1596. The Most Frequently Ordered Products for Each Customer ¶ SQL 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21WITH RankedOrders AS ( SELECT Orders.customer_id, Orders.product_id, Products.product_name, RANK() OVER( PARTITION BY Orders.customer_id ORDER BY COUNT(Orders.product_id) DESC ) AS `rank` FROM Orders INNER JOIN Products USING (product_id) GROUP BY 1, 2 ) SELECT customer_id, product_id, product_name FROM RankedOrders WHERE `rank` = 1;