**Architectural Logic**: Simple: SELECT email FROM users GROUP BY email HAVING COUNT(*) > 1. With context: SELECT email, COUNT(*) cnt, ARRAY_AGG(user_id) ids FROM users GROUP BY email HAVING COUNT(*) > 1. Window: SELECT DISTINCT email FROM (SELECT email, COUNT(*) OVER (PARTITION...
This medium-level SQL question appears frequently in data engineering interviews at companies like Daniel Wellington, Goldman Sachs, Swiggy. While less common, it tests deeper understanding that distinguishes strong candidates. Mastering the underlying concepts (partition, sql, window) will help you answer variations of this question confidently.
Break this problem into components. Identify the core trade-offs involved, then walk the interviewer through your reasoning step by step. Demonstrate awareness of edge cases and production considerations - this is what separates good answers from great ones.
Architectural Logic: Simple: SELECT email FROM users GROUP BY email HAVING COUNT() > 1. With context: SELECT email, COUNT() cnt, ARRAY_AGG(user_id) ids FROM users GROUP BY email HAVING COUNT() > 1. Window: SELECT DISTINCT email FROM (SELECT email, COUNT() OVER (PARTITION BY email) cnt FROM users) t WHERE cnt > 1. Why: GROUP BY + HAVING is standard; window useful if you need other columns per duplicate. Scalability: GROUP BY requires full scan; email index can help for small result sets. Cost: All require full scan; GROUP BY is typically most efficient. Best practice: After identification, add UNIQUE constraint and backfill; use MERGE for dedup.
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Analyze My Answer — FreeAccording to DataEngPrep.tech, this is one of the most frequently asked SQL interview questions, reported at 3 companies. DataEngPrep.tech maintains a curated database of 1,863+ real data engineering interview questions across 7 categories, verified by industry professionals.