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Product Sales Analysis I - Problem

Database Easy

You're working for a business intelligence team that needs to analyze product sales data. You have two important tables:

Sales Table: Contains transaction records with sale IDs, product IDs, years, quantities, and unit prices
Product Table: Contains product information mapping product IDs to product names

Your task is to create a comprehensive sales report that shows what products were sold, when they were sold, and at what price.

Goal: Write a SQL query to join these tables and return the product_name, year, and price for each sale record.

The sales manager wants to see all sales transactions with readable product names instead of cryptic product IDs.

Sales Table
sale_id (int)	product_id (int)
year (int)	quantity (int)
price (int)

Product Table
product_id (int)	product_name (varchar)

Input & Output

basic_example.sql — SQL Query

$ Input: Sales: [(1,100,2023,5,999), (2,200,2023,3,29), (3,100,2024,2,1099)] Product: [(100,'Laptop'), (200,'Mouse'), (300,'Keyboard')]

› Output: [('Laptop',2023,999), ('Mouse',2023,29), ('Laptop',2024,1099)]

💡 Note: Each sales record is joined with its corresponding product. Sale ID 1 (product 100) matches with 'Laptop', Sale ID 2 (product 200) matches with 'Mouse', and Sale ID 3 (product 100) matches with 'Laptop' again.

single_product.sql — SQL Query

$ Input: Sales: [(1,100,2023,10,1200)] Product: [(100,'Gaming Laptop')]

› Output: [('Gaming Laptop',2023,1200)]

💡 Note: Simple case with only one sales record and one product. The join produces a single result row with the product name, year, and price.

multiple_years.sql — SQL Query

$ Input: Sales: [(1,100,2022,1,800), (2,100,2023,1,900), (3,100,2024,1,1000)] Product: [(100,'Laptop')]

› Output: [('Laptop',2022,800), ('Laptop',2023,900), ('Laptop',2024,1000)]

💡 Note: Same product sold across different years at different prices. Each sales record maintains its individual year and price information after the join.

Constraints

1 ≤ Sales table rows ≤ 10⁵
1 ≤ Product table rows ≤ 10⁴
All product_id values in Sales table exist in Product table
Primary key constraint: (sale_id, year) is unique in Sales table
product_id is primary key in Product table

Visualization

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Understanding the Visualization

Table Preparation

Database loads both Sales and Product tables into memory

Hash Table Creation

Build hash table from smaller Product table using product_id as key

Sales Record Processing

For each Sales record, lookup matching Product using hash table

Result Generation

Combine matched records and output selected columns

Key Takeaway

🎯 Key Insight: SQL INNER JOIN leverages hash tables for O(1) lookups, making it far more efficient than Cartesian products for combining related data from multiple tables.

Asked in

a Amazon 45 ⊞ Microsoft 38 G Google 32 f Meta 28

The optimal solution uses SQL INNER JOIN to efficiently combine the Sales and Product tables. By joining on product_id, we can replace cryptic product IDs with readable product names. The key insight is leveraging the database engine's optimized join algorithms for O(n + m) performance with proper indexing.

Common Approaches

Approach	Time	Space	Notes
✓ Cross Product with WHERE Filter	O(n × m)	O(n × m)	Generate all possible combinations, then filter matching records
Inner Join (Optimal)	O(n + m)	O(min(n,m))	Use SQL INNER JOIN to efficiently match records in one operation

Cross Product with WHERE Filter — Algorithm Steps

Create all possible combinations of Sales and Product rows
Filter combinations where Sales.product_id = Product.product_id
Select required columns: product_name, year, price

Visualization

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Step-by-Step Walkthrough

Generate All Combinations

Create every possible pair of Sales and Product records

Apply WHERE Filter

Keep only pairs where product_ids match

Select Columns

Extract product_name, year, and price from matched pairs

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

char* solution() {
    static char query[] = """
    SELECT p.product_name, s.year, s.price
    FROM Sales s, Product p
    WHERE s.product_id = p.product_id;
    """;
    return query;
}

int main() {
    printf("%s\n", solution());
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

Creates n×m combinations where n is sales records and m is products

✓ Linear Growth

Space Complexity

O(n × m)

Temporarily stores all combinations before filtering

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Cross Product with WHERE Filter — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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