How do you solve Customers Who Bought All Products on LeetCode?

Customers Who Bought All Products (LeetCode #1045) 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: Understanding how to aggregate data using GROUP BY is crucial.

What is the time complexity of Customers Who Bought All Products?

The optimal solution for Customers Who Bought All Products runs in O(n) time and uses O(n) space. The complexity is O(n) because we only need to iterate through the Customer table and the Product table once each.

What is the brute force solution for Customers Who Bought All Products?

The brute force approach for Customers Who Bought All Products is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each customer to see if they have purchased every product. This is straightforward but can be inefficient, especially with many customers and products. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Customers Who Bought All Products?

Customers Who Bought All Products uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation.

What are the interview tips for Customers Who Bought All Products?

Always clarify the problem requirements before diving into coding. Think about edge cases, such as no customers or no products. Practice writing SQL queries as they can be tricky in interviews.

What are common mistakes when solving Customers Who Bought All Products?

Not considering duplicate purchases when counting products. Forgetting to handle NULL values in the database.

#1045

Customers Who Bought All Products

Medium

DatabaseHash MapAggregation

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 hash map to count purchases per customer and compares it to the total number of products. This reduces the number of checks needed.

⚙️

Algorithm

4 steps

1Step 1: Count the total number of unique products from the Product table.
2Step 2: Create a hash map to count how many unique products each customer has purchased.
3Step 3: Iterate through the Customer table, updating the hash map with product counts for each customer.
4Step 4: Select customers from the hash map whose count matches the total number of unique products.

solution.py8 lines

1# Full working Python code
2WITH ProductCount AS (SELECT COUNT(*) AS total_products FROM Product),
3CustomerProducts AS (SELECT customer_id, COUNT(DISTINCT product_key) AS product_count
4FROM Customer
5GROUP BY customer_id)
6SELECT customer_id
7FROM CustomerProducts
8WHERE product_count = (SELECT total_products FROM ProductCount);

ℹ

Complexity note: The complexity is O(n) because we only need to iterate through the Customer table and the Product table once each.

1Understanding how to aggregate data using GROUP BY is crucial.
2Using hash maps can significantly optimize counting problems.

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