feat: add sql solutions to lc problems

.prettierrc

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

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

@@ -50,7 +50,8 @@ day 1
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sh
-
+# Read from the file words.txt and output the word frequency list to stdout.
+cat words.txt | tr -s ' ' '\n' | sort | uniq -c | sort -nr | awk '{print $2, $1}'
 ```
 
 <!-- tabs:end -->

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

@@ -46,7 +46,8 @@ day 1
 ### **Bash**
 
 ```sh
-
+# Read from the file words.txt and output the word frequency list to stdout.
+cat words.txt | tr -s ' ' '\n' | sort | uniq -c | sort -nr | awk '{print $2, $1}'
 ```
 
 <!-- tabs:end -->

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

@@ -1,17 +1,2 @@
-#!/usr/bin/env bash
-
-# NF是当前行的field字段数；NR是正在处理的当前行数。
-awk '{
-  for(i=1;i<=NF;i++){
-    res[$i]++
-  }
-}
-END{
-  for(k in res){
-    print k,res[k]
-  }
-}' words.txt | sort -nr -k2
-
-#or:
-
-cat words.txt | tr -s ' ' '\n' | sort | uniq -c | sort -r | awk '{print $2" "$1}'
+# Read from the file words.txt and output the word frequency list to stdout.
+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

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

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

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

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

@@ -0,0 +1,27 @@
+# Write your MySQL query statement below
+WITH
+    P AS (
+        SELECT 'Android' AS platform
+        UNION
+        SELECT 'IOS'
+        UNION
+        SELECT 'Web'
+    ),
+    Exp AS (
+        SELECT 'Reading' AS experiment_name
+        UNION
+        SELECT 'Sports'
+        UNION
+        SELECT 'Programming'
+    ),
+    T AS (
+        SELECT *
+        FROM
+            P,
+            Exp
+    )
+SELECT platform, experiment_name, count(experiment_id) AS num_experiments
+FROM
+    T AS t
+    LEFT JOIN Experiments USING (platform, experiment_name)
+GROUP BY 1, 2;

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

@@ -89,7 +89,15 @@ Streams table:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT count(DISTINCT sub.account_id) AS accounts_count
+FROM
+    Subscriptions AS sub
+    LEFT JOIN Streams AS ss ON sub.account_id = ss.account_id
+WHERE
+    year(start_date) <= 2021
+    AND year(end_date) >= 2021
+    AND (year(stream_date) != 2021 OR stream_date > end_date);
 ```
 
 <!-- tabs:end -->

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

@@ -86,7 +86,15 @@ User 11 did not subscribe in 2021.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT count(DISTINCT sub.account_id) AS accounts_count
+FROM
+    Subscriptions AS sub
+    LEFT JOIN Streams AS ss ON sub.account_id = ss.account_id
+WHERE
+    year(start_date) <= 2021
+    AND year(end_date) >= 2021
+    AND (year(stream_date) != 2021 OR stream_date > end_date);
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,9 @@
+# Write your MySQL query statement below
+SELECT count(DISTINCT sub.account_id) AS accounts_count
+FROM
+    Subscriptions AS sub
+    LEFT JOIN Streams AS ss ON sub.account_id = ss.account_id
+WHERE
+    year(start_date) <= 2021
+    AND year(end_date) >= 2021
+    AND (year(stream_date) != 2021 OR stream_date > end_date);

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

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

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

@@ -94,7 +94,14 @@ Rounds table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT candidate_id
+FROM
+    Candidates AS c
+    LEFT JOIN Rounds AS r ON c.interview_id = r.interview_id
+WHERE years_of_exp >= 2
+GROUP BY c.interview_id
+HAVING sum(score) > 15;
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,8 @@
+# Write your MySQL query statement below
+SELECT candidate_id
+FROM
+    Candidates AS c
+    LEFT JOIN Rounds AS r ON c.interview_id = r.interview_id
+WHERE years_of_exp >= 2
+GROUP BY c.interview_id
+HAVING sum(score) > 15;

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

@@ -135,7 +135,23 @@ Purchases 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    m.member_id,
+    name,
+    CASE
+        WHEN count(v.visit_id) = 0 THEN 'Bronze'
+        WHEN 100 * count(charged_amount) / count(
+            v.visit_id
+        ) >= 80 THEN 'Diamond'
+        WHEN 100 * count(charged_amount) / count(v.visit_id) >= 50 THEN 'Gold'
+        ELSE 'Silver'
+    END AS category
+FROM
+    Members AS m
+    LEFT JOIN Visits AS v ON m.member_id = v.member_id
+    LEFT JOIN Purchases AS p ON v.visit_id = p.visit_id
+GROUP BY member_id;
 ```
 
 <!-- tabs:end -->

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

@@ -129,7 +129,23 @@ Purchases table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    m.member_id,
+    name,
+    CASE
+        WHEN count(v.visit_id) = 0 THEN 'Bronze'
+        WHEN 100 * count(charged_amount) / count(
+            v.visit_id
+        ) >= 80 THEN 'Diamond'
+        WHEN 100 * count(charged_amount) / count(v.visit_id) >= 50 THEN 'Gold'
+        ELSE 'Silver'
+    END AS category
+FROM
+    Members AS m
+    LEFT JOIN Visits AS v ON m.member_id = v.member_id
+    LEFT JOIN Purchases AS p ON v.visit_id = p.visit_id
+GROUP BY member_id;
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,17 @@
+# Write your MySQL query statement below
+SELECT
+    m.member_id,
+    name,
+    CASE
+        WHEN count(v.visit_id) = 0 THEN 'Bronze'
+        WHEN 100 * count(charged_amount) / count(
+            v.visit_id
+        ) >= 80 THEN 'Diamond'
+        WHEN 100 * count(charged_amount) / count(v.visit_id) >= 50 THEN 'Gold'
+        ELSE 'Silver'
+    END AS category
+FROM
+    Members AS m
+    LEFT JOIN Visits AS v ON m.member_id = v.member_id
+    LEFT JOIN Purchases AS p ON v.visit_id = p.visit_id
+GROUP BY member_id;

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

@@ -80,7 +80,16 @@ Transactions 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    account_id,
+    day,
+    sum(if(type = 'Deposit', amount, -amount)) OVER (
+        PARTITION BY account_id
+        ORDER BY day
+    ) AS balance
+FROM Transactions
+ORDER BY 1, 2;
 ```
 
 <!-- tabs:end -->

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

@@ -72,7 +72,16 @@ Account 2:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    account_id,
+    day,
+    sum(if(type = 'Deposit', amount, -amount)) OVER (
+        PARTITION BY account_id
+        ORDER BY day
+    ) AS balance
+FROM Transactions
+ORDER BY 1, 2;
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,10 @@
+# Write your MySQL query statement below
+SELECT
+    account_id,
+    day,
+    sum(if(type = 'Deposit', amount, -amount)) OVER (
+        PARTITION BY account_id
+        ORDER BY day
+    ) AS balance
+FROM Transactions
+ORDER BY 1, 2;

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

@@ -80,7 +80,18 @@ Orders table:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT DISTINCT customer_id
+        FROM Orders
+        WHERE order_type = 0
+    )
+SELECT *
+FROM Orders AS o
+WHERE
+    order_type = 0
+    OR NOT EXISTS (SELECT 1 FROM T AS t WHERE t.customer_id = o.customer_id);
 ```
 
 <!-- tabs:end -->

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

@@ -76,7 +76,18 @@ Customer 4 has two orders of type 1. We return both of them.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT DISTINCT customer_id
+        FROM Orders
+        WHERE order_type = 0
+    )
+SELECT *
+FROM Orders AS o
+WHERE
+    order_type = 0
+    OR NOT EXISTS (SELECT 1 FROM T AS t WHERE t.customer_id = o.customer_id);
 ```
 
 <!-- tabs:end -->