解答
201
SQL文(クエリ)
WITH master AS ( SELECT user_id , product_id FROM sample.sales GROUP BY user_id , product_id ) SELECT prod_combi , COUNT(DISTINCT user_id) AS users FROM ( SELECT m1.user_id , m1.product_id AS prod1 , m2.product_id AS prod2 , concat(m1.product_id, "-", m2.product_id) AS prod_combi FROM master AS m1 INNER JOIN master AS m2 ON m1.user_id = m2.user_id AND m1.product_id < m2.product_id ) GROUP BY prod_combi ORDER BY 2 DESC LIMIT 3