How do you solve Product Sales Analysis I on LeetCode?

Product Sales Analysis I (LeetCode #1068) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using JOINs in SQL can significantly reduce the time complexity of queries by avoiding nested loops.

What is the time complexity of Product Sales Analysis I?

The optimal solution for Product Sales Analysis I runs in O(n) time and uses O(n) space. The complexity is O(n) because we are processing each sale exactly once in a single JOIN operation.

What algorithm or data structure is used to solve Product Sales Analysis I?

Product Sales Analysis I uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Product Sales Analysis I?

Always think about how you can reduce the number of operations in your solution. Practice writing SQL queries with JOINs to become comfortable with combining data from multiple tables. Be prepared to explain your thought process and the reasoning behind your chosen approach.

What are common mistakes when solving Product Sales Analysis I?

Not using JOINs effectively and instead relying on multiple queries or nested loops. Overlooking the importance of indexing on foreign keys which can speed up lookups.

Product Sales Analysis I — LeetCode #1068 (Easy)

Tags: Database

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

In the brute force approach, we can iterate through each sale in the Sales table and then for each sale, look up the corresponding product name in the Product table. This is straightforward but inefficient as it requires multiple lookups.

This complexity arises because for each sale (n), we are performing a lookup in the Product table (also n), resulting in O(n * n).

Step 1: For each row in the Sales table, get the product_id.
Step 2: For each product_id, perform a lookup in the Product table to find the product_name.
Step 3: Collect the sale_id, product_name, year, and price for each sale.

1. Sale ID 1: product_id 100 -> lookup Product table -> product_name 'Nokia'. Result: (1, 'Nokia', 2008, 5000). 2. Sale ID 2: product_id 100 -> lookup Product table -> product_name 'Nokia'. Result: (2, 'Nokia', 2009, 5000). 3. Sale ID 7: product_id 200 -> lookup Product table -> product_name 'Apple'. Result: (7, 'Apple', 2011, 9000).

Optimal Solution approach

Time complexity: O(n). Space complexity: O(n).

The optimal solution uses a JOIN operation to combine the Sales and Product tables in a single query. This avoids the need for multiple lookups and is much more efficient.

The complexity is O(n) because we are processing each sale exactly once in a single JOIN operation.

Step 1: Use a SQL JOIN to combine the Sales and Product tables based on the product_id.
Step 2: Select the required columns: sale_id, product_name, year, and price.
Step 3: Return the result set.

1. Combine Sales and Product: (1, 100, 2008, 5000, 'Nokia'), (2, 100, 2009, 5000, 'Nokia'), (7, 200, 2011, 9000, 'Apple'). 2. Result: (1, 'Nokia', 2008, 5000), (2, 'Nokia', 2009, 5000), (7, 'Apple', 2011, 9000).

Key Insights

Using JOINs in SQL can significantly reduce the time complexity of queries by avoiding nested loops.
Understanding the structure of your data (like foreign keys) is crucial for efficient querying.

Common Mistakes

Not using JOINs effectively and instead relying on multiple queries or nested loops.
Overlooking the importance of indexing on foreign keys which can speed up lookups.

Interview Tips

Always think about how you can reduce the number of operations in your solution.
Practice writing SQL queries with JOINs to become comfortable with combining data from multiple tables.
Be prepared to explain your thought process and the reasoning behind your chosen approach.

#1068

Product Sales Analysis I

Easy

Database↗Hash Map↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a JOIN operation to combine the Sales and Product tables in a single query. This avoids the need for multiple lookups and is much more efficient.

⚙️

Algorithm

3 steps

1Step 1: Use a SQL JOIN to combine the Sales and Product tables based on the product_id.
2Step 2: Select the required columns: sale_id, product_name, year, and price.
3Step 3: Return the result set.

solution.py3 lines

1SELECT s.sale_id, p.product_name, s.year, s.price
2FROM Sales s
3JOIN Product p ON s.product_id = p.product_id;

ℹ

Complexity note: The complexity is O(n) because we are processing each sale exactly once in a single JOIN operation.

1Using JOINs in SQL can significantly reduce the time complexity of queries by avoiding nested loops.
2Understanding the structure of your data (like foreign keys) is crucial for efficient querying.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.