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#1050
Easy Database

Actors and directors who cooperated at least three times

Database
70.9% acceptance
Feb 27, 2026
779
57
Table: ActorDirector +-------------+---------+ | Column Name | Type | +-------------+---------+ | actor_id | int | | director_id | int | | timestamp | int | +-------------+---------+ timestamp is the primary key (column with unique values) for this table. Write a solution to find all the pairs (actor_id, director_id) where the actor has cooperated with the director at least three times. Return the result table in any order. The result format is in the following example.

Solution

SQL
LeetCode
solution.sql
#
# Table: ActorDirector
# +-------------+---------+
# | Column Name | Type    |
# +-------------+---------+
# | actor_id    | int     |
# | director_id | int     |
# | timestamp   | int     |
# +-------------+---------+
# timestamp is the primary key (column with unique values) for this table.
# Write a solution to find all the pairs (actor_id, director_id) where the actor has cooperated with the director at least three times.
# Return the result table in any order.
# The result format is in the following example.
# Example 1:
# Input:
# ActorDirector table:
# +-------------+-------------+-------------+
# | actor_id    | director_id | timestamp   |
# +-------------+-------------+-------------+
# | 1           | 1           | 0           |
# | 1           | 1           | 1           |
# | 1           | 1           | 2           |
# | 1           | 2           | 3           |
# | 1           | 2           | 4           |
# | 2           | 1           | 5           |
# | 2           | 1           | 6           |
# +-------------+-------------+-------------+
# Output:
# +-------------+-------------+
# | actor_id    | director_id |
# +-------------+-------------+
# | 1           | 1           |
# +-------------+-------------+
# Explanation: The only pair is (1, 1) where they cooperated exactly 3 times.
#

# Write your MySQL query statement below

SELECT actor_id, director_id
FROM ActorDirector
GROUP BY actor_id, director_id
HAVING COUNT(*) >= 3;