should show as start_date the date of the first -- occurrence, and in end_date the date of the last occurrence of this status, -- and not the date of the beginning of the next status with t1 as (select order_id, status, orderdate as startdate, lead(orderdate) over (partition by orderid order by orderdate) as enddate, ifnull(lag(status) over (partition by order_id order by or"See full answer

Harshi B. - "SELECT DISTINCT title, ROUND(AVG(rating) over (partition by title),1) avg_rating, ROUND(AVG(rating) over (partition by genre),1) genre_rating FROM rating r JOIN movie m ON r.movieid=m.movieid ORDER by 1"See full answer

Lucas G. - "select sub.name subreddit_name, count(distinct us.userid) totalusers from user_subreddit as us left join subreddit as sub on us.subredditid = sub.subredditid group by us.subreddit_id having count(distinct us.user_id) > 3"See full answer

Lucas G. - "SELECT COUNT(*) unique_conversations FROM messenger_sends WHERE senderid < receiverid"See full answer

Gowtami K. - "select DISTINCT p.product_id, p.product_name , CASE when sale_date is null then 'Not Sold' else 'Sold' END as sale_status from products p left join sales s on p.productid= s.productid `"See full answer

Jayveer S. - "SELECT p1.player_name AS player1, p2.player_name AS player2, ABS(p1.level - p2.level) AS level_disparity FROM players p1 JOIN players p2 ON p1.playername < p2.playername WHERE ABS(p1.level - p2.level) <= 5 ORDER BY level_disparity ASC;"See full answer

Akshay D. - "SELECT name, type, CASE WHEN type = 'Electronic' THEN price-(0.10*price) WHEN type = 'Clothing' THEN price-(0.20*price) WHEN type = 'Grocery' THEN price-(0.05*price) WHEN type = 'Book' THEN price-(0.15*price) ELSE price END AS discounted_price FROM products; `"See full answer

Erjan G. - "this task is misleading . i used lag(1) and lead(1) cuz it did not say "compare temperature from 2 days before and 1 day before" , it reads to me as if its asking "compare cur temperature to prev and future and see if it rose and fall""See full answer

Jayveer S. - "WITH suspicious_transactions AS ( SELECT c.first_name, c.last_name, t.receipt_number, COUNT(t.receiptnumber) OVER (PARTITION BY c.customerid) AS noofoffences FROM customers c JOIN transactions t ON c.customerid = t.customerid WHERE t.receipt_number LIKE '%999%' OR t.receipt_number LIKE '%1234%' OR t.receipt_number LIKE '%XYZ%' ) SELECT first_name, last_name, receipt_number, noofoffences FROM suspicious_transactions WHERE noofoffences >= 2;"See full answer

MnM - "SELECT c.customerid, c.customername FROM customers c WHERE rowid%3=0 `"See full answer

Alvin P. - "SELECT e1.empid AS manageremployee_id, e1.empname AS managername, COUNT(e2.empid) AS numberofdirectreports FROM employees AS e1 INNER JOIN employees AS e2 ON e2.managerid = e1.empid GROUP BY e1.emp_id HAVING COUNT(e2.emp_id) >= 2 ORDER BY numberofdirectreports DESC, managername ASC `"See full answer

Gowtami K. - "with cte as (select *, row_number() over(order by score desc) as rn from players) select player_name, score, rn as ranking from cte where rn= 4 or rn =6 or rn =11 `"See full answer