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feat: add sql solutions to lc problems (#1121)
* No.1990.Count the Number of Experiments * No.2020.Number of Accounts That Did Not Stream * No.2051.The Category of Each Member in the Store * No.2066.Account Balance * No.2084.Drop Type 1 Orders for Customers With Type 0 Orders * No.2112.The Airport With the Most Traffic
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25 files changed

+301
-33
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‎.prettierrc

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@@ -33,6 +33,7 @@
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"solution/1600-1699/1667.Fix Names in a Table/Solution.sql",
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"solution/1700-1799/1777.Product's Price for Each Store/Solution.sql",
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"solution/1900-1999/1972.First and Last Call On the Same Day/Solution.sql",
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"solution/2000-2099/2066.Account Balance/Solution.sql",
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"solution/2600-2699/2686.Immediate Food Delivery III/Solution.sql",
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"solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/Solution.sql"
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],

‎solution/0100-0199/0192.Word Frequency/README.md

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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sh
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# Read from the file words.txt and output the word frequency list to stdout.
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cat words.txt | tr -s ' ' '\n' | sort | uniq -c | sort -nr | awk '{print $2, $1}'
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```
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<!-- tabs:end -->

‎solution/0100-0199/0192.Word Frequency/README_EN.md

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### **Bash**
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```sh
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# Read from the file words.txt and output the word frequency list to stdout.
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cat words.txt | tr -s ' ' '\n' | sort | uniq -c | sort -nr | awk '{print $2, $1}'
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```
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<!-- tabs:end -->

‎solution/0100-0199/0192.Word Frequency/Solution.sh

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#!/usr/bin/env bash
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# NF是当前行的field字段数；NR是正在处理的当前行数。
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awk '{
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for(i=1;i<=NF;i++){
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res[$i]++
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}
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}
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END{
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for(k in res){
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print k,res[k]
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}
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}' words.txt | sort -nr -k2
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#or:
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cat words.txt | tr -s ' ' '\n' | sort | uniq -c | sort -r | awk '{print $2" "$1}'
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# Read from the file words.txt and output the word frequency list to stdout.
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cat words.txt | tr -s ' ' '\n' | sort | uniq -c | sort -nr | awk '{print $2, $1}'

‎solution/1900-1999/1990.Count the Number of Experiments/README.md

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@@ -79,7 +79,33 @@ Experiments table:
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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WITH
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P AS (
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SELECT 'Android' AS platform
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UNION
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SELECT 'IOS'
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UNION
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SELECT 'Web'
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),
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Exp AS (
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SELECT 'Reading' AS experiment_name
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UNION
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SELECT 'Sports'
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UNION
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SELECT 'Programming'
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),
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T AS (
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SELECT *
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FROM
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P,
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Exp
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)
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SELECT platform, experiment_name, count(experiment_id) AS num_experiments
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FROM
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T AS t
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LEFT JOIN Experiments USING (platform, experiment_name)
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GROUP BY 1, 2;
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```
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<!-- tabs:end -->

‎solution/1900-1999/1990.Count the Number of Experiments/README_EN.md

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@@ -71,7 +71,33 @@ On the platform &quot;Web&quot;, we had two &quot;Reading&quot; experiments and
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### **SQL**
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```sql
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# Write your MySQL query statement below
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WITH
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P AS (
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SELECT 'Android' AS platform
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UNION
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SELECT 'IOS'
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UNION
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SELECT 'Web'
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),
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Exp AS (
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SELECT 'Reading' AS experiment_name
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UNION
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SELECT 'Sports'
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UNION
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SELECT 'Programming'
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),
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T AS (
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SELECT *
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FROM
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P,
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Exp
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)
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SELECT platform, experiment_name, count(experiment_id) AS num_experiments
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FROM
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T AS t
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LEFT JOIN Experiments USING (platform, experiment_name)
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GROUP BY 1, 2;
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```
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<!-- tabs:end -->

‎solution/1900-1999/1990.Count the Number of Experiments/Solution.sql

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# Write your MySQL query statement below
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WITH
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P AS (
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SELECT 'Android' AS platform
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UNION
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SELECT 'IOS'
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UNION
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SELECT 'Web'
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),
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Exp AS (
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SELECT 'Reading' AS experiment_name
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UNION
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SELECT 'Sports'
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UNION
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SELECT 'Programming'
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),
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T AS (
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SELECT *
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FROM
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P,
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Exp
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)
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SELECT platform, experiment_name, count(experiment_id) AS num_experiments
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FROM
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T AS t
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LEFT JOIN Experiments USING (platform, experiment_name)
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GROUP BY 1, 2;

‎solution/2000-2099/2020.Number of Accounts That Did Not Stream/README.md

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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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SELECT count(DISTINCT sub.account_id) AS accounts_count
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FROM
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Subscriptions AS sub
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LEFT JOIN Streams AS ss ON sub.account_id = ss.account_id
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WHERE
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year(start_date) <= 2021
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AND year(end_date) >= 2021
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AND (year(stream_date) != 2021 OR stream_date > end_date);
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```
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<!-- tabs:end -->

‎solution/2000-2099/2020.Number of Accounts That Did Not Stream/README_EN.md

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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT count(DISTINCT sub.account_id) AS accounts_count
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FROM
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Subscriptions AS sub
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LEFT JOIN Streams AS ss ON sub.account_id = ss.account_id
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WHERE
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year(start_date) <= 2021
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AND year(end_date) >= 2021
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AND (year(stream_date) != 2021 OR stream_date > end_date);
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```
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<!-- tabs:end -->

‎solution/2000-2099/2020.Number of Accounts That Did Not Stream/Solution.sql

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# Write your MySQL query statement below
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SELECT count(DISTINCT sub.account_id) AS accounts_count
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FROM
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Subscriptions AS sub
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LEFT JOIN Streams AS ss ON sub.account_id = ss.account_id
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WHERE
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year(start_date) <= 2021
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AND year(end_date) >= 2021
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AND (year(stream_date) != 2021 OR stream_date > end_date);

‎solution/2000-2099/2041.Accepted Candidates From the Interviews/README.md

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@@ -101,7 +101,14 @@ Rounds table:
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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SELECT candidate_id
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FROM
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Candidates AS c
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LEFT JOIN Rounds AS r ON c.interview_id = r.interview_id
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WHERE years_of_exp >= 2
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GROUP BY c.interview_id
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HAVING sum(score) > 15;
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```
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<!-- tabs:end -->

‎solution/2000-2099/2041.Accepted Candidates From the Interviews/README_EN.md

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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT candidate_id
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FROM
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Candidates AS c
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LEFT JOIN Rounds AS r ON c.interview_id = r.interview_id
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WHERE years_of_exp >= 2
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GROUP BY c.interview_id
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HAVING sum(score) > 15;
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```
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<!-- tabs:end -->

‎solution/2000-2099/2041.Accepted Candidates From the Interviews/Solution.sql

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# Write your MySQL query statement below
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SELECT candidate_id
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FROM
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Candidates AS c
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LEFT JOIN Rounds AS r ON c.interview_id = r.interview_id
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WHERE years_of_exp >= 2
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GROUP BY c.interview_id
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HAVING sum(score) > 15;

‎solution/2000-2099/2051.The Category of Each Member in the Store/README.md

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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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SELECT
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m.member_id,
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name,
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CASE
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WHEN count(v.visit_id) = 0 THEN 'Bronze'
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WHEN 100 * count(charged_amount) / count(
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v.visit_id
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) >= 80 THEN 'Diamond'
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WHEN 100 * count(charged_amount) / count(v.visit_id) >= 50 THEN 'Gold'
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ELSE 'Silver'
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END AS category
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FROM
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Members AS m
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LEFT JOIN Visits AS v ON m.member_id = v.member_id
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LEFT JOIN Purchases AS p ON v.visit_id = p.visit_id
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GROUP BY member_id;
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```
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<!-- tabs:end -->

‎solution/2000-2099/2051.The Category of Each Member in the Store/README_EN.md

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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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m.member_id,
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name,
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CASE
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WHEN count(v.visit_id) = 0 THEN 'Bronze'
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WHEN 100 * count(charged_amount) / count(
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v.visit_id
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) >= 80 THEN 'Diamond'
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WHEN 100 * count(charged_amount) / count(v.visit_id) >= 50 THEN 'Gold'
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ELSE 'Silver'
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END AS category
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FROM
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Members AS m
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LEFT JOIN Visits AS v ON m.member_id = v.member_id
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LEFT JOIN Purchases AS p ON v.visit_id = p.visit_id
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GROUP BY member_id;
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```
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<!-- tabs:end -->

‎solution/2000-2099/2051.The Category of Each Member in the Store/Solution.sql

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# Write your MySQL query statement below
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SELECT
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m.member_id,
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name,
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CASE
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WHEN count(v.visit_id) = 0 THEN 'Bronze'
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WHEN 100 * count(charged_amount) / count(
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v.visit_id
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) >= 80 THEN 'Diamond'
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WHEN 100 * count(charged_amount) / count(v.visit_id) >= 50 THEN 'Gold'
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ELSE 'Silver'
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END AS category
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FROM
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Members AS m
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LEFT JOIN Visits AS v ON m.member_id = v.member_id
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LEFT JOIN Purchases AS p ON v.visit_id = p.visit_id
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GROUP BY member_id;

‎solution/2000-2099/2066.Account Balance/README.md

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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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SELECT
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account_id,
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day,
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sum(if(type = 'Deposit', amount, -amount)) OVER (
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PARTITION BY account_id
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ORDER BY day
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) AS balance
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FROM Transactions
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ORDER BY 1, 2;
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```
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<!-- tabs:end -->

‎solution/2000-2099/2066.Account Balance/README_EN.md

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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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account_id,
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day,
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sum(if(type = 'Deposit', amount, -amount)) OVER (
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PARTITION BY account_id
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ORDER BY day
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) AS balance
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FROM Transactions
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ORDER BY 1, 2;
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```
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<!-- tabs:end -->

‎solution/2000-2099/2066.Account Balance/Solution.sql

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# Write your MySQL query statement below
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SELECT
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account_id,
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day,
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sum(if(type = 'Deposit', amount, -amount)) OVER (
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PARTITION BY account_id
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ORDER BY day
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) AS balance
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FROM Transactions
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ORDER BY 1, 2;

‎solution/2000-2099/2084.Drop Type 1 Orders for Customers With Type 0 Orders/README.md

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<!-- 这里可写当前语言的特殊实现逻辑 -->
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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 DISTINCT customer_id
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FROM Orders
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WHERE order_type = 0
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)
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SELECT *
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FROM Orders AS o
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WHERE
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order_type = 0
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OR NOT EXISTS (SELECT 1 FROM T AS t WHERE t.customer_id = o.customer_id);
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```
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<!-- tabs:end -->

‎solution/2000-2099/2084.Drop Type 1 Orders for Customers With Type 0 Orders/README_EN.md

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### **SQL**
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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 DISTINCT customer_id
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FROM Orders
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WHERE order_type = 0
85+
)
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SELECT *
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FROM Orders AS o
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WHERE
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order_type = 0
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OR NOT EXISTS (SELECT 1 FROM T AS t WHERE t.customer_id = o.customer_id);
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```
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<!-- tabs:end -->

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