Leetcode SQL

586. Customer Placing the Largest Number of Orders

Query the customer_number from the orders table for the customer who has placed the largest number of orders.

It is guaranteed that exactly one customer will have placed more orders than any other customer.

The orders table is defined as follows:

| Column            | Type      |
|-------------------|-----------|
| order_number (PK) | int       |
| customer_number   | int       |
| order_date        | date      |
| required_date     | date      |
| shipped_date      | date      |
| status            | char(15)  |
| comment           | char(200) |

Sample Input

| order_number | customer_number | order_date | required_date | shipped_date | status | comment |
|--------------|-----------------|------------|---------------|--------------|--------|---------|
| 1            | 1               | 2017-04-09 | 2017-04-13    | 2017-04-12   | Closed |         |
| 2            | 2               | 2017-04-15 | 2017-04-20    | 2017-04-18   | Closed |         |
| 3            | 3               | 2017-04-16 | 2017-04-25    | 2017-04-20   | Closed |         |
| 4            | 3               | 2017-04-18 | 2017-04-28    | 2017-04-25   | Closed |         |

Sample Output

| customer_number |
|-----------------|
| 3               |

Explanation

The customer with number '3' has two orders, which is greater than either customer '1' or '2' because each of them  only has one order. 
So the result is customer_number '3'.

Follow up: What if more than one customer have the largest number of orders, can you find all the customer_number in this case?

Solution:

select customer_number from orders 
group by 1
order by count(distinct order_number) desc limit 1;

Solution for Follow-up:

select customer_number from orders
group by 1
having count(*) >= all(select count(customer_number) from orders
group by customer_number)

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