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leetcode-problem-solutions

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2025-04-12 09:38:24
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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] = \...
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...
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_in...
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 ...
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 on...
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\',\'d...
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. $$O(n)$$ --...
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 ac...
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:\...
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\']>...
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 ...
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 ap...
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 `...
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...
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 ...
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_...
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_d...
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. $$O(n)$$ --...
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. $$O(n)$$ --...
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. $$O(n)$$ --...
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...
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` cl...
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. $$O(n)$$ --...
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. $$O(n)$$ --...
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. $$O(n)$$ --...
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 ...
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...
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 her...
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 `...
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. $$O(n)$$ --...
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 frequenc...
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. $$O(n)$$ --...
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 t...
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 acto...
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