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Spotify

Top 5 Artists

Find the top 5 artists whose songs appear most frequently in the Top 10 of the global_song_rank table. Output the artist name and their rank.

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Execution Result

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Problem Context & Learning

đź’ˇWhy This Question Matters

This multi-table ranking problem reflects Spotify's music analytics and chart generation systems. Determining which artists dominate the charts requires joining multiple tables, filtering by rank thresholds, aggregating counts, and then ranking the results—a complex pipeline that tests your ability to chain SQL operations logically. Spotify uses this to assess candidates' skills in building the analytical queries that power their 'Wrapped' features and artist insights.

🔑Key SQL Concepts

Advanced concepts: multiple JOINs across dimension tables (artists, songs, rankings), WHERE filtering before aggregation, GROUP BY for artist-level aggregation, window functions (DENSE_RANK) for ranking aggregated results, and CTEs for query organization. Understanding DENSE_RANK vs RANK is crucial—DENSE_RANK ensures no gaps in rankings when there are ties.

🌍Real-World Applications

Spotify's data teams use similar queries to: generate weekly chart rankings and 'Top Artists' features, create personalized year-end Wrapped summaries, identify trending artists for playlist curation, calculate royalty distributions based on play counts, power artist analytics dashboards showing chart performance, and feed recommendation algorithms with popularity signals.

Interview Insights & Approach

Strategic Approach

When tackling this Spotify problem, the key is to understand the grain of the result. Are you returning one row per user, or one row per category? Always start by identifying your unique join keys and consider if filtered aggregations (CASE WHEN) are more efficient than multiple subqueries.

Common Pitfalls

Be careful with NULL values in your JOIN conditions or aggregate functions. In interview scenarios, datasets often include edge cases like zero-count categories or duplicate entries that can throw off a simple COUNT(*) if not handled with DISTINCT.

Discussion & Solutions

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