MYSQL SOLUTION WITH COMPLETE EXPLANATION !!!!!!
Intuition\n Describe your first thoughts on how to solve this problem. \nThe query aims to find customer IDs who have purchased all the products from the Custom
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe query aims to find customer IDs who have purchased all the products from the Customer table. It does so by counting the distinct product keys purchased by each customer and comparing that count to the total number of products in the `Product` table.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. The inner subquery groups the customers by their IDs and counts the distinct product keys purchased by each customer using `COUNT(DISTINCT product_key)`.\n\n2. The `HAVING` clause filters out only those customers who have purchased as many distinct products as the total number of products in the `Product` table. This ensures that the customer has bought all the available products.\n\n3. The outer query selects the `customer_id` from the result of the inner subquery, which gives us the customer IDs who meet the criteria.\n\n\n# Complexity\n## Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- The time complexity of this query depends on the size of the `Customer` table and the `Product` table.\n\n- The inner subquery involves grouping and counting for each customer, which can be considered O(N), where N is the number of rows in the `Customer` table.\n- The subquery `(SELECT COUNT(*) FROM Product)` is a constant-time operation, as it calculates the total number of products, which doesn\'t depend on the size of the tables.\n\nSo, the overall time complexity is approximately **`O(N)`**, where N is the **number of customers**.\n\n## Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n- The space complexity is primarily determined by the temporary result set created by the inner subquery. This result set stores the `customer_id` and the count of distinct product keys for each customer.\n\n- **Assuming M is the number of distinct customers who meet the criteria (those who have bought all products), the space complexity is `O(M)`**.\n\nIn terms of space complexity, it\'s essential to consider the size of the result set that needs to be stored in memory.\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT customer_id \nFROM(\n    SELECT customer_id , COUNT(DISTINCT product_key) AS cnt\n    FROM Customer\n    GROUP BY customer_id\n    HAVING cnt = (SELECT COUNT(*) FROM Product)\n) Temp;\n```\n\n## Do Upvote if you got it! :)
['MySQL']