Complete breakdown | Beats 91% ✅ | Easy to understand
Steps: \n\n1. Number of Products in the Product Table\n2. Group Customers and Count the Number of Distinct Products\n3. Customer IDs That Bought All Products sq
# Steps: \n\n1. Number of Products in the Product Table\n2. Group Customers and Count the Number of Distinct Products\n3. Customer IDs That Bought All Products sql\n\n# Complexity: \n\n- Time complexity: $$O(N+M)$$\n- Space complexity: $$O(N)$$\n\n_Try coming up with a justification for these complexities on your own. Explanation in detail after code._ \n\n# Code\n```\n# Write your MySQL query statement below\nSELECT customer_id FROM Customer \nGROUP BY customer_id\nHAVING COUNT(DISTINCT product_key) = (SELECT COUNT(product_key) FROM Product)\n```\n\n## Step 1: Count the Number of Products in the Product Table\n\n```\nSELECT \n    COUNT(product_key) AS total_products\nFROM \n    Product;\n```\n\n## Result:\n\n```\n+---------------+\n| total_products|\n+---------------+\n| 2             |\n+---------------+\n```\n\n## Step 2: Group Customers and Count the Number of Distinct Products They Bought\n\n```\nSELECT \n    customer_id,\n    COUNT(DISTINCT product_key) AS distinct_products_bought\nFROM \n    Customer\nGROUP BY \n    customer_id;\n```\n\n## Result:\n\n```\n+-------------+------------------------+\n| customer_id | distinct_products_bought|\n+-------------+------------------------+\n| 1           | 2                      |\n| 2           | 1                      |\n| 3           | 2                      |\n+-------------+------------------------+\n```\n\n## Step 3: Select Customer IDs That Bought All Products\n```\nSELECT \n    customer_id\nFROM \n    Customer\nGROUP BY \n    customer_id\nHAVING \n    COUNT(DISTINCT product_key) = (SELECT COUNT(product_key) FROM Product);\n```\n\n## Result:\n\n```\n+-------------+\n| customer_id |\n+-------------+\n| 1           |\n| 3           |\n+-------------+\n```\n\n\n---\n\n\n# Complexity breakdown: \n\n1.  Time complexity: \n\n- Let, \n**N**: Number of rows in the Customer table, \n**M**: Number of rows in the Product table.\n\n- `Grouping by Customer ID`:  $$O(N)$$ - Scans through all rows in the Customer table, grouping by customer_id.\n\n- `Counting Distinct Product Keys`: $$O(N)$$ - Performs COUNT(DISTINCT product_key) for each grouped customer.\n\n- `Subquery for Total Product Keys`: $$O(M)$$ - Subquery (SELECT COUNT(product_key) FROM Product) counts the **Total number of product keys in the Product table.**\n\n- `HAVING Clause`: $$O(N)$$ - Number of customers containing all distinct product keys.\n\n- Overall Time Complexity: $$O(N + M)$$ \n\n2. Space complexity:\n\n- The query most likely uses a hash map or similar data structure to group rows by customer_id and store distinct product keys for each customer. \n\nLet, **N: Number of rows in the Customer table**\n\nSpace Complexity: $$O(N)$$\n\n\n
['MySQL']