leetcode_gym/_problems/sql_facebook_practice.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	SQL Facebook Practice Questions		sql		Attempted

sql fiddle

Problem

{% highlight sql %}

create table salesperson ( id int, name varchar, age int, salary int );

create table customer ( id int, name varchar, city varchar, industrytype varchar );

create table orders ( number int, order_date date, cust_id int, salesperson_id int, amount int );

insert into salesperson values (1, 'Abe', 61, 140000); insert into salesperson values (2, 'Bob', 34, 44000); insert into salesperson values (5, 'Chris', 34, 40000); insert into salesperson values (7, 'Dan', 41, 52000); insert into salesperson values (8, 'Ken', 57, 115000); insert into salesperson values (11, 'Joe', 38, 38000);

insert into customer values (4, 'Samsonic', 'pleasant', 'J'); insert into customer values (6, 'Panusung', 'oaktown', 'J'); insert into customer values (7, 'Samony', 'jackson', 'B'); insert into customer values (9, 'Orange', 'jackson', 'B');

insert into orders values (10, '1996-8-2', 4, 2, 540); insert into orders values (20, '1999-1-30', 4, 8, 1800); insert into orders values (30, '1995-7-14', 9, 1, 460 ); insert into orders values (40, '1998-1-29', 7, 2, 2400 ); insert into orders values (50, '1998-2-3', 6, 7, 600 ); insert into orders values (60, '1998-3-2', 6, 7, 720 ); insert into orders values (70, '1992-5-6', 9, 7, 150 );

{% endhighlight %}

Solution

{% highlight sql %}

-- a. The names of all salespeople that have an order with Samsonic. select name from salesperson where id in ( select o.salesperson_id from orders o join customer c on o.cust_id=c.id where c.name='Samsonic' )

-- b. The names of all salespeople that do not have any order with Samsonic. select distinct s.name from orders o join salesperson s on o.salesperson_id=s.id where cust_id not in ( select id from customer where name='Samsonic' )

-- c. The names of salespeople that have 2 or more orders. select name from salesperson where id in ( -- use group/having to get ids of salespeople with count>=2 select o.salesperson_id from salesperson s join orders o on s.id=o.salesperson_id group by o.salesperson_id having count(*)>=2 );

-- d. The names and ages of all salespersons must having a salary of 100,000 or greater. select name, age from salesperson where salary>100000;

-- e. What sales people have sold more than 1400 total units? select distinct s.name from salesperson s join orders o on s.id=o.salesperson_id where o.salesperson_id in ( select salesperson_id from orders group by salesperson_id having sum(amount) > 1400 );

-- f. When was the earliest and latest order made to Samsonite?

select number from orders where cust_id=4 -- join and get customer name if you have to and order_date in (

-- earliest order select min(order_date) from orders o join customer c on c.id=o.cust_id where c.name='Samsonic'

union

-- latest order select max(order_date) from orders o join customer c on c.id=o.cust_id where c.name='Samsonic' )

{% endhighlight %}