Create 1045-customers_who_bought_all_products.sql

Author: milangrahovac

1045-customers_who_bought_all_products.sql

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+-- 1045 - Customers who bought all products
+  
+-- Table: Customer
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | customer_id | int     |
+-- | product_key | int     |
+-- +-------------+---------+
+
+-- This table may contain duplicates rows. 
+-- customer_id is not NULL.
+-- product_key is a foreign key (reference column) to Product table.
+ 
+
+-- Table: Product
+
+-- +-------------+---------+
+-- | Column Name | Type    |
+-- +-------------+---------+
+-- | product_key | int     |
+-- +-------------+---------+
+
+-- product_key is the primary key (column with unique values) for this table.
+
+-- Write a solution to report the customer ids from the Customer table that bought all the products in the Product table.
+
+-- Return the result table in any order.
+
+-- The result format is in the following example.
+
+ 
+
+-- Example 1:
+
+-- Input: 
+-- Customer table:
+-- +-------------+-------------+
+-- | customer_id | product_key |
+-- +-------------+-------------+
+-- | 1           | 5           |
+-- | 2           | 6           |
+-- | 3           | 5           |
+-- | 3           | 6           |
+-- | 1           | 6           |
+-- +-------------+-------------+
+
+-- Product table:
+-- +-------------+
+-- | product_key |
+-- +-------------+
+-- | 5           |
+-- | 6           |
+-- +-------------+
+
+-- Output: 
+-- +-------------+
+-- | customer_id |
+-- +-------------+
+-- | 1           |
+-- | 3           |
+-- +-------------+
+
+-- Explanation: 
+-- The customers who bought all the products (5 and 6) are customers with IDs 1 and 3.
+
+
+SELECT  
+    customer_id 
+FROM 
+    Customer 
+GROUP BY 
+    customer_id
+HAVING 
+    COUNT(distinct product_key) = (SELECT COUNT(product_key) FROM Product)
+;
+
+-- OR
+
+SELECT
+    customer_id
+FROM
+    (SELECT 
+        customer_id,
+        COUNT(DISTINCT product_key) AS total_unique_products
+    FROM 
+        Customer
+    GROUP BY 
+        customer_id) c
+JOIN
+    (SELECT COUNT(*) AS total FROM Product AS total) p
+WHERE c.total_unique_products = p.total
+;