# 1050. 合作过至少三次的演员和导演
# https://leetcode-cn.com/problems/actors-and-directors-who-cooperated-at-least-three-times/

# SQL架构
Create table If Not Exists ActorDirector (actor_id int, director_id int, timestamp int);
Truncate table ActorDirector;
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '1', '0');
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '1', '1');
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '1', '2');
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '2', '3');
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '2', '4');
insert into ActorDirector (actor_id, director_id, timestamp) values ('2', '1', '5');
insert into ActorDirector (actor_id, director_id, timestamp) values ('2', '1', '6');

# Write your MySQL query statement below
select
    actor_id,
    director_id
from
    ActorDirector
group by
    actor_id,
    director_id
having
    count(*) >= 3;

# clean-up
drop table ActorDirector;