MySQL Solution for List the Products Ordered In a Period Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe query aims to find products that had significant sales (100 or more units) within a
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe query aims to find products that had significant sales (100 or more units) within a specific date range.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. The query performs a left join between the "Orders" table (aliased as "o") and the "Products" table (aliased as "p") using the common "product_id" column.\n1. The join condition links the product IDs between the two tables.\n1. The WHERE clause filters the results based on the "order_date" column, restricting it to records between \'2020-02-01\' and \'2020-02-29\'.\n1. The GROUP BY clause groups the results by the "product_id".\n1. The SELECT clause retrieves the product name and calculates the sum of the "unit" column (total units sold) for each product.\n1. The HAVING clause filters the groups based on the sum of the "unit" column, only including groups with a sum of 100 or more units.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- The time complexity of the query depends on the size of the "Orders" table and the number of records within the specified date range.\n- If there are appropriate indexes on the "product_id" column in both tables, the join operation can be performed efficiently, resulting in a time complexity of O(N + M), where N is the number of rows in the "Orders" table and M is the number of rows in the "Products" table.\n- The grouping operation also has a time complexity of O(N), where N is the number of rows within the specified date range.\n- Therefore, the overall time complexity is O(N + M).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n- The space complexity is determined by the size of the result set and the memory required to store temporary data during query execution.\n- The space complexity is typically proportional to the number of distinct product IDs within the specified date range.\n- If there are no additional factors (such as subqueries or large data structures), the space complexity can be considered as O(1) or O(K), where K is the number of distinct product IDs within the specified date range.\n\n# Code\n```\n# Write your MySQL query statement below\nselect product_name, sum(unit) as unit from Orders as o left join Products as p\non o.product_id = p.product_id where order_date between \'2020-02-01\' and \'2020-02-29\'\ngroup by o.product_id having sum(unit) >= 100\n```
['MySQL']