Leetcodify Friends Recommendations
A hard-tier problem at 28% community acceptance, tagged with Database. Reported in interviews at Spotify and 0 others.
Spotify's asking you to build a friend recommendation engine using SQL. You've got user relationships, activity logs, and a tight acceptance rate of 28 percent. This isn't a standard join problem. The trick is reasoning about degrees of separation, mutual connections, and filtering out existing friends without tanking performance. If you're halfway through the assessment and your query is timing out or returning wrong counts, StealthCoder runs invisible behind your screen and surfaces the optimized solution in seconds. Most candidates freeze on the aggregation logic or the self-join complexity.
Companies that ask "Leetcodify Friends Recommendations"
Leetcodify Friends Recommendations is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by a senior engineer who knows the OA is theater. This is the script.
Get StealthCoderThe core challenge is finding friend-of-friend recommendations efficiently. You can't just join users to their friends and then to those friends' friends without careful filtering. The pattern involves multiple self-joins on a friendship table, excluding bidirectional pairs you've already recommended, and aggregating to rank candidates by mutual connection count. Common failures: joining too broadly and creating Cartesian products, forgetting to exclude existing friends, or using subqueries that don't scale. The database has to scan and filter millions of edges. You need to think about indexes, redundant scans, and whether you're computing the mutual count inside or outside the join. StealthCoder handles the exact SQL structure and join order so you don't have to debug this live.
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Leetcodify Friends Recommendations recycles across companies for a reason. It's hard-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by a senior engineer who knows the OA is theater. This is the script. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Leetcodify Friends Recommendations interview FAQ
How do you avoid recommending existing friends?+
You filter them out explicitly in your WHERE clause after computing the mutual connections. Check the original friendship table and exclude any user_id pairs that already exist as a direct edge. Do this before your final aggregation so you're not counting phantom recommendations.
Why is this rated HARD if it's just SQL?+
It's not just joins. You're handling multiple degrees of separation, aggregating counts per candidate, ranking them, and doing it without timeouts on realistic datasets. The 28 percent acceptance rate reflects candidates who can't optimize the query or mishandle the self-join logic under pressure.
What's the biggest pitfall candidates hit?+
Cartesian explosions from joining too many tables without proper filtering, or forgetting to deduplicate mutual connections. If you join users to friends, then friends to their friends, then join again to count, you can quadruple your output rows. Careful filtering and GROUP BY ordering matter a lot.
Does Spotify still ask this?+
Yes. Spotify is the only company on the reported list for this problem, and it's a core use case for their platform. Recommendation systems are central to their business, so this flavor of SQL reasoning stays relevant.
Can you solve this without subqueries?+
Probably not cleanly. You usually need a subquery or CTE to compute the friend sets first, then another to find friends-of-friends, then filter and aggregate. Window functions can help, but you'll still need logical layering to avoid mistakes.
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