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feat: add sql solutions to lc problems: No.2984+ (doocs#2172)
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‎solution/2900-2999/2984.Find Peak Calling Hours for Each City/README.md

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@@ -69,7 +69,29 @@ Output table is ordered by peak_calling_hour and city in descending order.</pre>
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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
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*,
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RANK() OVER (
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PARTITION BY city
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ORDER BY cnt DESC
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) AS rk
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FROM
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(
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SELECT
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city,
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HOUR(call_time) AS h,
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COUNT(1) AS cnt
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FROM Calls
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GROUP BY 1, 2
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) AS t
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)
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SELECT city, h AS peak_calling_hour, cnt AS number_of_calls
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FROM T
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WHERE rk = 1
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ORDER BY 2 DESC, 1 DESC;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2984.Find Peak Calling Hours for Each City/README_EN.md

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@@ -63,7 +63,29 @@ Output table is ordered by peak_calling_hour and city in descending order.</pre>
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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
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*,
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RANK() OVER (
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PARTITION BY city
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ORDER BY cnt DESC
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) AS rk
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FROM
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(
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SELECT
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city,
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HOUR(call_time) AS h,
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COUNT(1) AS cnt
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FROM Calls
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GROUP BY 1, 2
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) AS t
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)
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SELECT city, h AS peak_calling_hour, cnt AS number_of_calls
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FROM T
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WHERE rk = 1
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ORDER BY 2 DESC, 1 DESC;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2984.Find Peak Calling Hours for Each City/Solution.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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*,
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RANK() OVER (
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PARTITION BY city
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ORDER BY cnt DESC
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) AS rk
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FROM
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(
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SELECT
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city,
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HOUR(call_time) AS h,
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COUNT(1) AS cnt
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FROM Calls
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GROUP BY 1, 2
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) AS t
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)
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SELECT city, h AS peak_calling_hour, cnt AS number_of_calls
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FROM T
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WHERE rk = 1
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ORDER BY 2 DESC, 1 DESC;

‎solution/2900-2999/2985.Calculate Compressed Mean/README.md

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@@ -63,7 +63,13 @@ The calculation is as follows:
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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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ROUND(
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SUM(item_count * order_occurrences) / SUM(order_occurrences),
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2
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) AS average_items_per_order
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FROM Orders;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2985.Calculate Compressed Mean/README_EN.md

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@@ -57,7 +57,13 @@ The calculation is as follows:
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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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ROUND(
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SUM(item_count * order_occurrences) / SUM(order_occurrences),
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2
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) AS average_items_per_order
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FROM Orders;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2985.Calculate Compressed Mean/Solution.sql

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# Write your MySQL query statement below
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SELECT
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ROUND(
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SUM(item_count * order_occurrences) / SUM(order_occurrences),
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2
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) AS average_items_per_order
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FROM Orders;

‎solution/2900-2999/2986.Find Third Transaction/README.md

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@@ -70,7 +70,32 @@ Output table is ordered by user_id in ascending order.
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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
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*,
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RANK() OVER (
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PARTITION BY user_id
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ORDER BY transaction_date
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) AS rk,
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spend > (
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LAG(spend) OVER (
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PARTITION BY user_id
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ORDER BY transaction_date
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)
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)
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AND spend > (
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LAG(spend, 2) OVER (
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PARTITION BY user_id
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ORDER BY transaction_date
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)
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) AS st
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FROM Transactions
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)
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SELECT user_id, spend AS third_transaction_spend, transaction_date AS third_transaction_date
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FROM T
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WHERE rk = 3 AND st = 1;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2986.Find Third Transaction/README_EN.md

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@@ -64,7 +64,32 @@ Output table is ordered by user_id in ascending order.
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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
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*,
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RANK() OVER (
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PARTITION BY user_id
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ORDER BY transaction_date
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) AS rk,
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spend > (
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LAG(spend) OVER (
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PARTITION BY user_id
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ORDER BY transaction_date
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)
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)
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AND spend > (
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LAG(spend, 2) OVER (
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PARTITION BY user_id
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ORDER BY transaction_date
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)
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) AS st
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FROM Transactions
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)
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SELECT user_id, spend AS third_transaction_spend, transaction_date AS third_transaction_date
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FROM T
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WHERE rk = 3 AND st = 1;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2986.Find Third Transaction/Solution.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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*,
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RANK() OVER (
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PARTITION BY user_id
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ORDER BY transaction_date
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) AS rk,
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spend > (
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LAG(spend) OVER (
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PARTITION BY user_id
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ORDER BY transaction_date
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)
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)
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AND spend > (
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LAG(spend, 2) OVER (
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PARTITION BY user_id
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ORDER BY transaction_date
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)
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) AS st
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FROM Transactions
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)
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SELECT user_id, spend AS third_transaction_spend, transaction_date AS third_transaction_date
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FROM T
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WHERE rk = 3 AND st = 1;

‎solution/2900-2999/2987.Find Expensive Cities/README.md

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@@ -75,7 +75,12 @@ Only Chicago and Los Angeles have average home prices exceeding the national ave
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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 city
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FROM Listings
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GROUP BY city
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HAVING AVG(price) > (SELECT AVG(price) FROM Listings)
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ORDER BY 1;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2987.Find Expensive Cities/README_EN.md

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@@ -69,7 +69,12 @@ Only Chicago and Los Angeles have average home prices exceeding the national ave
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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 city
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FROM Listings
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GROUP BY city
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HAVING AVG(price) > (SELECT AVG(price) FROM Listings)
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ORDER BY 1;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2987.Find Expensive Cities/Solution.sql

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# Write your MySQL query statement below
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SELECT city
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FROM Listings
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GROUP BY city
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HAVING AVG(price) > (SELECT AVG(price) FROM Listings)
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ORDER BY 1;

‎solution/2900-2999/2988.Manager of the Largest Department/README.md

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@@ -72,7 +72,19 @@ Output table is ordered by dep_id in ascending order.
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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 dep_id, COUNT(1) AS cnt
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FROM Employees
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GROUP BY 1
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)
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SELECT emp_name AS manager_name, t.dep_id
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FROM
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T AS t
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JOIN Employees AS e ON t.dep_id = e.dep_id AND e.position = 'Manager'
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WHERE cnt = (SELECT MAX(cnt) FROM T)
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ORDER BY 2;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2988.Manager of the Largest Department/README_EN.md

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@@ -66,7 +66,19 @@ Output table is ordered by dep_id in ascending order.
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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 dep_id, COUNT(1) AS cnt
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FROM Employees
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GROUP BY 1
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)
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SELECT emp_name AS manager_name, t.dep_id
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FROM
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T AS t
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JOIN Employees AS e ON t.dep_id = e.dep_id AND e.position = 'Manager'
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WHERE cnt = (SELECT MAX(cnt) FROM T)
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ORDER BY 2;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2988.Manager of the Largest Department/Solution.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 dep_id, COUNT(1) AS cnt
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FROM Employees
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GROUP BY 1
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)
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SELECT emp_name AS manager_name, t.dep_id
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FROM
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T AS t
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JOIN Employees AS e ON t.dep_id = e.dep_id AND e.position = 'Manager'
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WHERE cnt = (SELECT MAX(cnt) FROM T)
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ORDER BY 2;

‎solution/2900-2999/2989.Class Performance/README.md

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@@ -64,14 +64,23 @@ student_id 321 has the highest score of 230, while student_id 896 has the lowest
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：最大值最小值**
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我们可以使用 `MAX` 和 `MIN` 函数来分别获取 `assignment1`、`assignment2`、`assignment3` 的和的最大值和最小值，然后相减即可。
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<!-- tabs:start -->
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### **SQL**
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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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MAX(assignment1 + assignment2 + assignment3) - MIN(
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assignment1 + assignment2 + assignment3
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) AS difference_in_score
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FROM Scores;
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```
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<!-- tabs:end -->

‎solution/2900-2999/2989.Class Performance/README_EN.md

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## Solutions
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**Solution 1: Maximum and Minimum**
64+
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We can use the `MAX` and `MIN` functions to get the maximum and minimum sums of `assignment1`, `assignment2`, and `assignment3`, respectively. Then, subtract the minimum from the maximum.
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<!-- tabs:start -->
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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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MAX(assignment1 + assignment2 + assignment3) - MIN(
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assignment1 + assignment2 + assignment3
76+
) AS difference_in_score
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FROM Scores;
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```
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7180
<!-- tabs:end -->

‎solution/2900-2999/2989.Class Performance/Solution.sql

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# Write your MySQL query statement below
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SELECT
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MAX(assignment1 + assignment2 + assignment3) - MIN(
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assignment1 + assignment2 + assignment3
5+
) AS difference_in_score
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FROM Scores;

‎solution/2900-2999/2990.Loan Types/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：分组求和**
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67+
我们可以对 `Loans` 表按照 `user_id` 进行分组，找出既包含 `Refinance` 又包含 `Mortgage` 的用户，然后按照 `user_id` 进行排序。
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<!-- tabs:start -->
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### **SQL**
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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
77+
SELECT user_id
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FROM Loans
79+
GROUP BY 1
80+
HAVING SUM(loan_type = 'Refinance') > 0 AND SUM(loan_type = 'Mortgage') > 0
81+
ORDER BY 1;
7382
```
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7584
<!-- tabs:end -->

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