How do you solve Market Analysis I on LeetCode?

Market Analysis I (LeetCode #1158) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: Using aggregation functions like COUNT() can simplify counting operations.

What is the brute force solution for Market Analysis I?

The brute force approach for Market Analysis I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each user and counting their orders for the year 2019 by iterating through all orders. This method is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n + m).

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

Market Analysis I uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation, JOINs.

What are the interview tips for Market Analysis I?

Always consider edge cases, such as users with no orders. Explain your thought process clearly while writing queries. Practice writing SQL queries to become familiar with JOINs and GROUP BY.

What are common mistakes when solving Market Analysis I?

Not using LEFT JOIN, which can lead to missing users with zero orders. Forgetting to filter orders by the correct year.

#1158

Market Analysis I

Medium

DatabaseHash MapAggregationJOINs

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n + m)Space O(n)

The optimal solution uses a single pass through the orders to create a count of orders per user for 2019, then joins this with the Users table. This significantly reduces the number of operations needed.

⚙️

Algorithm

3 steps

1Step 1: Create a temporary table that counts orders per buyer_id for the year 2019.
2Step 2: Join this temporary table with the Users table to get the user details.
3Step 3: Select user_id, join_date, and the order count from the joined result.

solution.py7 lines

1SELECT u.user_id, u.join_date, IFNULL(order_count, 0) AS order_count 
2FROM Users u 
3LEFT JOIN (SELECT buyer_id, COUNT(order_id) AS order_count 
4            FROM Orders 
5            WHERE YEAR(order_date) = 2019 
6            GROUP BY buyer_id) o 
7ON u.user_id = o.buyer_id;

ℹ

Complexity note: The complexity is linear because we make one pass through the Orders table (m) to count orders and one pass through the Users table (n) to join results.

1Using aggregation functions like COUNT() can simplify counting operations.
2Joining tables effectively can reduce the need for nested loops.

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