Solution Explanation for LeetCode Problem 2082: The Number of Rich Customers

This problem requires finding the number of distinct customers who have at least one bill with an amount greater than 500. The provided Store table contains information about bills, including bill_id, customer_id, and amount.

Approach

The most efficient approach involves directly querying the database using SQL. We can achieve this by:

1. Filtering the data: Select only those rows from the Store table where the amount is strictly greater than 500. This filters out any bills with amounts less than or equal to 500.

2. Counting distinct customers: Count the number of unique customer_id values from the filtered data. Using COUNT(DISTINCT customer_id) ensures that each customer is counted only once, even if they have multiple bills exceeding 500.

3. Aliasing the result: Alias the count as rich_count to match the required output format.

SQL Solution (MySQL)

SELECT
    COUNT(DISTINCT customer_id) AS rich_count
FROM Store
WHERE amount > 500;

This SQL query directly implements the described approach. The WHERE clause filters the rows, COUNT(DISTINCT customer_id) counts the unique customers, and AS rich_count assigns the desired alias.

Time and Space Complexity Analysis

• Time Complexity: The time complexity of this SQL query is dominated by the filtering and counting operations. In a well-indexed database (with an index on amount and potentially customer_id), the filtering would likely have a logarithmic or linearithmic time complexity (depending on the specific database engine's implementation). The counting operation also has a linear complexity relative to the number of rows after filtering. Overall, the time complexity can be considered approximately O(N log N) or O(N), where N is the number of rows in the Store table, depending on the database index. In the worst case (no index), it is O(N).

• Space Complexity: The space complexity is primarily determined by the space required to store the intermediate results. This space is generally proportional to the number of distinct customers with amounts greater than 500. Therefore, the space complexity is considered O(M), where M is the number of such customers. In most cases, M is much smaller than N. The space used by the SQL engine's internal operations is also considered constant relative to the input size.

This solution is highly efficient because it leverages the database's built-in optimization capabilities for filtering and aggregation. It avoids the need for any client-side processing, making it suitable for handling large datasets.