MySQL Solution | Beats 43.72% | Sub-Query
Code\n\nSELECT                                      \n    customer_id\nFROM Customer\nGROUP BY customer_id\nHAVING SUM(DISTINCT product_key) = (\n    SELECT\n  
# Code\n```\nSELECT                                      \n    customer_id\nFROM Customer\nGROUP BY customer_id\nHAVING SUM(DISTINCT product_key) = (\n    SELECT\n        SUM(product_key)\n    FROM Product\n); \n```
['MySQL']