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Data Scientist SQL Interview Questions

Review this list of 26 SQL Data Scientist interview questions and answers verified by hiring managers and candidates.
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    Video answer for 'Analyze Monthly Customer Transactions'
    Data Scientist
    SQL
    +3 more
  • 13 answers
    +10

    "SELECT upsellcampaignid, COUNT(DISTINCT trans.userid) AS eligibleusers FROM campaign JOIN "transaction" AS trans ON transactiondate BETWEEN datestart AND date_end JOIN user ON trans.userid = user.userid WHERE iseligibleforupsellcampaign = 1 GROUP BY upsellcampaignid `"

    Alina G. - "SELECT upsellcampaignid, COUNT(DISTINCT trans.userid) AS eligibleusers FROM campaign JOIN "transaction" AS trans ON transactiondate BETWEEN datestart AND date_end JOIN user ON trans.userid = user.userid WHERE iseligibleforupsellcampaign = 1 GROUP BY upsellcampaignid `"See full answer

    Data Scientist
    SQL
    +3 more
  • Deloitte logoAsked at Deloitte 
    3 answers

    "BETWEEN and HAVING clauses in SQL serve different purposes: 1. BETWEEN Clause Used to filter rows based on a range of values. Works with numeric, date, or text values. Can be used with WHERE or HAVING clauses. The range includes both lower and upper bounds. Example: Filtering employees with salaries between 30,000 and 50,000 `SELECT * FROM Employees WHERE salary BETWEEN 30000 AND 50000;` 2. HAVING Clause Used to filter **groups"

    Meenakshi D. - "BETWEEN and HAVING clauses in SQL serve different purposes: 1. BETWEEN Clause Used to filter rows based on a range of values. Works with numeric, date, or text values. Can be used with WHERE or HAVING clauses. The range includes both lower and upper bounds. Example: Filtering employees with salaries between 30,000 and 50,000 `SELECT * FROM Employees WHERE salary BETWEEN 30000 AND 50000;` 2. HAVING Clause Used to filter **groups"See full answer

    Data Scientist
    SQL
    +4 more
  • 7 answers
    +4

    "-- Write your query here select u.userid as userid, IFNULL(sum(purchase_value), 0) AS LTV FROM user_sessions u JOIN attribution a ON u.sessionid = a.sessionid group by user_id order by LTV desc ; Needs a full join. Wondering why cant we do a left outer join here. All the sessions should have complete data."

    Aneesha K. - "-- Write your query here select u.userid as userid, IFNULL(sum(purchase_value), 0) AS LTV FROM user_sessions u JOIN attribution a ON u.sessionid = a.sessionid group by user_id order by LTV desc ; Needs a full join. Wondering why cant we do a left outer join here. All the sessions should have complete data."See full answer

    Data Scientist
    SQL
    +3 more
  • Airbnb logoAsked at Airbnb 
    2 answers

    "Clarification questions What is the purpose of connecting the DB? Do we expect high-volumes of traffic to hit the DB Do we have scalability or reliability concerns? Format Code -> DB Code -> Cache -> DB API -> Cache -> DB - APIs are built for a purpose and have a specified protocol (GET, POST, DELETE) to speak to the DB. APIs can also use a contract to retrieve information from a DB much faster than code. Load balanced APIs -> Cache -> DB **Aut"

    Aaron W. - "Clarification questions What is the purpose of connecting the DB? Do we expect high-volumes of traffic to hit the DB Do we have scalability or reliability concerns? Format Code -> DB Code -> Cache -> DB API -> Cache -> DB - APIs are built for a purpose and have a specified protocol (GET, POST, DELETE) to speak to the DB. APIs can also use a contract to retrieve information from a DB much faster than code. Load balanced APIs -> Cache -> DB **Aut"See full answer

    Data Scientist
    SQL
    +6 more
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  • 6 answers
    +3

    " SELECT * FROM( SELECT c.first_name, c.last_name, t.receipt_number, COUNT(*) OVER (PARTITION BY c.customerid) AS noof_offences 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%' ) sub WHERE noofoffences > 1 ; `"

    Harshitha B. - " SELECT * FROM( SELECT c.first_name, c.last_name, t.receipt_number, COUNT(*) OVER (PARTITION BY c.customerid) AS noof_offences 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%' ) sub WHERE noofoffences > 1 ; `"See full answer

    Data Scientist
    SQL
    +3 more
Showing 21-26 of 26
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