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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.
1const sqlite3 = require('sqlite3').verbose();
2
3let db = new sqlite3.Database(':memory:');
4
5// Assuming that database and Orders table are already set up
6let sql = `SELECT customer_number FROM Orders GROUP BY customer_number ORDER BY COUNT(order_number) DESC LIMIT 1;`;
7
8db.get(sql, [], (err, row) => {
9 if (err) {
10 throw err;
11 }
12 console.log(`Customer Number: ${row.customer_number}`);
13});
14
15db.close();This JavaScript code uses the SQLite library to connect to a database and execute a SQL command that groups orders by customer_number and returns the customer with the highest count of orders using JavaScript's asynchronous model.
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.