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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.
1#include <iostream>
2#include <sqlite3.h>
3
4void findLargestOrderCustomer(sqlite3* db) {
5 sqlite3_stmt* stmt;
6 const char* sql = "SELECT customer_number FROM Orders GROUP BY customer_number ORDER BY COUNT(order_number) DESC LIMIT 1;";
7
8 if (sqlite3_prepare_v2(db, sql, -1, &stmt, 0) == SQLITE_OK) {
9 if (sqlite3_step(stmt) == SQLITE_ROW) {
10 std::cout << "Customer Number: " << sqlite3_column_int(stmt, 0) << std::endl;
11 }
12 sqlite3_finalize(stmt);
13 }
14}
15
16int main() {
17 sqlite3* db;
18 sqlite3_open(":memory:", &db);
19 // Assume table creation and data population.
20 findLargestOrderCustomer(db);
21 sqlite3_close(db);
22 return 0;
23}This C++ solution connects to an SQLite database and executes a SQL query to determine the customer with the maximum number of orders using GROUP BY and ORDER BY with a LIMIT clause for efficiency.
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.