Sponsored
This approach leverages the SQL capabilities of grouping and aggregation to solve the problem efficiently.
First, we perform a GROUP BY operation on the customer_number to count the number of orders per customer. Then, we use ORDER BY in descending order and LIMIT 1 to retrieve only the customer with the maximum count of orders.
Time Complexity: O(n), where n is the number of orders.
Space Complexity: O(1), as we are not using any additional space outside of the database's space.
1import sqlite3
2
3def get_largest_order_customer():
4 connection = sqlite3.connect(':memory:')
5 cursor = connection.cursor()
6 # Assumed table creation and data insertion here.
7 query = '''
8 SELECT customer_number
9 FROM Orders
10 GROUP BY customer_number
11 ORDER BY COUNT(order_number) DESC
12 LIMIT 1;
13 '''
14 cursor.execute(query)
15 result = cursor.fetchone()
16 if result:
17 print(f"Customer Number: {result[0]}")
18 connection.close()
19
20get_largest_order_customer()This Python snippet connects to an SQLite database, assumes the existence of a table called Orders, and executes a query to find the customer with the most orders using Python's sqlite3 module.
This approach involves manually iterating over the orders and using a data structure like a hash map (or dictionary) to count the number of orders per customer. Once counted, we can iterate over this collection to find the customer with the maximum orders.
Time Complexity: O(n), where n is the number of orders.
Space Complexity: O(m), where m is the number of unique customers.
1def find_customer_with_most_orders(orders):
2 order_count = {}
3 # Count orders per customer
This Python code manually counts customer orders using a dictionary to map customer numbers to their respective order counts. Finally, it identifies the customer with the maximum count.