Table: Products
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| product_id | int |
| low_fats | enum |
| recyclable | enum |
+-------------+---------+
product_id is the primary key (column with unique values) for this table.
low_fats is an ENUM (category) of type ('Y', 'N') where 'Y' means this product is low fat and 'N' means it is not.
recyclable is an ENUM (category) of types ('Y', 'N') where 'Y' means this product is recyclable and 'N' means it is not.
Write a solution to find the ids of products that are both low fat and recyclable.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Products table: +-------------+----------+------------+ | product_id | low_fats | recyclable | +-------------+----------+------------+ | 0 | Y | N | | 1 | Y | Y | | 2 | N | Y | | 3 | Y | Y | | 4 | N | N | +-------------+----------+------------+ Output: +-------------+ | product_id | +-------------+ | 1 | | 3 | +-------------+ Explanation: Only products 1 and 3 are both low fat and recyclable.
In this approach, we use a SQL query to select the desired products directly from the database. We utilize the powerful SELECT statement combined with a WHERE clause to filter the rows that meet our conditions. Specifically, we look for rows where both the low_fats column and the recyclable column are equal to 'Y'. This is a simple yet efficient method to retrieve the data required.
The SQL solution involves a straightforward SELECT statement. We select the product_id from the table Products where the columns low_fats and recyclable both have values 'Y'. Simply, we filter our table to get only those entries that meet both conditions.
The time complexity of this SQL query is O(n), where n is the number of rows in the table, because it needs to scan through each row to check the conditions. The space complexity is O(1), assuming the result set fits into memory.
Another approach is to simulate the operation in a procedural programming language. We would read a list of products representing rows from the table and filter this list programmatically. For each product, we check the conditions of being low fat and recyclable. This approach is helpful in scenarios where you deal with data outside databases or wish to manipulate and filter datasets directly in your code.
In the Python approach, we define a list of dictionaries where each dictionary represents a product. We then use a list comprehension to filter products that have both low_fats and recyclable set to 'Y'. We iterate over the list, apply our conditions, and collect the product IDs that fulfill them.
Time complexity is O(n), where n is the number of products, as it involves scanning through each product. The space complexity is also O(n), since we store a portion of the original list in memory when we filter it.
This C program uses a simple hash table to count the frequency of each element in a given set of keys. The hash function modulates the key by the table size to find an index. We then increment the count at the hashtable index corresponding to each key.
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Time Complexity: O(n), as we traverse the input data once.
Space Complexity: O(1) to O(n), as the space used by the hash table can be considered constant in terms of input size if limited by TABLE_SIZE.
This C program sorts the array using qsort and then counts frequencies of sorted elements. It prints the element and its count whenever a new number is encountered in the sorted list.
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Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) if in-place sorting is used, otherwise O(n) if additional space is required for sorting.
| Approach | Complexity |
|---|---|
| SQL Filtering Approach | The time complexity of this SQL query is O(n), where n is the number of rows in the table, because it needs to scan through each row to check the conditions. The space complexity is O(1), assuming the result set fits into memory. |
| Programmatic Filtering Approach | Time complexity is O(n), where n is the number of products, as it involves scanning through each product. The space complexity is also O(n), since we store a portion of the original list in memory when we filter it. |
| Approach 1: Using Hash Maps | Time Complexity: O(n), as we traverse the input data once. |
| Approach 2: Sorting and Counting | Time Complexity: O(n log n) due to sorting. |
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