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kaysss
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leetcode-problem-solutions

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replace-all-s-to-avoid-consecutive-repeating-characters
C++ beats 100%. use a, b or c to fill
c-beats-100-use-a-b-or-c-to-fill-by-cind-0q5v
\nclass Solution {\npublic:\n string modifyString(string s) {\n for (int i=0;i<s.size();i++){\n if (s[i] == \'?\'){\n if (i
cindy7b
NORMAL
2020-12-02T06:06:28.238054+00:00
2020-12-02T06:06:28.238098+00:00
97
false
```\nclass Solution {\npublic:\n string modifyString(string s) {\n for (int i=0;i<s.size();i++){\n if (s[i] == \'?\'){\n if (i > 0 && i < s.size()-1){\n if ((s[i+1] ==\'a\' || s[i+1] ==\'b\') && (s[i-1] ==\'a\' || s[i-1] ==\'b\')){\n s[i] = \'c\';\n continue;\n }\n }\n \n if (i > 0 && s[i-1] ==\'a\'){\n s[i] = \'b\';\n continue;\n }else if (i > 0 && s[i-1] ==\'b\'){\n s[i] = \'a\';\n continue;\n }\n if (i < s.size()-1 && s[i+1] ==\'a\'){\n s[i] = \'b\';\n }else{\n s[i] = \'a\';\n }\n \n }\n }\n return s;\n }\n};\n```
1
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
MY SQL || SOLUTION
my-sql-solution-by-_himanshu_12-rw7o
GROUP BY:\n\nselect actor_id, director_id \nfrom(\nselect actor_id,director_id, \ncount(timestamp) as cooperated \nfrom ActorDirector \ngroup by actor_id,direct
_himanshu_12
NORMAL
2022-10-01T10:58:16.218509+00:00
2022-10-01T11:12:39.520904+00:00
25,985
false
***GROUP BY:***\n```\nselect actor_id, director_id \nfrom(\nselect actor_id,director_id, \ncount(timestamp) as cooperated \nfrom ActorDirector \ngroup by actor_id,director_id) \ntable1\nwhere cooperated>=3;\n```\n\n***GROUP WITH HAVING CLAUSE:***\n```\nselect actor_id,director_id\nfrom ActorDirector \ngroup by actor_id,director_id\nHaving count(timestamp)>=3;\n```\n\n**EXPLANATION:**\n```\n+-------------+-------------+-------------+\n| actor_id | director_id | timestamp |\n+-------------+-------------+-------------+\n| 1 | 1 | 0 |\n| 1 | 1 | 1 |\n| 1 | 1 | 2 |\n| 1 | 2 | 3 |\n| 1 | 2 | 4 |\n| 2 | 1 | 5 |\n| 2 | 1 | 6 |\n+-------------+-------------+-------------+\n\nGroup by actor_id, director_id:\n+-------------+-------------+-------------+\n| actor_id | director_id | timestamp |\n+-------------+-------------+-------------+\n| 1 | 1 | 0, 1, 2 |\n| 1 | 2 | 3, 4 |\n| 2 | 1 | 5. 6 |\n+-------------+-------------+-------------+\n\nGroup by actor_id, director_id\nHaving count(timestamp)>=3\n\n| actor_id | director_id | count(timestamp) |\n+-------------+-------------+-------------+\n| 1 | 1 | 3 |\n| 1 | 2 | 2 |\n| 2 | 1 | 2 |\n+-------------+-------------+-------------+\n\nselect actor_id, director_id from:\n\n| actor_id | director_id \n+-------------+---------+\n| 1 | 1 |\n+-------------+---------+\n\n```\n\n***PLEASE UPVOTE IF YOU FIND IT A LITTLE BIT HELPFUL, MEANS A LOT ;)***
266
0
['MySQL']
19
actors-and-directors-who-cooperated-at-least-three-times
Concise MySQL Solution Using HAVING Clause
concise-mysql-solution-using-having-clau-2ty8
SQL\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(1) >= 3\n\n
zac4
NORMAL
2019-05-23T04:38:31.066287+00:00
2019-07-24T05:41:08.920158+00:00
17,342
false
```SQL\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(1) >= 3\n```\n
58
2
[]
5
actors-and-directors-who-cooperated-at-least-three-times
Simple Solution with GROUP BY
simple-solution-with-group-by-by-simanti-yd04
\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp)>=3\n
simantiri
NORMAL
2021-01-19T18:35:08.144298+00:00
2021-01-19T18:35:08.144363+00:00
5,536
false
```\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp)>=3\n```
31
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
Pandas vs SQL | Elegant & Short | All 30 Days of Pandas solutions ✅
pandas-vs-sql-elegant-short-all-30-days-ggdsj
Complexity\n- Time complexity: O(n)\n- Space complexity: O(n)\n\n# Code\nPython []\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n
Kyrylo-Ktl
NORMAL
2023-08-04T14:58:02.907714+00:00
2023-08-06T17:21:53.154178+00:00
3,783
false
# Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```Python []\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n stats = actor_director.groupby(\n [\'actor_id\', \'director_id\'],\n ).agg(\n count=(\'director_id\', \'count\'),\n ).reset_index()\n return stats[stats[\'count\'] >= 3][[\'actor_id\', \'director_id\']]\n```\n```SQL []\nSELECT actor_id,\n director_id\n FROM ActorDirector\n GROUP BY actor_id,\n director_id\nHAVING count(*) >= 3;\n```\n\n# Important!\n###### If you like the solution or find it useful, feel free to **upvote** for it, it will support me in creating high quality solutions)\n\n# 30 Days of Pandas solutions\n\n### Data Filtering \u2705\n- [Big Countries](https://leetcode.com/problems/big-countries/solutions/3848474/pandas-elegant-short-1-line/)\n- [Recyclable and Low Fat Products](https://leetcode.com/problems/recyclable-and-low-fat-products/solutions/3848500/pandas-elegant-short-1-line/)\n- [Customers Who Never Order](https://leetcode.com/problems/customers-who-never-order/solutions/3848527/pandas-elegant-short-1-line/)\n- [Article Views I](https://leetcode.com/problems/article-views-i/solutions/3867192/pandas-elegant-short-1-line/)\n\n\n### String Methods \u2705\n- [Invalid Tweets](https://leetcode.com/problems/invalid-tweets/solutions/3849121/pandas-elegant-short-1-line/)\n- [Calculate Special Bonus](https://leetcode.com/problems/calculate-special-bonus/solutions/3867209/pandas-elegant-short-1-line/)\n- [Fix Names in a Table](https://leetcode.com/problems/fix-names-in-a-table/solutions/3849167/pandas-elegant-short-1-line/)\n- [Find Users With Valid E-Mails](https://leetcode.com/problems/find-users-with-valid-e-mails/solutions/3849177/pandas-elegant-short-1-line/)\n- [Patients With a Condition](https://leetcode.com/problems/patients-with-a-condition/solutions/3849196/pandas-elegant-short-1-line-regex/)\n\n\n### Data Manipulation \u2705\n- [Nth Highest Salary](https://leetcode.com/problems/nth-highest-salary/solutions/3867257/pandas-elegant-short-1-line/)\n- [Second Highest Salary](https://leetcode.com/problems/second-highest-salary/solutions/3867278/pandas-elegant-short/)\n- [Department Highest Salary](https://leetcode.com/problems/department-highest-salary/solutions/3867312/pandas-elegant-short-1-line/)\n- [Rank Scores](https://leetcode.com/problems/rank-scores/solutions/3872817/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Delete Duplicate Emails](https://leetcode.com/problems/delete-duplicate-emails/solutions/3849211/pandas-elegant-short/)\n- [Rearrange Products Table](https://leetcode.com/problems/rearrange-products-table/solutions/3849226/pandas-elegant-short-1-line/)\n\n\n### Statistics \u2705\n- [The Number of Rich Customers](https://leetcode.com/problems/the-number-of-rich-customers/solutions/3849251/pandas-elegant-short-1-line/)\n- [Immediate Food Delivery I](https://leetcode.com/problems/immediate-food-delivery-i/solutions/3872719/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Count Salary Categories](https://leetcode.com/problems/count-salary-categories/solutions/3872801/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n\n\n### Data Aggregation \u2705\n- [Find Total Time Spent by Each Employee](https://leetcode.com/problems/find-total-time-spent-by-each-employee/solutions/3872715/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Game Play Analysis I](https://leetcode.com/problems/game-play-analysis-i/solutions/3863223/pandas-elegant-short-1-line/)\n- [Number of Unique Subjects Taught by Each Teacher](https://leetcode.com/problems/number-of-unique-subjects-taught-by-each-teacher/solutions/3863239/pandas-elegant-short-1-line/)\n- [Classes More Than 5 Students](https://leetcode.com/problems/classes-more-than-5-students/solutions/3863249/pandas-elegant-short/)\n- [Customer Placing the Largest Number of Orders](https://leetcode.com/problems/customer-placing-the-largest-number-of-orders/solutions/3863257/pandas-elegant-short-1-line/)\n- [Group Sold Products By The Date](https://leetcode.com/problems/group-sold-products-by-the-date/solutions/3863267/pandas-elegant-short-1-line/)\n- [Daily Leads and Partners](https://leetcode.com/problems/daily-leads-and-partners/solutions/3863279/pandas-elegant-short-1-line/)\n\n\n### Data Aggregation \u2705\n- [Actors and Directors Who Cooperated At Least Three Times](https://leetcode.com/problems/actors-and-directors-who-cooperated-at-least-three-times/solutions/3863309/pandas-elegant-short/)\n- [Replace Employee ID With The Unique Identifier](https://leetcode.com/problems/replace-employee-id-with-the-unique-identifier/solutions/3872822/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Students and Examinations](https://leetcode.com/problems/students-and-examinations/solutions/3872699/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Managers with at Least 5 Direct Reports](https://leetcode.com/problems/managers-with-at-least-5-direct-reports/solutions/3872861/pandas-elegant-short/)\n- [Sales Person](https://leetcode.com/problems/sales-person/solutions/3872712/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n\n\n
28
0
['Python', 'Python3', 'MySQL', 'Pandas']
2
actors-and-directors-who-cooperated-at-least-three-times
[Pandas] || 2-step simple code with comments...
pandas-2-step-simple-code-with-comments-fht54
\n\n# Code\n\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n # Group the data by actor_id and director_id,
sriganesh777
NORMAL
2023-08-22T18:10:26.167604+00:00
2023-08-22T18:10:26.167633+00:00
2,810
false
\n\n# Code\n```\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n # Group the data by actor_id and director_id, and count the number of cooperations\n grouped = actor_director.groupby([\'actor_id\', \'director_id\']).size().reset_index(name=\'cooperation_count\')\n \n # Filter the pairs where the cooperation count is at least three\n filtered_pairs = grouped[grouped[\'cooperation_count\'] >= 3]\n \n return filtered_pairs[[\'actor_id\', \'director_id\']]\n\n```
15
0
['Pandas']
0
actors-and-directors-who-cooperated-at-least-three-times
Simple and fast MySQL solution in one line
simple-and-fast-mysql-solution-in-one-li-mdfy
\nselect actor_id,director_id from ActorDirector group by 1,2 having count(*)>=3;\n\n\nNote: If you find it helpful then upvote. Cheers!!
harshitdwivedi01
NORMAL
2022-08-15T04:43:15.744584+00:00
2022-08-15T04:43:15.744623+00:00
2,238
false
```\nselect actor_id,director_id from ActorDirector group by 1,2 having count(*)>=3;\n```\n\nNote: If you find it helpful then upvote. Cheers!!
14
0
['MySQL']
2
actors-and-directors-who-cooperated-at-least-three-times
Easy and Simple Approach😊
easy-and-simple-approach-by-professional-clhg
\n# Code\n\nselect actor_id , director_id\nfrom ActorDirector\ngroup by actor_id,director_id\nhaving count(timestamp)>=3\n\n\n
ProfessionalMonk
NORMAL
2023-08-02T02:56:49.741096+00:00
2023-08-02T02:56:49.741113+00:00
873
false
\n# Code\n```\nselect actor_id , director_id\nfrom ActorDirector\ngroup by actor_id,director_id\nhaving count(timestamp)>=3\n```\n![1271c6fe-345b-4f3d-959e-d13e791111d7_1677726484.8832572.jpeg](https://assets.leetcode.com/users/images/2079f1fc-d489-45a8-a099-ed1e5832c642_1690944996.0136197.jpeg)\n
12
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Another Solution Using Window Function
another-solution-using-window-function-b-yttj
SELECT DISTINCT(actor_id), director_id\nFROM(\nSELECT actor_id, director_id , COUNT(timestamp) OVER (Partition by actor_id, director_id) as number_of_times\nFRO
joyjiang
NORMAL
2022-04-11T00:56:14.044735+00:00
2022-04-11T00:56:14.044765+00:00
644
false
SELECT DISTINCT(actor_id), director_id\nFROM(\nSELECT actor_id, director_id , COUNT(timestamp) OVER (Partition by actor_id, director_id) as number_of_times\nFROM ActorDirector) AS temp\nWHERE number_of_times >= 3
9
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
Solution Using MySQL GROUP BY and HAVING
solution-using-mysql-group-by-and-having-18uu
\nSELECT ACTOR_ID, DIRECTOR_ID\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(*) > 2\n
huangguoxin0512
NORMAL
2019-05-25T21:43:14.910879+00:00
2019-05-25T21:43:54.902710+00:00
5,317
false
```\nSELECT ACTOR_ID, DIRECTOR_ID\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(*) > 2\n```
9
0
[]
1
actors-and-directors-who-cooperated-at-least-three-times
MYSQL SOLN.
mysql-soln-by-hemant_1451-wbta
\n# Code\n\n# Write your MySQL query statement below\nSELECT ACTOR_ID, DIRECTOR_ID FROM ACTORDIRECTOR GROUP BY ACTOR_ID,DIRECTOR_ID HAVING COUNT(TIMESTAMP)>2;\n
Hemant_1451
NORMAL
2023-01-10T01:41:53.888213+00:00
2023-01-10T01:41:53.888250+00:00
3,152
false
\n# Code\n```\n# Write your MySQL query statement below\nSELECT ACTOR_ID, DIRECTOR_ID FROM ACTORDIRECTOR GROUP BY ACTOR_ID,DIRECTOR_ID HAVING COUNT(TIMESTAMP)>2;\n```
8
0
['Database', 'MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
[MySQL] || GROUP BY || Beginner Friendly
mysql-group-by-beginner-friendly-by-coma-49uc
Write your MySQL query statement below\n\n\nSELECT actor_id , director_id\nFROM ActorDirector \nGROUP BY actor_id , director_id\nHAVING COUNT(timestamp)>=3;\n
comauro7511
NORMAL
2022-10-13T18:17:46.437931+00:00
2022-10-14T15:56:48.604253+00:00
2,506
false
# Write your MySQL query statement below\n```\n\nSELECT actor_id , director_id\nFROM ActorDirector \nGROUP BY actor_id , director_id\nHAVING COUNT(timestamp)>=3;\n```
7
0
['MySQL']
2
actors-and-directors-who-cooperated-at-least-three-times
1050. Actors and Directors Who Cooperated At Least Three Times
1050-actors-and-directors-who-cooperated-5csm
```\nSELECT actor_id, director_id FROM ActorDirector\nGROUP BY actor_id, director_id HAVING COUNT(timestamp)>2;\n
Spaulding_
NORMAL
2022-09-07T18:33:52.999003+00:00
2022-09-07T18:33:52.999045+00:00
979
false
```\nSELECT actor_id, director_id FROM ActorDirector\nGROUP BY actor_id, director_id HAVING COUNT(timestamp)>2;\n
7
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Easy solution with HAVING COUNT(a =b)
easy-solution-with-having-counta-b-by-mm-3d4b
SELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(actor_id = director_id) >= 3;
mma4
NORMAL
2020-07-15T18:07:18.291568+00:00
2020-07-15T18:07:18.291599+00:00
1,207
false
SELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(actor_id = director_id) >= 3;
7
2
[]
2
actors-and-directors-who-cooperated-at-least-three-times
Easy my~SQL sol || think of group by / having / count
easy-mysql-sol-think-of-group-by-having-1p5gm
\nselect actor_id , director_id \nfrom actordirector \ngroup by actor_id ,director_id\nhaving count(*)>=3 ; \n\nThanks
abhi-kuks
NORMAL
2022-09-12T11:12:59.749406+00:00
2022-09-12T11:12:59.749443+00:00
1,132
false
```\nselect actor_id , director_id \nfrom actordirector \ngroup by actor_id ,director_id\nhaving count(*)>=3 ; \n```\n***Thanks***
6
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
MySQL - Slightly Optimized Solution
mysql-slightly-optimized-solution-by-xor-ls5y
\nSELECT actor_id as ACTOR_ID, director_id as DIRECTOR_ID\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING(COUNT(DISTINCT ActorDirector.timestamp) >=
xorobotxo
NORMAL
2019-09-08T19:38:33.938459+00:00
2019-09-08T19:38:33.938528+00:00
2,184
false
```\nSELECT actor_id as ACTOR_ID, director_id as DIRECTOR_ID\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING(COUNT(DISTINCT ActorDirector.timestamp) >= 3);\n```\n\nMy original solution included a ```COUNT(*)``` which produced a result faster than only 5% of other submissions.\nAfter switching to counting only the distinct timestamps, the result proved faster than 42% of other submissions - which is still not ideal, but an interesting testiment to incremental query optimization.
6
0
[]
3
actors-and-directors-who-cooperated-at-least-three-times
✅ 100% EASY || FAST 🔥|| CLEAN SOLUTION 🌟
100-easy-fast-clean-solution-by-kartik_k-l6f6
\n\n# Code\n\n/* Write your PL/SQL query statement below */\nSELECT actor_id, director_id FROM ActorDirector GROUP BY \n\nactor_id, director_id HAVING COUNT(dir
kartik_ksk7
NORMAL
2024-06-03T09:51:10.211090+00:00
2024-06-03T09:51:10.211108+00:00
522
false
\n\n# Code\n```\n/* Write your PL/SQL query statement below */\nSELECT actor_id, director_id FROM ActorDirector GROUP BY \n\nactor_id, director_id HAVING COUNT(director_id) >=3 ;\n```\n![5kej8w.jpg](https://assets.leetcode.com/users/images/639e5323-3f53-44ab-9781-325497bdb78b_1717408268.9256334.jpeg)\n
5
0
['Database', 'Oracle']
0
actors-and-directors-who-cooperated-at-least-three-times
Easy solution || MySQL || Pandas || Beats 100% !!
easy-solution-mysql-pandas-beats-100-by-t83dk
Code\n\n# Write your MySQL query statement below\nSELECT actor_id,director_id\nFROM ActorDirector \nGROUP BY actor_id,director_id\nHAVING count(timestamp)>=3;\n
prathams29
NORMAL
2023-08-01T17:02:50.362654+00:00
2023-08-01T17:02:50.362681+00:00
1,245
false
# Code\n```\n# Write your MySQL query statement below\nSELECT actor_id,director_id\nFROM ActorDirector \nGROUP BY actor_id,director_id\nHAVING count(timestamp)>=3;\n```\n```\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n df = actor_director.groupby([\'actor_id\',\'director_id\']).agg(count =(\'director_id\',\'count\')).reset_index()\n df = df[df[\'count\'] >= 3]\n return df[[\'actor_id\',\'director_id\']]\n```
5
0
['MySQL', 'Pandas']
1
actors-and-directors-who-cooperated-at-least-three-times
Easy to understand || SQL
easy-to-understand-sql-by-yashwardhan24_-7x3y
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
yashwardhan24_sharma
NORMAL
2023-03-09T05:59:41.122335+00:00
2023-03-09T05:59:41.122368+00:00
2,421
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp)>=3;\n```
5
0
['Database', 'MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
📌📌 easy solution :D || MySql ⚡
easy-solution-d-mysql-by-maary-h51x
\nSELECT actor_id, director_id \nFROM\n (SELECT actor_id, director_id, COUNT(*) AS cnt\n FROM ActorDirector\n GROUP BY actor_id, director_id\n ) A
maary_
NORMAL
2023-03-05T00:49:07.037819+00:00
2023-03-05T22:43:15.245222+00:00
1,170
false
```\nSELECT actor_id, director_id \nFROM\n (SELECT actor_id, director_id, COUNT(*) AS cnt\n FROM ActorDirector\n GROUP BY actor_id, director_id\n ) AS t\nWHERE cnt >= 3;\n```
5
0
['MySQL', 'MS SQL Server']
2
actors-and-directors-who-cooperated-at-least-three-times
mysql, group by, having count(*)
mysql-group-by-having-count-by-nov05-riww
https://leetcode.com/submissions/detail/865539807/\n\n# Write your MySQL query statement below\nselect actor_id, director_id\nfrom ActorDirector\ngroup by actor
nov05
NORMAL
2022-12-26T03:56:19.670910+00:00
2022-12-26T03:56:19.670935+00:00
1,200
false
https://leetcode.com/submissions/detail/865539807/\n```\n# Write your MySQL query statement below\nselect actor_id, director_id\nfrom ActorDirector\ngroup by actor_id, director_id\nhaving count(*) >= 3\n```
5
0
[]
2
actors-and-directors-who-cooperated-at-least-three-times
Simple Group By
simple-group-by-by-shivamleet07-55z0
\nSELECT ACTOR_ID, DIRECTOR_ID\n\tFROM ActorDirector\nGROUP BY ACTOR_ID, DIRECTOR_ID\n\tHAVING COUNT(*) >= 3\n
shivamleet07
NORMAL
2022-07-30T17:57:13.344340+00:00
2022-07-30T17:57:13.344382+00:00
331
false
```\nSELECT ACTOR_ID, DIRECTOR_ID\n\tFROM ActorDirector\nGROUP BY ACTOR_ID, DIRECTOR_ID\n\tHAVING COUNT(*) >= 3\n```
5
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Optimised Approach || Group By
optimised-approach-group-by-by-vipul_05_-m7b1
IntuitionWe need to find actor-director pairs who have collaborated on 3 or more movies. By grouping by both actor_id and director_id, we can count the number o
Vipul_05_Jain
NORMAL
2025-01-23T07:37:44.224225+00:00
2025-01-23T07:37:44.224225+00:00
401
false
# Intuition We need to find actor-director pairs who have collaborated on 3 or more movies. By grouping by both actor_id and director_id, we can count the number of times each pair has worked together. Using HAVING COUNT(timestamp) >= 3, we filter the pairs with 3 or more collaborations. # Approach 1. Group by both actor_id and director_id to aggregate data by actor-director pairs. 2. Count the number of collaborations (using timestamp or any other field that indicates a collaboration). 3. Filter the pairs where the count of collaborations is 3 or more using HAVING. # Please upvote guys...!! # Code ```mysql [] # Write your MySQL query statement below SELECT actor_id, director_id FROM ActorDirector GROUP BY actor_id, director_id HAVING COUNT(timestamp)>=3 ```
4
0
['MySQL']
1
actors-and-directors-who-cooperated-at-least-three-times
MySQL Different Solution | With & Row_Number() Functions |
mysql-different-solution-with-row_number-do5g
\n# Code\nmysql []\nwith cte as (select *, row_number() over(partition by actor_id, director_id) as r from actordirector)\n\nselect distinct actor_id, director_
Kamal_Kumar123
NORMAL
2024-09-08T05:28:43.517651+00:00
2024-09-08T05:28:43.517672+00:00
196
false
\n# Code\n```mysql []\nwith cte as (select *, row_number() over(partition by actor_id, director_id) as r from actordirector)\n\nselect distinct actor_id, director_id from cte where r >= 3\n\n```
4
0
['MySQL']
1
actors-and-directors-who-cooperated-at-least-three-times
easy, clean pandas solution
easy-clean-pandas-solution-by-pooya_rost-4yjj
Intuition\neasy, clean pandas solution\n\n# Approach\ngroupby the data by combination of \'actor_id\' and \'director_id\' and count their occurance.\nnext, take
pooya_rostami_m
NORMAL
2023-11-20T14:14:01.682892+00:00
2023-11-20T14:14:01.682926+00:00
452
false
# Intuition\neasy, clean pandas solution\n\n# Approach\ngroupby the data by combination of \'actor_id\' and \'director_id\' and count their occurance.\nnext, take the ones that happen at least three times and show the \'actor_id\' and \'director_id\'\n\n# Complexity\n- Time complexity:\nnot sure\n\n- Space complexity:\nnot sure\n\n# Code\n```\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n return (\n actor_director\n .groupby(by=[\'actor_id\', \'director_id\'], as_index=False)\n .agg({\'timestamp\': \'count\'})\n .query(\'timestamp >= 3\')\n [[\'actor_id\', \'director_id\']]\n )\n```
4
0
['Pandas']
0
actors-and-directors-who-cooperated-at-least-three-times
🔥 Pandas Easy and Concise Solution For Beginners 💯
pandas-easy-and-concise-solution-for-beg-i6xl
\uD83D\uDD3C IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE \uD83D\uDC4D\n\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.Dat
pniraj657
NORMAL
2023-08-28T11:39:15.827313+00:00
2023-08-28T11:39:15.827336+00:00
284
false
**\uD83D\uDD3C IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE \uD83D\uDC4D**\n```\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n \n result = actor_director.groupby([\'actor_id\', \'director_id\']).count().reset_index()\n \n return result[result[\'timestamp\']>=3][[\'actor_id\', \'director_id\']]\n```\n**Thank you for reading! \uD83D\uDE04 Comment if you have any questions or feedback.**
4
0
['Python', 'Python3']
0
actors-and-directors-who-cooperated-at-least-three-times
MySQL Solution for Actors and Directors Who Cooperated At Least Three Times Problem
mysql-solution-for-actors-and-directors-et5q2
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the given solution is to find actor-director pairs that have worke
Aman_Raj_Sinha
NORMAL
2023-05-18T03:30:40.877203+00:00
2023-05-18T03:30:40.877302+00:00
966
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the given solution is to find actor-director pairs that have worked together in at least three movies. The approach involves grouping the rows in the ActorDirector table by actor_id and director_id, and then applying a condition in the HAVING clause to count the number of occurrences of each actor-director pair and filter out those with a count of less than three.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Group the rows in the ActorDirector table by actor_id and director_id.\n1. For each group, count the number of occurrences of the actor-director pair.\n1. Filter out the groups where the count is less than three.\n1. Return the actor_id and director_id values from the remaining groups.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this solution depends on the size of the ActorDirector table. The grouping operation requires scanning the ActorDirector table, which has a time complexity of O(n), where n is the number of rows in the ActorDirector table. Additionally, the count operation requires counting the occurrences of each actor-director pair, which also has a time complexity of O(n).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this solution depends on the number of distinct actor_id and director_id pairs that satisfy the condition. The query requires storing the grouped actor_id and director_id values temporarily, so the space complexity is determined by the cardinality of these values.\n\n# Code\n```\n# Write your MySQL query statement below\nselect actor_id, director_id from ActorDirector\ngroup by actor_id, director_id\nhaving count(*) >= 3\n```
4
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
concise mysql solution using concat
concise-mysql-solution-using-concat-by-9-h0gs
\nselect actor_id, director_id from actordirector group by actor_id, director_id having count(concat(actor_id, director_id)) >= 3;\n
96sayak
NORMAL
2022-11-27T07:08:42.190316+00:00
2022-11-27T07:08:42.190342+00:00
434
false
```\nselect actor_id, director_id from actordirector group by actor_id, director_id having count(concat(actor_id, director_id)) >= 3;\n```
4
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
MS SQL Server 99.5%
ms-sql-server-995-by-guseowhtjs-bkc3
\n/* Write your T-SQL query statement below */\nselect actor_id, director_id\n from actordirector\n group by actor_id, director_id\n having count(timestamp) >
guseowhtjs
NORMAL
2022-07-28T12:48:49.545463+00:00
2022-07-28T12:48:49.545507+00:00
703
false
```\n/* Write your T-SQL query statement below */\nselect actor_id, director_id\n from actordirector\n group by actor_id, director_id\n having count(timestamp) > 2\n```
4
0
['MS SQL Server']
0
actors-and-directors-who-cooperated-at-least-three-times
✅MySQL using "timestamp"
mysql-using-timestamp-by-ranbeer_singh-o1uj
\nSELECT\n actor_id,\n director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp) > 2\n
Ranbeer_Singh
NORMAL
2022-07-16T03:38:19.911703+00:00
2022-07-16T03:38:19.911748+00:00
563
false
```\nSELECT\n actor_id,\n director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp) > 2\n```
4
0
['MySQL']
1
actors-and-directors-who-cooperated-at-least-three-times
Simple and Clean
simple-and-clean-by-nkalash-82e7
SELECT ACTOR_ID, DIRECTOR_ID FROM ACTORDIRECTOR \nGROUP BY ACTOR_ID, DIRECTOR_ID \nHAVING COUNT(*) >= 3
nkalash
NORMAL
2022-05-24T07:45:02.931582+00:00
2022-05-24T07:45:02.931611+00:00
667
false
SELECT ACTOR_ID, DIRECTOR_ID FROM ACTORDIRECTOR \nGROUP BY ACTOR_ID, DIRECTOR_ID \nHAVING COUNT(*) >= 3
4
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
MS SQL, MySQL and Oracle solution with GROUP BY and HAVING
ms-sql-mysql-and-oracle-solution-with-gr-1uiw
\nSELECT\n actor_id,\n director_id\nFROM\n ActorDirector\nGROUP BY\n actor_id,\n director_id\nHAVING \n COUNT(1) > 2\n
vazhev
NORMAL
2022-04-23T20:40:09.170419+00:00
2022-04-23T20:40:09.170447+00:00
535
false
```\nSELECT\n actor_id,\n director_id\nFROM\n ActorDirector\nGROUP BY\n actor_id,\n director_id\nHAVING \n COUNT(1) > 2\n```
4
0
['MySQL', 'Oracle', 'MS SQL Server']
1
actors-and-directors-who-cooperated-at-least-three-times
Simple Query ✅ | With Proper Explanation 🤝
simple-query-with-proper-explanation-by-r0not
IntuitionThe problem requires us to find all pairs of actors and directors who have worked together on at least three movies. This means we need to identify fre
NadeemMohammed
NORMAL
2025-03-06T04:32:12.495492+00:00
2025-03-06T04:32:12.495492+00:00
212
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> The problem requires us to find all pairs of actors and directors who have worked together on at least three movies. This means we need to identify frequent collaborations between actors and directors by counting how many times each pair appears in the ActorDirector table. # Approach <!-- Describe your approach to solving the problem. --> 1. Group the data: We use GROUP BY actor_id, director_id to group records based on the actor_id and director_id, ensuring that each unique pair is considered. 2. Count the occurrences: Using COUNT(*), we count how many times each actor-director pair appears in the table, which represents the number of movies they have worked on together. 3. Filter out pairs with fewer than 3 collaborations: The HAVING clause ensures that only pairs where COUNT(*) >= 3 are included in the result. 4. Select the required columns: Finally, we retrieve actor_id and director_id for the qualifying pairs. # Complexity - Time complexity: <!-- Add your time complexity here, e.g. $$O(n)$$ --> - Space complexity: <!-- Add your space complexity here, e.g. $$O(n)$$ --> If you found my solution helpful, I’d really appreciate it if you could give it an upvote! 🙌 Your support keeps me motivated to share more simple and efficient solutions. Thank you! 😊 # Code ```postgresql [] -- Write your PostgreSQL query statement below SELECT actor_id, director_id FROM ActorDirector GROUP BY actor_id, director_id HAVING COUNT(*) >= 3 ```
3
0
['PostgreSQL']
1
actors-and-directors-who-cooperated-at-least-three-times
sql solution using case
sql-solution-using-case-by-sachin_kumar_-70iz
IntuitionApproachComplexity Time complexity: Space complexity: Code
Sachin_Kumar_Sharma
NORMAL
2025-01-19T10:15:55.177903+00:00
2025-01-19T10:15:55.177903+00:00
84
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> # Approach <!-- Describe your approach to solving the problem. --> # Complexity - Time complexity: <!-- Add your time complexity here, e.g. $$O(n)$$ --> - Space complexity: <!-- Add your space complexity here, e.g. $$O(n)$$ --> # Code ```mysql [] select actor_id,director_id from ActorDirector group by actor_id,director_id having sum( case when (actor_id/director_id) then 1 else 0 end ) >= 3 ```
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Simple solution beats 99.9%
simple-solution-beats-999-by-rshikharev-49hw
Intuition\nThe task requires us to find pairs of actor_id and director_id who have collaborated at least three times. The initial thought is to group the data b
rshikharev
NORMAL
2024-11-05T15:56:06.347046+00:00
2024-11-05T15:57:26.636779+00:00
485
false
# Intuition\nThe task requires us to find pairs of `actor_id` and `director_id` who have collaborated at least three times. The initial thought is to group the data by each unique actor-director pair and then filter out the pairs that meet the collaboration threshold.\n\n# Approach\n1. **GROUP BY**: Use `GROUP BY actor_id, director_id` to aggregate rows for each actor-director pair.\n2. **COUNT & HAVING**: We use `COUNT(timestamp)` to determine the number of collaborations for each pair, and `HAVING COUNT(timestamp) >= 3` to retain only those pairs with at least three collaborations.\n\nThis approach is efficient, leveraging SQL\'s aggregation and filtering capabilities to retrieve the required data.\n\n# Complexity\n- **Time complexity**: $$O(n)$$, where $$n$$ is the number of rows in the `ActorDirector` table. The grouping and counting operations depend linearly on the size of the input.\n- **Space complexity**: $$O(k)$$, where $$k$$ is the number of unique actor-director pairs, as the grouping requires storage for each unique pair.\n\n# Code\n```mysql []\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp) >= 3;\n```
3
0
['Database', 'MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Beginner Friendly Solution🌻
beginner-friendly-solution-by-ravithemor-x4hr
Intuition:\nThe SQL query retrieves pairs of actor and director IDs from the ActorDirector table where the same actor has worked with the same director at least
ravithemore
NORMAL
2023-11-17T06:33:45.050843+00:00
2023-11-17T06:33:45.050895+00:00
286
false
**Intuition:**\nThe SQL query retrieves pairs of actor and director IDs from the `ActorDirector` table where the same actor has worked with the same director at least three times, based on the count of timestamps.\n\n**Approach:**\n1. The query groups the records in the `ActorDirector` table by pairs of `actor_id` and `director_id`.\n2. It then uses the `HAVING` clause to filter the grouped results, considering only those pairs where the count of timestamps (`count(timestamp)`) is greater than or equal to three.\n\n**Complexity:**\n- **Time complexity:** The time complexity is influenced by the grouping operation and the subsequent filtering based on the `HAVING` clause. Assuming appropriate indexing, the time complexity is likely to be O(n log n), where n is the number of rows in the `ActorDirector` table.\n\n- **Space complexity:** The space complexity is determined by the temporary storage required for the grouped results. Assuming m is the number of rows in the result set, the space complexity is O(m).\n\nKeep in mind that the actual performance may vary depending on the database engine, indexing, and the size of the `ActorDirector` table.\n\n# Code\n```\n# Write your MySQL query statement below\nselect actor_id, director_id from ActorDirector \ngroup by actor_id, director_id\nhaving count(timestamp)>=3;\n```
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Pandas | SQL | | Easy | Explained Step by Step | Actors & Directors Cooperated min Three Times
pandas-sql-easy-explained-step-by-step-a-f24v
see the successfully Accepted Submission\n\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n # Initially, we
Khosiyat
NORMAL
2023-09-19T17:42:56.711696+00:00
2023-10-01T17:37:52.725957+00:00
135
false
[see the successfully Accepted Submission](https://leetcode.com/submissions/detail/1053805243/)\n```\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n # Initially, we group \'actor_id\' and \'director_id\'.\n grouped_actor_director = actor_director.groupby([\'actor_id\', \'director_id\'])\n \n # In the next step, we count the number of \'actor_id\' and \'director_id\'.\n counted_actor_director = grouped_actor_director.size()\n \n # Then, we resets the index of the DataFrame.\n indexed_actor_director = counted_actor_director.reset_index(name=\'cooperated\')\n\n # After, that, we can filter the subquery to select rows where \'cooperated\' count is >= 3.\n filtered_actor_director = indexed_actor_director[indexed_actor_director[\'cooperated\'] >= 3]\n \n # Finally, we select and return a subset of \'actor_id\', \'director_id\' columns from the DataFrame .\n cooperated_actor_director = filtered_actor_director[[\'actor_id\', \'director_id\']]\n\n return cooperated_actor_director\n \n```\n\n**SQL**\n[see the successfully Accepted Submission](https://leetcode.com/submissions/detail/1061679792/)\n\n```\nSELECT actor_id, director_id \nFROM ActorDirector\n\nGROUP BY actor_id, director_id \nHAVING count(timestamp)>2;\n```\n\n\n```\n-- Select the "actor_id" and "director_id" columns from the "ActorDirector" table\n\n-- Specify the source table as "ActorDirector" from which data will be retrieved\nSELECT actor_id, director_id \n\nFROM ActorDirector\n\n-- Group the results by the combination of "actor_id" and "director_id"\nGROUP BY actor_id, director_id \n\n-- Apply a filter condition to the grouped results:\n-- We are selecting groups (combinations of actor and director) where the count of timestamps is greater than 2\nHAVING count(timestamp) > 2;\n\n```\n\n![image](https://assets.leetcode.com/users/images/13b1b1b3-377d-4e7b-a5b1-b8ea34bcfaad_1695145367.64976.jpeg)\n\n\n# Pandas & SQL | SOLVED & EXPLAINED LIST\n\n**EASY**\n\n- [Combine Two Tables](https://leetcode.com/problems/combine-two-tables/solutions/4051076/pandas-sql-easy-combine-two-tables/)\n\n- [Employees Earning More Than Their Managers](https://leetcode.com/problems/employees-earning-more-than-their-managers/solutions/4051991/pandas-sql-easy-employees-earning-more-than-their-managers/)\n\n- [Duplicate Emails](https://leetcode.com/problems/duplicate-emails/solutions/4055225/pandas-easy-duplicate-emails/)\n\n- [Customers Who Never Order](https://leetcode.com/problems/customers-who-never-order/solutions/4055429/pandas-sql-easy-customers-who-never-order-easy-explained/)\n\n- [Delete Duplicate Emails](https://leetcode.com/problems/delete-duplicate-emails/solutions/4055572/pandas-sql-easy-delete-duplicate-emails-easy/)\n- [Rising Temperature](https://leetcode.com/problems/rising-temperature/solutions/4056328/pandas-sql-easy-rising-temperature-easy/)\n\n- [ Game Play Analysis I](https://leetcode.com/problems/game-play-analysis-i/solutions/4056422/pandas-sql-easy-game-play-analysis-i-easy/)\n\n- [Find Customer Referee](https://leetcode.com/problems/find-customer-referee/solutions/4056516/pandas-sql-easy-find-customer-referee-easy/)\n\n- [Classes More Than 5 Students](https://leetcode.com/problems/classes-more-than-5-students/solutions/4058381/pandas-sql-easy-classes-more-than-5-students-easy/)\n- [Employee Bonus](https://leetcode.com/problems/employee-bonus/solutions/4058430/pandas-sql-easy-employee-bonus/)\n\n- [Sales Person](https://leetcode.com/problems/sales-person/solutions/4058824/pandas-sql-easy-sales-person/)\n\n- [Biggest Single Number](https://leetcode.com/problems/biggest-single-number/solutions/4063950/pandas-sql-easy-biggest-single-number/)\n\n- [Not Boring Movies](https://leetcode.com/problems/not-boring-movies/solutions/4065350/pandas-sql-easy-not-boring-movies/)\n- [Swap Salary](https://leetcode.com/problems/swap-salary/solutions/4065423/pandas-sql-easy-swap-salary/)\n\n- [Actors & Directors Cooperated min Three Times](https://leetcode.com/problems/actors-and-directors-who-cooperated-at-least-three-times/solutions/4065511/pandas-sql-easy-explained-step-by-step-actors-directors-cooperated-min-three-times/)\n\n- [Product Sales Analysis I](https://leetcode.com/problems/product-sales-analysis-i/solutions/4065545/pandas-sql-easy-product-sales-analysis-i/)\n\n- [Project Employees I](https://leetcode.com/problems/project-employees-i/solutions/4065635/pandas-sql-easy-project-employees-i/)\n- [Sales Analysis III](https://leetcode.com/problems/sales-analysis-iii/solutions/4065755/sales-analysis-iii-pandas-easy/)\n\n- [Reformat Department Table](https://leetcode.com/problems/reformat-department-table/solutions/4066153/pandas-sql-easy-explained-step-by-step-reformat-department-table/)\n\n- [Top Travellers](https://leetcode.com/problems/top-travellers/solutions/4066252/top-travellers-pandas-easy-eaxplained-step-by-step/)\n\n- [Replace Employee ID With The Unique Identifier](https://leetcode.com/problems/replace-employee-id-with-the-unique-identifier/solutions/4066321/pandas-sql-easy-replace-employee-id-with-the-unique-identifier/)\n- [Group Sold Products By The Date](https://leetcode.com/problems/group-sold-products-by-the-date/solutions/4066344/pandas-sql-easy-explained-step-by-step-group-sold-products-by-the-date/)\n\n- [Customer Placing the Largest Number of Orders](https://leetcode.com/problems/customer-placing-the-largest-number-of-orders/solutions/4068822/pandas-sql-easy-explained-step-by-step-customer-placing-the-largest-number-of-orders/)\n\n- [Article Views I](https://leetcode.com/problems/article-views-i/solutions/4069777/pandas-sql-easy-article-views-i/)\n\n- [User Activity for the Past 30 Days I](https://leetcode.com/problems/user-activity-for-the-past-30-days-i/solutions/4069797/pandas-sql-easy-user-activity-for-the-past-30-days-i/)\n\n- [Find Users With Valid E-Mails](https://leetcode.com/problems/find-users-with-valid-e-mails/solutions/4069810/pandas-sql-easy-find-users-with-valid-e-mails/)\n\n- [Patients With a Condition](https://leetcode.com/problems/patients-with-a-condition/solutions/4069817/pandas-sql-easy-patients-with-a-condition/)\n\n- [Customer Who Visited but Did Not Make Any Transactions](https://leetcode.com/problems/customer-who-visited-but-did-not-make-any-transactions/solutions/4072542/pandas-sql-easy-customer-who-visited-but-did-not-make-any-transactions/)\n\n\n- [Bank Account Summary II](https://leetcode.com/problems/bank-account-summary-ii/solutions/4072569/pandas-sql-easy-bank-account-summary-ii/)\n\n- [Invalid Tweets](https://leetcode.com/problems/invalid-tweets/solutions/4072599/pandas-sql-easy-invalid-tweets/)\n\n- [Daily Leads and Partners](https://leetcode.com/problems/daily-leads-and-partners/solutions/4072981/pandas-sql-easy-daily-leads-and-partners/)\n\n- [Recyclable and Low Fat Products](https://leetcode.com/problems/recyclable-and-low-fat-products/solutions/4073003/pandas-sql-easy-recyclable-and-low-fat-products/)\n\n- [Rearrange Products Table](https://leetcode.com/problems/rearrange-products-table/solutions/4073028/pandas-sql-easy-rearrange-products-table/)\n- [Calculate Special Bonus](https://leetcode.com/problems/calculate-special-bonus/solutions/4073595/pandas-sql-easy-calculate-special-bonus/)\n\n- [Count Unique Subjects](https://leetcode.com/problems/number-of-unique-subjects-taught-by-each-teacher/solutions/4073666/pandas-sql-easy-count-unique-subjects/)\n\n- [Count Unique Subjects](https://leetcode.com/problems/number-of-unique-subjects-taught-by-each-teacher/solutions/4073666/pandas-sql-easy-count-unique-subjects/)\n\n- [Fix Names in a Table](https://leetcode.com/problems/fix-names-in-a-table/solutions/4073695/pandas-sql-easy-fix-names-in-a-table/)\n- [Primary Department for Each Employee](https://leetcode.com/problems/primary-department-for-each-employee/solutions/4076183/pandas-sql-easy-primary-department-for-each-employee/)\n\n- [The Latest Login in 2020](https://leetcode.com/problems/the-latest-login-in-2020/solutions/4076240/pandas-sql-easy-the-latest-login-in-2020/)\n\n- [Find Total Time Spent by Each Employee](https://leetcode.com/problems/find-total-time-spent-by-each-employee/solutions/4076313/pandas-sql-easy-find-total-time-spent-by-each-employee/)\n\n- [Find Followers Count](https://leetcode.com/problems/find-followers-count/solutions/4076342/pandas-sql-easy-find-followers-count/)\n- [Percentage of Users Attended a Contest](https://leetcode.com/problems/percentage-of-users-attended-a-contest/solutions/4077301/pandas-sql-easy-percentage-of-users-attended-a-contest/)\n\n- [Employees With Missing Information](https://leetcode.com/problems/employees-with-missing-information/solutions/4077308/pandas-sql-easy-employees-with-missing-information/)\n\n- [Average Time of Process per Machine](https://leetcode.com/problems/average-time-of-process-per-machine/solutions/4077402/pandas-sql-easy-average-time-of-process-per-machine/)\n
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Pandas returning index (columns not duplicated for grouping and selecting items)
pandas-returning-index-columns-not-dupli-5due
Pandas returning index\nI like Panda\'s possibility to avoid repeating listing actor_id, director_id both for grouping and then final selecting items to return
garncarz
NORMAL
2023-08-16T04:17:22.828501+00:00
2023-08-16T04:19:02.342820+00:00
327
false
# Pandas returning index\nI like Panda\'s possibility to avoid repeating listing `actor_id`, `director_id` both for grouping and then final selecting items to return (as seen in the solution below).\n\n```\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n return (actor_director\n .groupby([\'actor_id\', \'director_id\'])\n [\'timestamp\'].count()\n .loc[lambda x: x >= 3]\n .index.to_frame()\n )\n```\n\n# Pandas using query\n```\nimport pandas as pd\n\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n return (actor_director\n .groupby([\'actor_id\', \'director_id\'], as_index=False)\n [\'timestamp\'].count()\n .query(\'timestamp >= 3\')\n [[\'actor_id\', \'director_id\']]\n )\n```\n\n# MySQL\n```\nselect\n actor_id,\n director_id\nfrom ActorDirector\ngroup by actor_id, director_id\nhaving count(`timestamp`) >= 3\n```
3
0
['MySQL', 'Pandas']
0
actors-and-directors-who-cooperated-at-least-three-times
Simple MySQL Solution
simple-mysql-solution-by-abhradip_360-1d7y
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
abhradip_360
NORMAL
2023-06-07T12:47:30.591544+00:00
2023-06-07T12:47:30.591593+00:00
213
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect actor_id, director_id from ActorDirector \ngroup by actor_id, director_id\nhaving count(timestamp)>=3;\n\n```
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
MySQL Easy Solution ✅✅
mysql-easy-solution-by-shubhamjain287-rwbd
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
Shubhamjain287
NORMAL
2023-03-08T05:40:25.132064+00:00
2023-03-08T05:40:25.132108+00:00
946
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp)>=3\n```
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
MySQL Solution
mysql-solution-by-pranto1209-wknp
Code\n\n# Write your MySQL query statement below\nselect actor_id, director_id from ActorDirector \ngroup by actor_id, director_id having count(timestamp) >= 3;
pranto1209
NORMAL
2023-03-07T18:20:12.564667+00:00
2023-03-13T16:00:09.303804+00:00
848
false
# Code\n```\n# Write your MySQL query statement below\nselect actor_id, director_id from ActorDirector \ngroup by actor_id, director_id having count(timestamp) >= 3;\n```
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Using Group by and Having Clause || Actors and Directors Who Cooperated At Least Three Times
using-group-by-and-having-clause-actors-232cs
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
himshrivas
NORMAL
2023-01-26T08:44:53.468558+00:00
2023-01-26T08:44:53.468595+00:00
1,085
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect actor_id, director_id from ActorDirector group by actor_id,director_id having count(timestamp)>=3;\n```\nPlease upvote!!
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
✅✅Easy C++ Solution || Simple to Understand
easy-c-solution-simple-to-understand-by-0iv2f
\tselect actor_id, director_id\n\tfrom ActorDirector\n\tgroup by actor_id, director_id\n\thaving count(timestamp) >= 3;\nI hope that you\'ve found the solution
debav2
NORMAL
2022-10-07T19:25:07.181578+00:00
2022-10-08T06:59:20.637152+00:00
466
false
\tselect actor_id, director_id\n\tfrom ActorDirector\n\tgroup by actor_id, director_id\n\thaving count(timestamp) >= 3;\nI hope that you\'ve found the solution useful.\nIn that case, please do upvote. Happy Coding :)
3
0
['MySQL']
1
actors-and-directors-who-cooperated-at-least-three-times
Simple solution | Fast and easy
simple-solution-fast-and-easy-by-max_tar-40ae
\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY 1, 2\nHAVING COUNT(1) > 2\n
max_tar
NORMAL
2022-09-06T23:19:37.440725+00:00
2022-09-06T23:19:37.440756+00:00
1,003
false
```\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY 1, 2\nHAVING COUNT(1) > 2\n```
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Simple 3 line code | mysql
simple-3-line-code-mysql-by-rohan_fasale-iwpp
select actor_id,director_id from actordirector\ngroup by director_id,actor_id\nhaving count(director_id)>=3
Rohan_Fasale
NORMAL
2022-09-02T18:07:20.865359+00:00
2022-09-02T18:07:20.865406+00:00
629
false
select actor_id,director_id from actordirector\ngroup by director_id,actor_id\nhaving count(director_id)>=3
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
✅MySQL || Beginner level || Faster than 98%||Simple-Short -Solution✅
mysql-beginner-level-faster-than-98simpl-jqdv
Please upvote to motivate me in my quest of documenting all leetcode solutions. HAPPY CODING:)\nAny suggestions and improvements are always welcome.\n==========
Anos
NORMAL
2022-08-31T19:48:06.760521+00:00
2022-08-31T19:48:27.152623+00:00
276
false
**Please upvote to motivate me in my quest of documenting all leetcode solutions. HAPPY CODING:)\nAny suggestions and improvements are always welcome.***\n*====================================================================*\n\u2705 **MySQL Code :**\n Your runtime beats 98.12 % of mysql submissions.\n```\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp)>=3\n```\n**Runtime:** 299 ms\n**Memory Usage:** 0B\n________________________________\n__________________________________\n\nIf you like the solution, please upvote \uD83D\uDD3C\nFor any questions, or discussions, comment below. \uD83D\uDC47\uFE0F
3
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
MSSQL
mssql-by-lavinamall-feeb
\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(actor_id) > 2\n
lavinamall
NORMAL
2022-07-22T01:08:35.155701+00:00
2022-07-22T01:08:35.155742+00:00
704
false
```\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(actor_id) > 2\n```
3
0
['MS SQL Server']
1
actors-and-directors-who-cooperated-at-least-three-times
Group by the pair
group-by-the-pair-by-user5091-94d7
\nselect actor_id, director_id\nfrom ActorDirector\ngroup by actor_id, director_id\nhaving count(timestamp)>2;\n
user5091
NORMAL
2021-11-13T18:05:58.729504+00:00
2021-11-13T18:05:58.729534+00:00
408
false
```\nselect actor_id, director_id\nfrom ActorDirector\ngroup by actor_id, director_id\nhaving count(timestamp)>2;\n```
3
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
Fast MS SQL solution
fast-ms-sql-solution-by-momo1621-ikzi
\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHaving count(distinct timestamp) >= 3\n
Momo1621
NORMAL
2019-11-22T07:01:19.206096+00:00
2020-01-18T07:01:42.519627+00:00
784
false
```\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHaving count(distinct timestamp) >= 3\n```
3
2
[]
2
actors-and-directors-who-cooperated-at-least-three-times
SELECT actor_id, director_id FROM ActorDirector GROUP BY actor_id, director_id HAVING COUNT(*)>=3;
select-actor_id-director_id-from-actordi-rqqv
null
Vishal1431
NORMAL
2025-02-04T19:09:39.811050+00:00
2025-02-04T19:09:39.811050+00:00
161
false
```mysql [] SELECT actor_id, director_id FROM ActorDirector GROUP BY actor_id, director_id HAVING COUNT(*)>=3; ```
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Group by clause in Postgres
group-by-clause-in-postgres-by-amoghamai-av1h
Code
AmoghaMayya
NORMAL
2025-01-21T07:02:56.021501+00:00
2025-01-21T07:02:56.021501+00:00
162
false
# Code ```postgresql [] -- Write your PostgreSQL query statement below select actor_id , director_id from actordirector group by actor_id , director_id having count(*) > 2; ```
2
0
['PostgreSQL']
0
actors-and-directors-who-cooperated-at-least-three-times
SQL
sql-by-kanvapuri_sai_praneetha-bss9
Code\n\nSELECT actor_id, director_id \nFROM ActorDirector \nGROUP BY actor_id, director_id \nHAVING count(*)>= 3;\n
kanvapuri_sai_praneetha
NORMAL
2024-08-10T02:59:19.129511+00:00
2024-08-10T02:59:19.129549+00:00
486
false
# Code\n```\nSELECT actor_id, director_id \nFROM ActorDirector \nGROUP BY actor_id, director_id \nHAVING count(*)>= 3;\n```
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
(Pandas) 1-line
pandas-1-line-by-qeetcode-id63
\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n return actor_director.groupby(by=["actor_id", "director_id"], as_index=False).siz
qeetcode
NORMAL
2023-08-17T20:19:20.000460+00:00
2023-08-17T20:19:20.000511+00:00
321
false
```\ndef actors_and_directors(actor_director: pd.DataFrame) -> pd.DataFrame:\n return actor_director.groupby(by=["actor_id", "director_id"], as_index=False).size().query("size >= 3").drop(labels="size", axis=1); \n```
2
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
Explained | Group By With Having | SQL and Pandas
explained-group-by-with-having-sql-and-p-vy3w
My SQL\nYou can add conditions to group by clause by using having clause.\nAdd actor_id, director_id to group by because they are being selected for showing. An
hridoy100
NORMAL
2023-08-12T06:58:21.816075+00:00
2023-08-12T06:58:21.816098+00:00
1,311
false
# My SQL\nYou can add conditions to `group by` clause by using `having` clause.\nAdd `actor_id, director_id` to group by because they are being selected for showing. Anything that\'s not inside an aggregate function (`count()`, `avg()`, `max()`, etc.) and needs to be shown as output must be written inside `GROUP BY` clause.\n\n``` SQL []\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING count(timestamp)>=3\n```\n\n# Pandas\n\nIf you understand the SQL clearly, you will be able to come up with the following solution.\n\n``` python3 []\nimport pandas as pd\n\ndef actors_and_directors(ad: pd.DataFrame) -> pd.DataFrame:\n df = ad.groupby([\'actor_id\', \'director_id\']).count().reset_index()\n return df[df[\'timestamp\']>=3][[\'actor_id\', \'director_id\']]\n```
2
0
['MySQL', 'Pandas']
0
actors-and-directors-who-cooperated-at-least-three-times
MySQL Solution
mysql-solution-by-triyambkeshtiwari-f3v1
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
triyambkeshtiwari
NORMAL
2023-08-10T15:55:55.331508+00:00
2023-08-10T15:55:55.331536+00:00
423
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect actor_id , director_id\nfrom ActorDirector\ngroup by actor_id,director_id\nhaving count(timestamp)>=3\n```
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
✔MySQL Simple Solution✔
mysql-simple-solution-by-arth_anaya-oal2
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
Arth_Anaya
NORMAL
2023-04-20T14:17:14.471908+00:00
2023-04-20T14:17:14.471940+00:00
373
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\n\nselect\n actor_id, director_id\nfrom actordirector\ngroup by actor_id, director_id\nhaving count(actor_id=director_id) >= 3\n```
2
0
['MySQL']
2
actors-and-directors-who-cooperated-at-least-three-times
Easy Solution | MySQL👀🥷
easy-solution-mysql-by-dipesh_12-0pnc
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
dipesh_12
NORMAL
2023-04-12T15:32:16.944482+00:00
2023-04-12T15:32:16.944516+00:00
4,202
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect actor_id,director_id from ActorDirector group by actor_id,director_id having count(*)>=3;\n```
2
0
['MySQL']
1
actors-and-directors-who-cooperated-at-least-three-times
SQL Server CLEAN & EASY
sql-server-clean-easy-by-rhazem13-29oq
\nSELECT ad.actor_id, ad.director_id\nFROM ActorDirector ad\nJOIN (\n SELECT actor_id, director_id\n FROM ActorDirector\n GROUP BY actor_id, director_id\n H
rhazem13
NORMAL
2023-03-17T08:37:18.457507+00:00
2023-03-17T08:37:18.457540+00:00
1,845
false
```\nSELECT ad.actor_id, ad.director_id\nFROM ActorDirector ad\nJOIN (\n SELECT actor_id, director_id\n FROM ActorDirector\n GROUP BY actor_id, director_id\n HAVING COUNT(*) >= 3\n) ad2 ON ad.actor_id = ad2.actor_id AND ad.director_id = ad2.director_id\nGROUP BY ad.actor_id, ad.director_id\n\n```
2
0
[]
1
actors-and-directors-who-cooperated-at-least-three-times
MYSQL || Easy approach
mysql-easy-approach-by-mrigank_2003-pga0
\n# Code\n\n# Write your MySQL query statement below\nselect actor_id, director_id from ActorDirector group by actor_id, director_id having count(timestamp)>=3;
mrigank_2003
NORMAL
2023-03-14T04:20:37.121636+00:00
2023-03-14T04:20:37.121682+00:00
3,207
false
\n# Code\n```\n# Write your MySQL query statement below\nselect actor_id, director_id from ActorDirector group by actor_id, director_id having count(timestamp)>=3;\n```
2
0
['MySQL']
1
actors-and-directors-who-cooperated-at-least-three-times
MySQL Solution
mysql-solution-by-abstractconnoisseurs-tc7h
Code\n\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(1) >= 3\n
abstractConnoisseurs
NORMAL
2023-01-27T09:07:51.616774+00:00
2023-01-27T09:07:51.616806+00:00
702
false
# Code\n```\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(1) >= 3\n```
2
0
['Database', 'MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Group by & Having
group-by-having-by-bejoysekhar-5edb
\nselect actor_id, director_id from actordirector group by actor_id, director_id having count(*) >= 3 \n
bejoysekhar
NORMAL
2022-10-07T02:08:26.106611+00:00
2022-10-07T02:08:26.106657+00:00
483
false
```\nselect actor_id, director_id from actordirector group by actor_id, director_id having count(*) >= 3 \n```
2
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
Easy & Concise MySQL Solution using HAVING Clause
easy-concise-mysql-solution-using-having-4z9p
Do upvote the solution to keep me motivated , incase you liked it ;)\n\n\nSELECT actor_id,director_id \nFROM ActorDirector \nGROUP BY actor_id,director_id\nHAVI
vartika23_02
NORMAL
2022-09-01T05:38:48.754473+00:00
2022-09-01T05:39:01.563709+00:00
58
false
**Do upvote the solution to keep me motivated , incase you liked it ;)**\n\n```\nSELECT actor_id,director_id \nFROM ActorDirector \nGROUP BY actor_id,director_id\nHAVING count(actor_id=director_id)>=3;\n```
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
MS SQL | Easy
ms-sql-easy-by-nikhil_katoch-6v02
\nselect actor_id, director_id\nfrom ActorDirector\ngroup by director_id,actor_id\nhaving count(director_id)>=3\n
nikhil_katoch
NORMAL
2022-08-28T13:17:01.522615+00:00
2022-08-28T13:17:01.522645+00:00
577
false
```\nselect actor_id, director_id\nfrom ActorDirector\ngroup by director_id,actor_id\nhaving count(director_id)>=3\n```
2
0
['MS SQL Server']
1
actors-and-directors-who-cooperated-at-least-three-times
Easy To Understand Solution
easy-to-understand-solution-by-inferno08-r113
\nSELECT DISTINCT t.actor_id, t.director_id FROM ActorDirector t\nGROUP BY t.actor_id , t.director_id\nHAVING COUNT(t.timestamp) >= 3\n
inferno080
NORMAL
2022-08-08T07:18:27.292902+00:00
2022-08-08T07:18:27.292940+00:00
169
false
```\nSELECT DISTINCT t.actor_id, t.director_id FROM ActorDirector t\nGROUP BY t.actor_id , t.director_id\nHAVING COUNT(t.timestamp) >= 3\n```
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
2 MySQL Solutions: GROUP BY & RANK()
2-mysql-solutions-group-by-rank-by-farya-1h5t
1. Using GROUP BY and HAVING clause (the simplest one)\n\nSELECT a.actor_id, a.director_id\nFROM ActorDirector a\nGROUP BY a.actor_id, a.director_id\nHAVING COU
faryar251
NORMAL
2022-07-26T10:52:19.452540+00:00
2022-07-26T10:54:12.445453+00:00
125
false
### 1. Using `GROUP BY` and `HAVING` clause (the simplest one)\n```\nSELECT a.actor_id, a.director_id\nFROM ActorDirector a\nGROUP BY a.actor_id, a.director_id\nHAVING COUNT(timestamp) > 2\n```\n\n### 2. Using `RANK() OVER` clause\n```\nSELECT a1.actor_id, a1.director_id\nFROM (\n SELECT \n a.actor_id, \n a.director_id,\n RANK() OVER (PARTITION BY a.actor_id, a.director_id\n ORDER BY a.timestamp) AS \'coop_count\'\n FROM ActorDirector a) a1\nWHERE a1.coop_count = 3\n```\n\nThanks for reading!\nIf there\'s any error or doubt, do comment. Feedbacks are always appreciated!!
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
[ MYSQL ] ✅✅ Simple MYSQL Solution Using Having Count Clause || Group By 🥳✌👍
mysql-simple-mysql-solution-using-having-1lmv
If You like the Solution, Don\'t Forget To UpVote Me, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n# Runtime: 376 ms, faster than 69.66% of MySQL online submissions
ashok_kumar_meghvanshi
NORMAL
2022-07-16T18:25:20.836308+00:00
2022-07-16T18:25:20.836345+00:00
227
false
# If You like the Solution, Don\'t Forget To UpVote Me, Please UpVote! \uD83D\uDD3C\uD83D\uDE4F\n# Runtime: 376 ms, faster than 69.66% of MySQL online submissions for Actors and Directors Who Cooperated At Least Three Times.\n# Memory Usage: 0B, less than 100.00% of MySQL online submissions for Actors and Directors Who Cooperated At Least Three Times.\n\n\tSELECT ACTOR_ID, DIRECTOR_ID\n\tFROM ACTORDIRECTOR\n\tGROUP BY ACTOR_ID, DIRECTOR_ID\n\tHAVING COUNT(TIMESTAMP) > 2
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Simplest
simplest-by-suraj21193-cf09
```\n/ Write your T-SQL query statement below /\nSELECT actor_id, director_id\nFROM ActorDirector \n GROUP BY actor_id, director_id\n Having COUNT(ti
suraj21193
NORMAL
2022-06-23T14:21:29.311327+00:00
2022-06-23T14:23:18.940728+00:00
163
false
```\n/* Write your T-SQL query statement below */\nSELECT actor_id, director_id\nFROM ActorDirector \n GROUP BY actor_id, director_id\n Having COUNT(timestamp) >=3
2
0
[]
1
actors-and-directors-who-cooperated-at-least-three-times
Using GROUP BY and HAVING
using-group-by-and-having-by-supratim_co-trpi
\nSELECT actor_id, director_id from ActorDirector GROUP BY actor_id, director_id HAVING COUNT(*)>2;\n
supratim_code
NORMAL
2022-05-29T16:17:28.563434+00:00
2022-05-29T16:17:28.563472+00:00
305
false
```\nSELECT actor_id, director_id from ActorDirector GROUP BY actor_id, director_id HAVING COUNT(*)>2;\n```
2
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
GROUP BY and HAVING | Very Easy Solution
group-by-and-having-very-easy-solution-b-crq0
\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(*) >= 3;\n
ajinkya2000
NORMAL
2022-05-18T10:23:18.321399+00:00
2022-05-18T10:23:18.321439+00:00
178
false
```\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(*) >= 3;\n```
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
mysql groupby and having
mysql-groupby-and-having-by-sulaymon-dev-xodz
\nselect actor_id,director_id from ActorDirector group by actor_id,director_id having count(*)>=3\n
Sulaymon-Dev20
NORMAL
2022-04-21T18:47:17.152177+00:00
2022-04-21T18:47:17.152219+00:00
107
false
```\nselect actor_id,director_id from ActorDirector group by actor_id,director_id having count(*)>=3\n```
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
simple
simple-by-du0118-wimp
select actor_id, director_id\nfrom actordirector\ngroup by actor_id, director_id\nhaving count(actor_id = director_id) >= 3\n;
DU0118
NORMAL
2022-03-05T20:51:42.484594+00:00
2022-03-05T20:51:42.484626+00:00
184
false
select actor_id, director_id\nfrom actordirector\ngroup by actor_id, director_id\nhaving count(actor_id = director_id) >= 3\n;
2
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
mySQL | having, group by | explained
mysql-having-group-by-explained-by-anhtr-7919
`\nselect actor_id, director_id\nfrom ActorDirector \n#return pair of (actor_id, director_id) \ngroup by actor_id, director_id\nhaving count(*) >= 3\n#having: c
anhtran114
NORMAL
2022-01-12T21:12:03.739409+00:00
2022-01-12T21:12:20.804532+00:00
468
false
```\nselect actor_id, director_id\nfrom ActorDirector \n#return pair of (actor_id, director_id) \ngroup by actor_id, director_id\nhaving count(*) >= 3\n#having: condition \n#count(*) means count all include NULL cases\n;\n``\n
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Group By and Having ()
group-by-and-having-by-yzhou2018-knnj
select actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nhaving count(timestamp)>=3;
yzhou2018
NORMAL
2021-11-29T02:31:26.671563+00:00
2021-11-29T02:31:26.671595+00:00
348
false
select actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nhaving count(timestamp)>=3;
2
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
Simple MySql Solution ,faster than 93.77% of MySQL online submissions
simple-mysql-solution-faster-than-9377-o-ij4n
\nSelect actor_id,director_id from actordirector\ngroup by actor_id,director_id\nhaving count(director_id)>=3\n
user8447f
NORMAL
2021-10-20T12:55:34.160217+00:00
2021-10-20T12:55:34.160246+00:00
384
false
```\nSelect actor_id,director_id from actordirector\ngroup by actor_id,director_id\nhaving count(director_id)>=3\n```
2
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
MySQL
mysql-by-vincentt126-6xc0
run time 342s\nselect actor_id, director_id\nfrom actordirector\ngroup by actor_id, director_id\nhaving count(*)>=3\n\nrun time 448s\n\nselect actor_id, directo
vincentt126
NORMAL
2021-06-18T18:08:35.769363+00:00
2021-06-18T18:09:22.743008+00:00
218
false
run time 342s\n```select actor_id, director_id\nfrom actordirector\ngroup by actor_id, director_id\nhaving count(*)>=3\n```\nrun time 448s\n```\nselect actor_id, director_id\nfrom actordirector\ngroup by actor_id, director_id\nhaving count(timestamp)>=3\n```\n
2
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
EASIEST SOLUTION IN MySQL
easiest-solution-in-mysql-by-hitarthshah-q1zt
IntuitionUPVOTE IF YOU FIND IT HELPFUL.ApproachComplexity Time complexity: Space complexity: Code
HITARTHSHAH2211
NORMAL
2025-04-01T04:29:57.633519+00:00
2025-04-01T04:29:57.633519+00:00
115
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> UPVOTE IF YOU FIND IT HELPFUL. # Approach <!-- Describe your approach to solving the problem. --> # Complexity - Time complexity: <!-- Add your time complexity here, e.g. $$O(n)$$ --> - Space complexity: <!-- Add your space complexity here, e.g. $$O(n)$$ --> # Code ```mysql [] # Write your MySQL query statement below select actor_id,director_id from ActorDirector group by actor_id,director_id having count(timestamp)>=3; ```
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Easy Solution | Using Group by and Having
easy-solution-using-group-by-and-having-83lhc
Code
Aminour_Islam
NORMAL
2025-01-15T13:44:20.830085+00:00
2025-01-15T13:44:20.830085+00:00
33
false
# Code ```mysql [] # Write your MySQL query statement below select actor_id, director_id from ActorDirector group by actor_id, director_id having count(*) >= 3; ```
1
0
['Database', 'MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
👉🏻FAST AND EASY TO UNDERSTAND SOLUTION || MySQL
fast-and-easy-to-understand-solution-mys-3je0
IntuitionApproachComplexity Time complexity: Space complexity: Code
djain7700
NORMAL
2025-01-08T18:05:14.577196+00:00
2025-01-08T18:05:14.577196+00:00
111
false
# Intuition <!-- Describe your first thoughts on how to solve this problem. --> # Approach <!-- Describe your approach to solving the problem. --> # Complexity - Time complexity: <!-- Add your time complexity here, e.g. $$O(n)$$ --> - Space complexity: <!-- Add your space complexity here, e.g. $$O(n)$$ --> # Code ```mysql [] # Write your MySQL query statement below SELECT actor_id,director_id FROM ActorDirector GROUP BY actor_id,director_id HAVING COUNT(*)>=3 ```
1
0
['Database', 'MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Easy Solution
easy-solution-by-vittalnesaragi-a3dw
\n\n# Code\noraclesql []\n/* Write your PL/SQL query statement below */\nSELECT ACTOR_ID,DIRECTOR_ID FROM ACTORDIRECTOR GROUP BY ACTOR_ID,DIRECTOR_ID HAVING COU
VittalNesaragi
NORMAL
2024-12-02T09:16:30.070593+00:00
2024-12-02T09:16:30.070621+00:00
219
false
\n\n# Code\n```oraclesql []\n/* Write your PL/SQL query statement below */\nSELECT ACTOR_ID,DIRECTOR_ID FROM ACTORDIRECTOR GROUP BY ACTOR_ID,DIRECTOR_ID HAVING COUNT(*)>=3;\n```
1
0
['MySQL', 'Oracle']
0
actors-and-directors-who-cooperated-at-least-three-times
Beginner-Friendly Solution Using SQL Server || Clear
beginner-friendly-solution-using-sql-ser-43ao
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
truongtamthanh2004
NORMAL
2024-06-03T14:48:50.375640+00:00
2024-06-03T14:48:50.375669+00:00
331
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/* Write your T-SQL query statement below */\nselect actor_id, director_id\nfrom ActorDirector\ngroup by actor_id, director_id\nhaving count(timestamp) >= 3\n```
1
0
['MS SQL Server']
0
actors-and-directors-who-cooperated-at-least-three-times
Ezzy Pzzy ✌️ 🤞|| Oracle Solution || BEGINNER FRIENDLY ||Complete Explanation|| Group by with count
ezzy-pzzy-oracle-solution-beginner-frien-sm0r
Intuition\nAs the problem requires pair of two column and its count. Therefore,my first thought is to use Group by. \n\n# Approach\nTo use group by on two colum
Ninja_
NORMAL
2023-09-07T11:03:56.178979+00:00
2023-09-07T11:03:56.179004+00:00
63
false
# Intuition\nAs the problem requires pair of two column and its count. Therefore,my first thought is to use Group by. \n\n# Approach\nTo use group by on two columns with count we can use:\ngroup by(col1, col2) having count(1)>2\nThis will first collect all the pairs of actors and directors and then count their frequency and return the rows where the frequency is greater then 2 or atleast 3\n\n\n# Code\n```\n/* Write your PL/SQL query statement below */\n\nselect actor_id, director_id from ActorDirector group by actor_id, director_id having count(1)>2;\n\n\n\n```\n![upvote.webp](https://assets.leetcode.com/users/images/e82af2e4-81a3-47ce-8159-65f821e3185d_1694084265.033327.webp)\n\n
1
0
['Oracle']
0
actors-and-directors-who-cooperated-at-least-three-times
PLSQL solution using having count()
plsql-solution-using-having-count-by-dib-rsnd
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
DibyaJyoti_Sarmah
NORMAL
2023-06-06T07:50:20.039113+00:00
2023-06-06T07:50:20.039151+00:00
263
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/* Write your PL/SQL query statement below */\n\nSelect actor_id , director_id from ActorDirector \nhaving count(1) >= 3\ngroup by actor_id , director_id\n```
1
0
['Oracle']
1
actors-and-directors-who-cooperated-at-least-three-times
Simple Code using HAVING : Actors and Directors Who Cooperated At Least Three Times
simple-code-using-having-actors-and-dire-6zy1
Code\n\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp) >=3;
kchheda97
NORMAL
2023-05-24T18:06:12.912909+00:00
2023-05-24T18:07:06.198097+00:00
34
false
# Code\n```\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(timestamp) >=3;\n```
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Simple solution with GROUP BY
simple-solution-with-group-by-by-iliaavd-2i4f
Approach\nGROUP BY + HAVING\n\n# Code\n\nSELECT actor_id, director_id FROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(*) >= 3\n
IliaAvdeev
NORMAL
2023-04-30T17:42:16.922351+00:00
2023-04-30T17:42:16.922411+00:00
5
false
# Approach\nGROUP BY + HAVING\n\n# Code\n```\nSELECT actor_id, director_id FROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(*) >= 3\n```
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
1050. Actors and Directors Who Cooperated At Least Three Times | MySQL
1050-actors-and-directors-who-cooperated-1i4l
\n Add your space complexity here, e.g. O(n) \n\n# Code\n\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY
codewithusama
NORMAL
2023-03-26T13:05:13.176331+00:00
2023-03-26T13:05:13.176366+00:00
77
false
\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(*) >= 3;\n```
1
0
['C++', 'Java', 'MySQL', 'Oracle', 'MS SQL Server']
0
actors-and-directors-who-cooperated-at-least-three-times
MySQL || GROUP BY || BEGINNER FRIENDLY || EASY SOLUTION
mysql-group-by-beginner-friendly-easy-so-qq2c
\nSELECT actor_id,director_id FROM ActorDirector \nGROUP BY actor_id,director_id \nhaving count(*) >= 3\n
mohitsatija
NORMAL
2022-10-18T19:16:59.079059+00:00
2022-10-18T19:16:59.079100+00:00
51
false
```\nSELECT actor_id,director_id FROM ActorDirector \nGROUP BY actor_id,director_id \nhaving count(*) >= 3\n```
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
MySQL solution in 4 lines
mysql-solution-in-4-lines-by-shubhamyada-2aar
\nselect actor_id, director_id\nfrom actordirector\ngroup by actor_id, director_id\nhaving count(1)>= 3\n
shubhamyadav32100
NORMAL
2022-10-14T16:25:48.187449+00:00
2022-10-14T16:25:48.187478+00:00
277
false
```\nselect actor_id, director_id\nfrom actordirector\ngroup by actor_id, director_id\nhaving count(1)>= 3\n```
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Can't be more simple than this one
cant-be-more-simple-than-this-one-by-div-lrag
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \ngrup by and count\n\n#
Diving_in_Ocean
NORMAL
2022-10-01T18:43:54.288709+00:00
2022-10-01T18:43:54.288749+00:00
162
false
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\ngrup by and count\n\n# Code\n```\n# Write your MySQL query statement below\n\nselect actor_id,director_id from (select actor_id , director_id , count(director_id) as tt from ActorDirector group by actor_id,director_id) t where t.tt>=3;\n```
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
GROUP BY HAVING
group-by-having-by-kajal_shukla124-cd35
SELECT actor_id,director_id FROM ActorDirector\nGROUP BY actor_id,director_id HAVING count(timestamp)>2;
kajal_shukla124
NORMAL
2022-09-25T07:28:41.181319+00:00
2022-09-25T07:28:41.181353+00:00
52
false
SELECT actor_id,director_id FROM ActorDirector\nGROUP BY actor_id,director_id HAVING count(timestamp)>2;
1
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
mysql soln
mysql-soln-by-quater_nion-xlbo
\nselect actor_id, director_id from ActorDirector\ngroup by actor_id, director_id \nhaving count(actor_id)>=3\n
quater_nion
NORMAL
2022-09-16T15:46:11.068570+00:00
2022-09-16T15:46:11.068616+00:00
262
false
```\nselect actor_id, director_id from ActorDirector\ngroup by actor_id, director_id \nhaving count(actor_id)>=3\n```
1
0
[]
1
actors-and-directors-who-cooperated-at-least-three-times
Actors and Directors who cooperated at least three times. Efficient solution
actors-and-directors-who-cooperated-at-l-ejks
select actor_id,director_id from ActorDirector group by actor_id,director_id\nhaving count(*)>=3;
Vinay_Ghildiyal
NORMAL
2022-09-11T07:15:08.683728+00:00
2022-09-11T07:15:08.683761+00:00
174
false
select actor_id,director_id from ActorDirector group by actor_id,director_id\nhaving count(*)>=3;
1
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
easy mysql solution
easy-mysql-solution-by-invincible2511-nmsa
\nselect distinct actor_id, director_id from ActorDirector \ngroup by actor_id, director_id\nhaving count(*)>2\n
Invincible2511
NORMAL
2022-09-09T13:57:25.790842+00:00
2022-09-09T13:57:25.790884+00:00
148
false
```\nselect distinct actor_id, director_id from ActorDirector \ngroup by actor_id, director_id\nhaving count(*)>2\n```
1
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
Easy
easy-by-naomi_tqf-gnrz
select actor_id, director_id from ActorDirector \ngroup by actor_id, director_id\nhaving count(*) > 2
Naomi_TQF
NORMAL
2022-09-04T03:41:02.719187+00:00
2022-09-04T03:41:02.719210+00:00
76
false
select actor_id, director_id from ActorDirector \ngroup by actor_id, director_id\nhaving count(*) > 2
1
0
[]
1
actors-and-directors-who-cooperated-at-least-three-times
MySQL solution using [HAVING] | Easy Step-by-Step Explanation
mysql-solution-using-having-easy-step-by-vks1
\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(director_id)>=3;\n\n\
MohammedHamza
NORMAL
2022-09-01T18:07:09.533179+00:00
2022-09-01T18:07:37.334226+00:00
86
false
```\n# Write your MySQL query statement below\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(director_id)>=3;\n```\n\n**Explanation:**\n```\nSELECT actor_id, director_id\nFROM ActorDirector\n```\n* First I have selected **actor_id and director_id**.\n\n```\nGROUP BY actor_id, director_id\n```\n* Then I have **GROUP BY** them by **actor_id and director_id** as we need both to fulfill the condition.\n\n```\nHAVING COUNT(director_id)>=3;\n```\n* I have used **HAVING** clause on **COUNT** of **director_id**.\n* If it is greater than equal to 3 than filter out the result.\n* You can also use **actor_id**.
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
SQL! 📸📸
sql-by-sc0d3r-jhcn
\nselect actor_id,\ndirector_id\nfrom ActorDirector\ngroup by actor_id,director_id\nhaving count(*) >= 3\n
SC0d3r
NORMAL
2022-08-30T12:36:22.903419+00:00
2022-08-30T12:36:22.903483+00:00
52
false
```\nselect actor_id,\ndirector_id\nfrom ActorDirector\ngroup by actor_id,director_id\nhaving count(*) >= 3\n```
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
MYSQL SOLUTION | GROUPBY && HAVING STATMENT | EASY UNDERSTANDING
mysql-solution-groupby-having-statment-e-d83g
\nSELECT a.actor_id , a.director_id \nFROM ActorDirector a\nGROUP BY a.actor_id , a.director_id \nHAVING COUNT(*) >= 3\n\n\n### kindly vote up if it usefull
gom3a98
NORMAL
2022-08-25T09:39:33.716397+00:00
2022-08-25T09:39:33.716438+00:00
136
false
```\nSELECT a.actor_id , a.director_id \nFROM ActorDirector a\nGROUP BY a.actor_id , a.director_id \nHAVING COUNT(*) >= 3\n```\n\n### **kindly vote up if it usefull**
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Faster than 72.91% | MySQL Solution | Easy to Understand
faster-than-7291-mysql-solution-easy-to-blc2k
```\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(*)>=3;
arghapaul002
NORMAL
2022-08-21T00:58:43.158909+00:00
2022-08-21T00:58:43.158956+00:00
131
false
```\nSELECT actor_id, director_id\nFROM ActorDirector\nGROUP BY actor_id, director_id\nHAVING COUNT(*)>=3;
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
mysql faster solution
mysql-faster-solution-by-anshika_soni-et8w
```# Write your MySQL query statement below\nselect actor_id,director_id from ActorDirector group by actor_id,director_id having count(*)>=3
anshika_soni
NORMAL
2022-08-14T14:24:55.597240+00:00
2022-08-14T14:24:55.597267+00:00
107
false
```# Write your MySQL query statement below\nselect actor_id,director_id from ActorDirector group by actor_id,director_id having count(*)>=3
1
0
['MySQL']
0
actors-and-directors-who-cooperated-at-least-three-times
Faster than 97% of solutions with CONCAT, GROUP BY and HAVING
faster-than-97-of-solutions-with-concat-hwhwo
select A.actor_id , A.director_id\nfrom ActorDirector A\nGROUP BY CONCAT(A.actor_id ,"_", A.director_id)\nHAVING COUNT(*) >= 3
sourabh_python
NORMAL
2022-08-13T12:57:54.821440+00:00
2022-08-13T12:57:54.821486+00:00
61
false
select A.actor_id , A.director_id\nfrom ActorDirector A\nGROUP BY CONCAT(A.actor_id ,"_", A.director_id)\nHAVING COUNT(*) >= 3
1
0
[]
0
actors-and-directors-who-cooperated-at-least-three-times
Simple Solution faster than 98.77%
simple-solution-faster-than-9877-by-user-g4eh
\n/* Write your T-SQL query statement below */\n\nselect actor_id , director_id FROM ActorDirector \nGROUP BY actor_id , director_id\nHAVING COUNT(*) > 2\n
user6736jw
NORMAL
2022-08-07T09:37:25.253221+00:00
2022-08-07T09:37:45.564346+00:00
100
false
```\n/* Write your T-SQL query statement below */\n\nselect actor_id , director_id FROM ActorDirector \nGROUP BY actor_id , director_id\nHAVING COUNT(*) > 2\n```
1
0
[]
0