Recyclable and Low Fat Products solution and explanation
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The problem
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.
Approach overview
Problem Overview: You are given a Products table containing product_id, low_fats, and recyclable. The task is straightforward: return the IDs of products that are marked as both low fat ('Y') and recyclable ('Y'). This is primarily a filtering problem where you scan records and keep only those matching both conditions.
Approach 1: SQL Filtering Approach (O(n) time, O(1) space)
The most direct solution uses a SQL WHERE clause to filter rows that satisfy both constraints. The query scans the Products table and checks whether low_fats = 'Y' and recyclable = 'Y'. Databases are optimized for this type of filtering, so the engine performs a single pass through the dataset (or uses indexes if available). This approach is ideal when solving the problem directly in SQL or preparing for database-focused interviews involving database queries.
Approach 2: Programmatic Filtering Approach (O(n) time, O(1) space)
If the dataset is represented as objects or rows in application code, iterate through the collection once and check the two attributes for each product. For every entry, evaluate whether both flags equal 'Y'. If the condition holds, append the product_id to the result list. This approach mirrors the SQL filter but is implemented in languages like Python or Java. It relies on a simple linear scan, making it efficient and easy to implement.
Approach 3: Using Hash Maps (O(n) time, O(n) space)
A hash map can store product IDs grouped by their attributes. As you iterate through the dataset, insert entries based on their low_fats and recyclable status. After building the structure, retrieve the IDs that fall into the category where both values are 'Y'. This method is usually unnecessary for this problem but demonstrates how attribute-based grouping can be implemented with constant-time lookups.
Approach 4: Sorting and Counting (O(n log n) time, O(1) extra space)
Another theoretical approach sorts records based on their attribute values so that similar combinations appear together. After sorting, scan the array and collect IDs belonging to the ('Y','Y') group. Sorting introduces an extra O(n log n) cost, so it is less efficient than direct filtering. However, it can be useful when working with datasets that must be ordered for other operations or when applying techniques from sorting algorithms.
Recommended for interviews: The SQL filtering approach is the expected solution for database problems. It demonstrates that you understand how to use conditional filtering in queries and how relational databases process data. The programmatic linear scan is the equivalent approach when solving the problem in a general-purpose language. Brute-force scanning shows basic reasoning, but recognizing that a single-pass filter solves the problem optimally demonstrates practical engineering judgment.
Approach 1: SQL Filtering Approach
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.
SELECT product_id FROM Products WHERE low_fats = 'Y' AND recyclable = 'Y';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.
Approach 2: Programmatic Filtering Approach
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.
products = [
{'product_id': 0, 'low_fats': 'Y', 'recyclable': 'N'},
{'product_id': 1, 'low_fats': 'Y', 'recyclable': 'Y'},
{'product_id': 2, 'low_fats': 'N', 'recyclable': 'Y'},
{'product_id': 3, 'low_fats': 'Y', 'recyclable': 'Y'},
{'product_id': 4, 'low_fats': 'N', 'recyclable': 'N'}
]
result = [product['product_id'] for product in products if product['low_fats'] == 'Y' and product['recyclable'] == 'Y']
print(result)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.
Approach 3: Approach 1: Using Hash Maps
def frequency_counter(arr):
freq_map = {}
for num in arr:
if num in freq_map:
freq_map[num] += 1
else:
freq_map[num] = 1
return freq_map
arr = [1, 2, 3, 4, 2, 3, 3]
freq_map = frequency_counter(arr)
print(freq_map)import java.util.HashMap;
public class Main {
public static void main(String[] args) {
HashMap<Integer, Integer> map = new HashMap<>();
int[] nums = {1, 2, 2, 3, 4, 3};
for (int num : nums) {
map.put(num, map.getOrDefault(num, 0) + 1);
}
map.forEach((key, value) -> System.out.println("Key: " + key + ", Value: " + value));
}
}#include <iostream>
#include <unordered_map>
using namespace std;
int main() {
unordered_map<int, int> map;
int arr[] = {1, 2, 2, 3, 3, 3};
for (int n : arr) {
map[n]++;
}
for (auto& pair : map) {
cout << "Element: " << pair.first << " Frequency: " << pair.second << endl;
}
return 0;
}const arr = [1, 2, 3, 4, 4, 5, 5, 5];
const freqMap = {};
arr.forEach(num => {
if (freqMap[num]) {
freqMap[num] += 1;
} else {
freqMap[num] = 1;
}
});
console.log(freqMap);#include <stdio.h>
#include <stdlib.h>
#define TABLE_SIZE 1000
int hashFunction(int key) {
return key % TABLE_SIZE;
}
void insert(int* hashtable, int key) {
int index = hashFunction(key);
hashtable[index]++;
}
int main() {
int hashtable[TABLE_SIZE] = {0};
int keys[] = {1, 2, 2, 3, 4, 5};
for (int i = 0; i < 6; i++) {
insert(hashtable, keys[i]);
}
// Further implementation can vary depending on the specific task, e.g., check frequency
return 0;
}using System;
using System.Collections.Generic;
class Program {
static void Main() {
Dictionary<int, int> dict = new Dictionary<int, int>();
int[] nums = { 1, 2, 3, 4, 2, 2, 5 };
foreach (int num in nums) {
if (dict.ContainsKey(num)) {
dict[num]++;
} else {
dict[num] = 1;
}
}
foreach (var pair in dict) {
Console.WriteLine($"Key: {pair.Key}, Value: {pair.Value}");
}
}
}This Python script creates a frequency counter for elements in an array using a dictionary. As it iterates over the array, it increments the count for each element seen, or initializes it to 1 if it is the first occurrence.
Time Complexity: O(n).
Space Complexity: O(n), where the dictionary stores key-value pairs for each unique element.
Approach 4: Approach 2: Sorting and Counting
def frequency_counter(arr):
arr.sort()
count = 1
for i in range(1, len(arr)):
if arr[i] == arr[i-1]:
count += 1
else:
print(f'Element: {arr[i-1]}, Frequency: {count}')
count = 1
print(f'Element: {arr[-1]}, Frequency: {count}')
arr = [2, 3, 1, 4, 4, 2, 1]
frequency_counter(arr)import java.util.Arrays;
public class Main {
public static void main(String[] args) {
int[] nums = {4, 2, 2, 3, 3, 1, 1};
Arrays.sort(nums);
int current = nums[0];
int count = 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] == current) {
count++;
} else {
System.out.println("Element: " + current + ", Frequency: " + count);
current = nums[i];
count = 1;
}
}
System.out.println("Element: " + current + ", Frequency: " + count);
}
}#include <iostream>
#include <vector>
#include <algorithm>
int main() {
std::vector<int> vec = {3, 3, 2, 1, 1, 2};
std::sort(vec.begin(), vec.end());
int current = vec[0];
int count = 1;
for (std::size_t i = 1; i < vec.size(); ++i) {
if (vec[i] == current) {
++count;
} else {
std::cout << "Element: " << current << " Frequency: " << count << std::endl;
current = vec[i];
count = 1;
}
}
std::cout << "Element: " << current << " Frequency: " << count << std::endl;
return 0;
}let arr = [9, 8, 7, 9, 7, 9];
arr.sort((a, b) => a - b);
let count = 1;
for (let i = 1; i < arr.length; i++) {
if (arr[i] === arr[i - 1]) {
count++;
} else {
console.log(`Element: ${arr[i - 1]}, Frequency: ${count}`);
count = 1;
}
}
console.log(`Element: ${arr[arr.length - 1]}, Frequency: ${count}`);#include <stdio.h>
#include <stdlib.h>
int compare(const void* a, const void* b) {
return (*(int*)a - *(int*)b);
}
void countFrequency(int* arr, int n) {
qsort(arr, n, sizeof(int), compare);
int count = 1;
for (int i = 1; i < n; i++) {
if (arr[i] == arr[i-1]) {
count++;
} else {
printf("Element: %d, Frequency: %d\n", arr[i-1], count);
count = 1;
}
}
printf("Element: %d, Frequency: %d\n", arr[n-1], count);
}
int main() {
int arr[] = {3, 1, 2, 2, 3, 1};
int n = sizeof(arr) / sizeof(arr[0]);
countFrequency(arr, n);
return 0;
}using System;
using System.Linq;
class Program {
static void Main() {
int[] nums = { 6, 5, 4, 6, 4 };
Array.Sort(nums);
int count = 1;
for (int i = 1; i < nums.Length; i++) {
if (nums[i] == nums[i - 1]) {
count++;
} else {
Console.WriteLine($"Element: {nums[i - 1]}, Frequency: {count}");
count = 1;
}
}
Console.WriteLine($"Element: {nums[nums.Length - 1]}, Frequency: {count}");
}
}This Python solution sorts the input array and counts frequencies of each unique element by comparing adjacent elements, printing the frequency when a different element is encountered.
Time Complexity: O(n log n).
Space Complexity: O(n) due to storage for the sorted array.
Complexity comparison
| 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). |
| Approach 2: Sorting and Counting | Time Complexity: O(n log n). |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| SQL Filtering Approach | O(n) | O(1) | Best for database queries where filtering rows with conditions is required |
| Programmatic Filtering | O(n) | O(1) | When processing product records in application code |
| Hash Map Grouping | O(n) | O(n) | Useful when grouping products by attributes for multiple lookups |
| Sorting and Counting | O(n log n) | O(1) | When records must be ordered before processing |
Frequently asked questions
Is Recyclable and Low Fat Products easy or hard?
Recyclable and Low Fat Products is classified as an Easy problem on LeetCode with a high acceptance rate. It mainly tests basic SQL filtering or simple conditional iteration in code rather than complex algorithms.
Recyclable and Low Fat Products Python/Java solution
In Python or Java, store the records in a list or array and iterate through them once. For each entry, check whether both attributes are 'Y'. If true, append the product_id to the output list. This maintains O(n) time and constant extra space.
How to solve Recyclable and Low Fat Products in O(n)?
Perform a single pass through the dataset and check two conditions for every record: low_fats == 'Y' and recyclable == 'Y'. In SQL, this is implemented using a WHERE clause. In programming languages, iterate through the list and collect product IDs that satisfy both conditions.
What is the best approach for Recyclable and Low Fat Products?
The SQL filtering approach is the most efficient and expected solution. A simple query with a WHERE clause checks whether low_fats = 'Y' and recyclable = 'Y'. This scans the table once, giving O(n) time complexity with constant extra space.
Is Recyclable and Low Fat Products asked at Google/Amazon/Meta?
Problems of this type commonly appear in interviews focused on SQL fundamentals and database querying. Companies such as Amazon, Google, and Meta frequently test candidates on filtering, aggregation, and relational query logic using simple table schemas.
What data structure is used in Recyclable and Low Fat Products?
The core solution does not require advanced data structures. It relies on scanning rows in a database table or iterating through a list of records. Optional implementations may use arrays or hash maps when processing the data programmatically.
What is the time complexity of Recyclable and Low Fat Products?
The optimal solution runs in O(n) time because each row in the Products table is evaluated once. The space complexity is O(1) since only a small result set of product IDs is stored without additional data structures.
