feat: update solutions to lc problems (#1788)

* No.0182.Duplicate Emails
* No.1050.Actors and Directors Who Cooperated At Least Three Times
* No.1495.Friendly Movies Streamed Last Month
* No.1511.Customer Order Frequency
* No.1587.Bank Account Summary II
* No.1693.Daily Leads and Partners
* No.1699.Number of Calls Between Two Persons
* No.2562.Find the Array Concatenation Value

Author: yanglbme

.prettierignore

@@ -16,7 +16,6 @@ node_modules/
 /solution/0100-0199/0177.Nth Highest Salary/Solution.sql
 /solution/1400-1499/1454.Active Users/Solution.sql
 /solution/1400-1499/1484.Group Sold Products By The Date/Solution.sql
-/solution/1500-1599/1511.Customer Order Frequency/Solution.sql
 /solution/1600-1699/1613.Find the Missing IDs/Solution.sql
 /solution/1600-1699/1635.Hopper Company Queries I/Solution.sql
 /solution/1600-1699/1651.Hopper Company Queries III/Solution.sql

.prettierrc

@@ -34,8 +34,10 @@
                 "solution/1400-1499/1445.Apples & Oranges/Solution.sql",
                 "solution/1400-1499/1479.Sales by Day of the Week/Solution.sql",
                 "solution/1500-1599/1501.Countries You Can Safely Invest In/Solution.sql",
+                "solution/1500-1599/1511.Customer Order Frequency/Solution.sql",
                 "solution/1500-1599/1555.Bank Account Summary/Solution.sql",
                 "solution/1600-1699/1667.Fix Names in a Table/Solution.sql",
+                "solution/1600-1699/1699.Number of Calls Between Two Persons/Solution.sql",
                 "solution/1700-1799/1777.Product's Price for Each Store/Solution.sql",
                 "solution/1900-1999/1934.Confirmation Rate/Solution.sql",
                 "solution/1900-1999/1972.First and Last Call On the Same Day/Solution.sql",

solution/0100-0199/0182.Duplicate Emails/README.md

@@ -55,15 +55,24 @@ Person 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：分组统计**
+
+我们可以使用 `GROUP BY` 语句，按照 `email` 字段进行分组，然后使用 `HAVING` 语句，筛选出现次数大于 $1$ 的 `email`。
+
+**方法二：自连接**
+
+我们可以使用自连接的方法，将 `Person` 表自身连接一次，然后筛选出 `id` 不同，但 `email` 相同的记录。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT Email
+# Write your MySQL query statement below
+SELECT email
 FROM Person
-GROUP BY Email
-HAVING count(Email) > 1;
+GROUP BY 1
+HAVING count(1) > 1;
 ```
 
 ```sql

solution/0100-0199/0182.Duplicate Emails/README_EN.md

@@ -49,15 +49,24 @@ Person table:
 
 ## Solutions
 
+**Solution 1: Group By + Having**
+
+We can use the `GROUP BY` statement to group the data by the `email` field, and then use the `HAVING` statement to filter out the `email` addresses that appear more than once.
+
+**Solution 2: Self-Join**
+
+We can use a self-join to join the `Person` table with itself, and then filter out the records where the `id` is different but the `email` is the same.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT Email
+# Write your MySQL query statement below
+SELECT email
 FROM Person
-GROUP BY Email
-HAVING count(Email) > 1;
+GROUP BY 1
+HAVING count(1) > 1;
 ```
 
 ```sql

solution/0100-0199/0182.Duplicate Emails/Solution.sql

@@ -1,4 +1,5 @@
-SELECT Email
+# Write your MySQL query statement below
+SELECT email
 FROM Person
-GROUP BY Email
-HAVING count(Email) > 1;
+GROUP BY 1
+HAVING count(1) > 1;

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/README.md

@@ -54,9 +54,9 @@ ActorDirector 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
-**方法一：使用 `GROUP BY` + `HAVING`**
+**方法一：分组统计**
 
-我们将 `ActorDirector` 表按照 `actor_id` 和 `director_id` 进行分组，然后使用 `HAVING` 过滤出合作次数大于等于 $3$ 次的组。
+我们可以使用 `GROUP BY` 语句，按照 `actor_id` 和 `director_id` 字段进行分组，然后使用 `HAVING` 语句，筛选出现次数大于等于 $3$ 的 `actor_id` 和 `director_id`。
 
 <!-- tabs:start -->
 

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/README_EN.md

@@ -53,7 +53,9 @@ ActorDirector table:
 
 ## Solutions
 
-Use `GROUP BY` & `HAVING`.
+**Solution 1: Group By + Having**
+
+We can use the `GROUP BY` statement to group the data by the `actor_id` and `director_id` fields, and then use the `HAVING` statement to filter out the `actor_id` and `director_id` that appear at least three times.
 
 <!-- tabs:start -->
 

solution/1400-1499/1495.Friendly Movies Streamed Last Month/README.md

@@ -91,29 +91,24 @@ TVProgram</code> 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：等值连接 + 条件筛选**
+
+我们可以先通过等值连接将两张表按照 `content_id` 字段连接起来，然后再通过条件筛选出在 $2020$ 年 $6$ 月份播放的儿童适宜电影。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT DISTINCT
-    title
+# Write your MySQL query statement below
+SELECT DISTINCT title
 FROM
-    Content
-    INNER JOIN TVProgram ON Content.content_id = TVProgram.content_id
+    TVProgram
+    JOIN Content USING (content_id)
 WHERE
-    content_type = 'Movies'
+    date_format(program_date, '%Y%m') = '202006'
     AND kids_content = 'Y'
-    AND program_date BETWEEN '2020-06-01' AND '2020-06-30';
-```
-
-```sql
-SELECT DISTINCT
-    title
-FROM
-    Content
-    INNER JOIN TVProgram ON Content.content_id = TVProgram.content_id
-WHERE kids_content = 'Y' AND (MONTH(program_date), YEAR(program_date)) = (6, 2020);
+    AND content_type = 'Movies';
 ```
 
 <!-- tabs:end -->

solution/1400-1499/1495.Friendly Movies Streamed Last Month/README_EN.md

@@ -87,27 +87,22 @@ Content table:
 
 ## Solutions
 
+**Solution 1: Equi-Join + Conditional Filtering**
+
+We can first use an equi-join to join the two tables based on the `content_id` field, and then use conditional filtering to select the child-friendly movies that were played in June 2020.
+
 <!-- tabs:start -->
 
 ```sql
-SELECT DISTINCT
-    title
+# Write your MySQL query statement below
+SELECT DISTINCT title
 FROM
-    Content
-    INNER JOIN TVProgram ON Content.content_id = TVProgram.content_id
+    TVProgram
+    JOIN Content USING (content_id)
 WHERE
-    content_type = 'Movies'
+    date_format(program_date, '%Y%m') = '202006'
     AND kids_content = 'Y'
-    AND program_date BETWEEN '2020-06-01' AND '2020-06-30';
-```
-
-```sql
-SELECT DISTINCT
-    title
-FROM
-    Content
-    INNER JOIN TVProgram ON Content.content_id = TVProgram.content_id
-WHERE kids_content = 'Y' AND (MONTH(program_date), YEAR(program_date)) = (6, 2020);
+    AND content_type = 'Movies';
 ```
 
 <!-- tabs:end -->

solution/1400-1499/1495.Friendly Movies Streamed Last Month/Solution.sql

@@ -1,16 +1,9 @@
-SELECT DISTINCT
-    title
+# Write your MySQL query statement below
+SELECT DISTINCT title
 FROM
-    Content
-    INNER JOIN TVProgram ON Content.content_id = TVProgram.content_id
+    TVProgram
+    JOIN Content USING (content_id)
 WHERE
-    content_type = 'Movies'
+    date_format(program_date, '%Y%m') = '202006'
     AND kids_content = 'Y'
-    AND program_date BETWEEN '2020-06-01' AND '2020-06-30';
-
-SELECT DISTINCT
-    title
-FROM
-    Content
-    INNER JOIN TVProgram ON Content.content_id = TVProgram.content_id
-WHERE kids_content = 'Y' AND (MONTH(program_date), YEAR(program_date)) = (6, 2020);
+    AND content_type = 'Movies';

solution/1500-1599/1511.Customer Order Frequency/README.md

@@ -117,23 +117,26 @@ Moustafa 在2020年6月花费了110 (10 * 2 + 45 * 2), 在7月花费了0.</pre>
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：等值连接 + 分组求和**
+
+我们可以使用 `JOIN` 语句，连接 `Orders` 和 `Product` 表，再连接 `Customers` 表，筛选出 `order_date` 在 $2020$ 年的记录，然后使用 `GROUP BY` 语句，按照 `customer_id` 分组，使用 `HAVING` 语句，筛选出 $6$ 月和 $7$ 月花费大于等于 $100$ 的客户。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    c.customer_id AS CUSTOMER_ID,
-    c.name AS NAME
+SELECT customer_id, name
 FROM
-    Customers c, Product p, Orders o
-WHERE
-    c.customer_id = o.customer_id
-AND p.product_id = o.product_id
-GROUP BY c.customer_id
-HAVING sum(if(month(o.order_date)=6, price*quantity, 0)) >= 100
-AND sum(if(month(o.order_date)=7, price*quantity, 0)) >= 100;
+    Orders
+    JOIN Product USING (product_id)
+    JOIN Customers USING (customer_id)
+WHERE year(order_date) = 2020
+GROUP BY 1
+HAVING
+    sum(if(month(order_date) = 6, quantity * price, 0)) >= 100
+    AND sum(if(month(order_date) = 7, quantity * price, 0)) >= 100;
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1511.Customer Order Frequency/README_EN.md

@@ -112,23 +112,26 @@ Moustafa spent 110 (10 * 2 + 45 * 2) in June and 0 in July 2020.
 
 ## Solutions
 
+**Solution 1: Join + Group By + Having**
+
+We can use the `JOIN` statement to join the `Orders` table and the `Product` table, and then join the result with the `Customers` table. We can filter out the records where the `order_date` is not in the year $2020$, and then use the `GROUP BY` statement to group the data by `customer_id`. Finally, we can use the `HAVING` statement to filter out the customers whose spending in June and July is greater than or equal to $100$.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    c.customer_id AS CUSTOMER_ID,
-    c.name AS NAME
+SELECT customer_id, name
 FROM
-    Customers c, Product p, Orders o
-WHERE
-    c.customer_id = o.customer_id
-AND p.product_id = o.product_id
-GROUP BY c.customer_id
-HAVING sum(if(month(o.order_date)=6, price*quantity, 0)) >= 100
-AND sum(if(month(o.order_date)=7, price*quantity, 0)) >= 100;
+    Orders
+    JOIN Product USING (product_id)
+    JOIN Customers USING (customer_id)
+WHERE year(order_date) = 2020
+GROUP BY 1
+HAVING
+    sum(if(month(order_date) = 6, quantity * price, 0)) >= 100
+    AND sum(if(month(order_date) = 7, quantity * price, 0)) >= 100;
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1511.Customer Order Frequency/Solution.sql

@@ -1,12 +1,11 @@
 # Write your MySQL query statement below
-SELECT
-    c.customer_id AS CUSTOMER_ID,
-    c.name AS NAME
+SELECT customer_id, name
 FROM
-    Customers c, Product p, Orders o
-WHERE
-    c.customer_id = o.customer_id
-AND p.product_id = o.product_id
-GROUP BY c.customer_id
-HAVING sum(if(month(o.order_date)=6, price*quantity, 0)) >= 100
-AND sum(if(month(o.order_date)=7, price*quantity, 0)) >= 100;
+    Orders
+    JOIN Product USING (product_id)
+    JOIN Customers USING (customer_id)
+WHERE year(order_date) = 2020
+GROUP BY 1
+HAVING
+    sum(if(month(order_date) = 6, quantity * price, 0)) >= 100
+    AND sum(if(month(order_date) = 7, quantity * price, 0)) >= 100;

solution/1500-1599/1587.Bank Account Summary II/README.md

@@ -90,33 +90,26 @@ Charlie 的余额为(6000 + 6000 - 4000) = 8000.
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：等值连接 + 分组求和**
+
+我们可以使用等值连接，将 `Users` 和 `Transactions` 表按照 `account` 列连接起来，然后按照 `account` 列分组求和，最后筛选出余额大于 $10000$ 的用户。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
-```sql
-SELECT
-    u.name,
-    SUM(t.amount) AS balance
-FROM
-    users AS u
-    JOIN transactions AS t ON u.account = t.account
-GROUP BY name
-HAVING SUM(t.amount) > 10000;
-```
-
 ```sql
 # Write your MySQL query statement below
 SELECT
     name,
     sum(amount) AS balance
 FROM
-    Users AS u
-    LEFT JOIN Transactions AS t ON u.account = t.account
-GROUP BY u.account
-HAVING sum(amount) > 10000;
+    Users
+    JOIN Transactions USING (account)
+GROUP BY account
+HAVING balance > 10000;
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1587.Bank Account Summary II/README_EN.md

@@ -84,31 +84,24 @@ Charlie's balance is (6000 + 6000 - 4000) = 8000.
 
 ## Solutions
 
+**Solution 1: Equi-Join + Group By + Sum**
+
+We can use an equi-join to join the `Users` table and the `Transactions` table on the condition of `account`, and then group by `account` to calculate the balance for each account using the `SUM` function. Finally, we can filter out the users whose balance is less than or equal to $10000$.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
-```sql
-SELECT
-    u.name,
-    SUM(t.amount) AS balance
-FROM
-    users AS u
-    JOIN transactions AS t ON u.account = t.account
-GROUP BY name
-HAVING SUM(t.amount) > 10000;
-```
-
 ```sql
 # Write your MySQL query statement below
 SELECT
     name,
     sum(amount) AS balance
 FROM
-    Users AS u
-    LEFT JOIN Transactions AS t ON u.account = t.account
-GROUP BY u.account
-HAVING sum(amount) > 10000;
+    Users
+    JOIN Transactions USING (account)
+GROUP BY account
+HAVING balance > 10000;
 ```
 
 <!-- tabs:end -->