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Commit 33b241b

Browse files
feat: add sql solutions to lc problems: No.0550,0570 (doocs#1185)
* feat: add sql solutions to lc problems: No.0550,0570 * No.0550.Game Play Analysis IV * No.0570.Managers with at Least 5 Direct Reports
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‎.github/pull_request_template.md‎

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<!--
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Below are template for copilot to generate CR message.
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Please DO NOT modify it.
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-->
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### Description
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copilot:summary
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### Explanation of Changes
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copilot:walkthrough
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copilot:walkthrough
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-->

‎.prettierrc‎

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"overrides": [
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{
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"files": [
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"solution/0500-0599/0550.Game Play Analysis IV/Solution.sql",
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"solution/0500-0599/0578.Get Highest Answer Rate Question/Solution.sql",
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"solution/0600-0699/0610.Triangle Judgement/Solution.sql",
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"solution/0600-0699/0618.Students Report By Geography/Solution.sql",

‎solution/0500-0599/0550.Game Play Analysis IV/README.md‎

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AND DATEDIFF(a.event_date, b.event_date) = -1;
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```
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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player_id,
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datediff(
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lead(event_date) OVER (
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PARTITION BY player_id
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ORDER BY event_date
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),
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event_date
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) AS diff,
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row_number() OVER (
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PARTITION BY player_id
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ORDER BY event_date
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) AS rk
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FROM Activity
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)
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SELECT
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round(
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count(DISTINCT if(diff = 1, player_id, NULL)) / count(
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DISTINCT player_id
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),
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2
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) AS fraction
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FROM T
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WHERE rk = 1;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0550.Game Play Analysis IV/README_EN.md‎

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AND DATEDIFF(a.event_date, b.event_date) = -1;
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```
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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player_id,
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datediff(
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lead(event_date) OVER (
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PARTITION BY player_id
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ORDER BY event_date
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),
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event_date
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) AS diff,
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row_number() OVER (
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PARTITION BY player_id
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ORDER BY event_date
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) AS rk
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FROM Activity
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)
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SELECT
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round(
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count(DISTINCT if(diff = 1, player_id, NULL)) / count(
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DISTINCT player_id
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),
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2
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) AS fraction
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FROM T
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WHERE rk = 1;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0570.Managers with at Least 5 Direct Reports/README.md‎

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ON e1.id = e2.managerId;
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```
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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managerId,
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count(1) OVER (PARTITION BY managerId) AS cnt
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FROM Employee
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)
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SELECT DISTINCT name
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FROM
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Employee AS e
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JOIN T AS t ON e.id = t.managerId
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WHERE cnt >= 5;
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```
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<!-- tabs:end -->

‎solution/0500-0599/0570.Managers with at Least 5 Direct Reports/README_EN.md‎

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ON e1.id = e2.managerId;
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```
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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managerId,
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count(1) OVER (PARTITION BY managerId) AS cnt
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FROM Employee
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)
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SELECT DISTINCT name
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FROM
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Employee AS e
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JOIN T AS t ON e.id = t.managerId
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WHERE cnt >= 5;
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```
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<!-- tabs:end -->
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# Write your MySQL query statement below
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SELECT
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name
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FROM
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Employee AS e1
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JOIN (
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WITH
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T AS (
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SELECT
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managerId
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managerId,
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count(1) OVER (PARTITION BY managerId) AS cnt
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FROM Employee
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WHERE managerId IS NOT NULL
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GROUP BY managerId
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HAVING count(1) >= 5
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) AS e2
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ON e1.id = e2.managerId;
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)
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SELECT DISTINCT name
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FROM
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Employee AS e
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JOIN T AS t ON e.id = t.managerId
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WHERE cnt >= 5;

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