Is Unique Orders and Customers Per Month easy or hard?

Unique Orders and Customers Per Month is categorized as an Easy database problem. The challenge focuses on understanding SQL aggregation functions such as GROUP BY and COUNT(DISTINCT), rather than complex algorithms or advanced data structures.

Unique Orders and Customers Per Month Python/Java solution

In Python data analysis workflows, Pandas solves the problem using groupby on the extracted month and nunique() for order_id and customer_id columns. In SQL-based systems like MySQL or PostgreSQL, the equivalent solution uses GROUP BY with COUNT(DISTINCT ...).

How to solve Unique Orders and Customers Per Month in O(n)?

Extract the month from the order_date column and group all rows by that value. Within each group, compute COUNT(DISTINCT order_id) and COUNT(DISTINCT customer_id). Because every row is processed once during aggregation, the total runtime remains O(n).

What is the best approach for Unique Orders and Customers Per Month?

The most efficient solution groups records by month and counts distinct order_id and customer_id values within each group. In SQL, this is implemented using GROUP BY with COUNT(DISTINCT ...). The query processes the dataset once, resulting in O(n) time complexity and minimal extra memory usage.

Is Unique Orders and Customers Per Month asked at Google/Amazon/Meta?

Monthly aggregation and distinct counting patterns appear frequently in SQL interview rounds at companies like Amazon, Google, and Meta. Problems like this reflect real analytics queries used for product metrics, reporting dashboards, and customer activity analysis.

What data structure is used in Unique Orders and Customers Per Month?

The underlying mechanism relies on hash-based grouping and distinct sets. SQL engines internally use hash tables or sorting strategies for GROUP BY and COUNT(DISTINCT). In Pandas, groupby combined with nunique() uses hash sets to track unique identifiers per group.

What is the time complexity of Unique Orders and Customers Per Month?

The grouped aggregation solution runs in O(n) time where n is the number of rows in the orders table. Each row is scanned once during grouping and distinct counting. Database engines optimize aggregation internally, so no nested iteration is required.

1565. Unique Orders and Customers Per Month - Practice Online

Table: Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| customer_id   | int     |
| invoice       | int     |
+---------------+---------+
order_id is the column with unique values for this table.
This table contains information about the orders made by customer_id.

Write a solution to find the number of unique orders and the number of unique customers with invoices > $20 for each different month.

Return the result table sorted in any order.

The result format is in the following example.

Example 1:

Input: 
Orders table:
+----------+------------+-------------+------------+
| order_id | order_date | customer_id | invoice    |
+----------+------------+-------------+------------+
| 1        | 2020-09-15 | 1           | 30         |
| 2        | 2020-09-17 | 2           | 90         |
| 3        | 2020-10-06 | 3           | 20         |
| 4        | 2020-10-20 | 3           | 21         |
| 5        | 2020-11-10 | 1           | 10         |
| 6        | 2020-11-21 | 2           | 15         |
| 7        | 2020-12-01 | 4           | 55         |
| 8        | 2020-12-03 | 4           | 77         |
| 9        | 2021-01-07 | 3           | 31         |
| 10       | 2021-01-15 | 2           | 20         |
+----------+------------+-------------+------------+
Output: 
+---------+-------------+----------------+
| month   | order_count | customer_count |
+---------+-------------+----------------+
| 2020-09 | 2           | 2              |
| 2020-10 | 1           | 1              |
| 2020-12 | 2           | 1              |
| 2021-01 | 1           | 1              |
+---------+-------------+----------------+
Explanation: 
In September 2020 we have two orders from 2 different customers with invoices > $20.
In October 2020 we have two orders from 1 customer, and only one of the two orders has invoice > $20.
In November 2020 we have two orders from 2 different customers but invoices < $20, so we don't include that month.
In December 2020 we have two orders from 1 customer both with invoices > $20.
In January 2021 we have two orders from 2 different customers, but only one of them with invoice > $20.