feat: add sql solutions to lc problems

solution/1100-1199/1179.Reformat Department Table/README.md

@@ -124,7 +124,7 @@ SELECT
         END
     ) AS Dec_Revenue
 FROM Department
-GROUP BY id;
+GROUP BY 1;
 ```
 
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solution/1100-1199/1179.Reformat Department Table/README_EN.md

@@ -125,7 +125,7 @@ SELECT
         END
     ) AS Dec_Revenue
 FROM Department
-GROUP BY id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1179.Reformat Department Table/Solution.sql

@@ -62,4 +62,4 @@ SELECT
         END
     ) AS Dec_Revenue
 FROM Department
-GROUP BY id;
+GROUP BY 1;

solution/1100-1199/1193.Monthly Transactions I/README.md

@@ -65,15 +65,15 @@ Transactions</code> table:
 
 ```sql
 # Write your MySQL query statement below
-SELECT  DATE_FORMAT(trans_date,'%Y-%m')      AS month
-       ,country
-       ,COUNT(*)                             AS trans_count
-       ,COUNT(IF(state = 'approved',1,NULL)) AS approved_count
-       ,SUM(amount)                          AS trans_total_amount
-       ,SUM(IF(state = 'approved',amount,0)) AS approved_total_amount
+SELECT
+    date_format(trans_date, '%Y-%m') AS month,
+    country,
+    count(1) AS trans_count,
+    sum(state = 'approved') AS approved_count,
+    sum(amount) AS trans_total_amount,
+    sum(if(state = 'approved', amount, 0)) AS approved_total_amount
 FROM Transactions
-GROUP BY  month
-         ,country
+GROUP BY 1, 2;
 ```
 
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solution/1100-1199/1193.Monthly Transactions I/README_EN.md

@@ -61,15 +61,15 @@ Transactions table:
 
 ```sql
 # Write your MySQL query statement below
-SELECT  DATE_FORMAT(trans_date,'%Y-%m')      AS month
-       ,country
-       ,COUNT(*)                             AS trans_count
-       ,COUNT(IF(state = 'approved',1,NULL)) AS approved_count
-       ,SUM(amount)                          AS trans_total_amount
-       ,SUM(IF(state = 'approved',amount,0)) AS approved_total_amount
+SELECT
+    date_format(trans_date, '%Y-%m') AS month,
+    country,
+    count(1) AS trans_count,
+    sum(state = 'approved') AS approved_count,
+    sum(amount) AS trans_total_amount,
+    sum(if(state = 'approved', amount, 0)) AS approved_total_amount
 FROM Transactions
-GROUP BY  month
-         ,country
+GROUP BY 1, 2;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1193.Monthly Transactions I/Solution.sql

@@ -1,10 +1,10 @@
 # Write your MySQL query statement below
 SELECT
-    DATE_FORMAT(trans_date, '%Y-%m') AS month,
+    date_format(trans_date, '%Y-%m') AS month,
     country,
-    COUNT(*) AS trans_count,
-    COUNT(IF(state = 'approved', 1, NULL)) AS approved_count,
-    SUM(amount) AS trans_total_amount,
-    SUM(IF(state = 'approved', amount, 0)) AS approved_total_amount
+    count(1) AS trans_count,
+    sum(state = 'approved') AS approved_count,
+    sum(amount) AS trans_total_amount,
+    sum(if(state = 'approved', amount, 0)) AS approved_total_amount
 FROM Transactions
-GROUP BY month, country;
+GROUP BY 1, 2;

solution/1200-1299/1204.Last Person to Fit in the Bus/README.md

@@ -81,4 +81,20 @@ ORDER BY a.turn DESC
 LIMIT 1;
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            person_name,
+            sum(weight) OVER (ORDER BY turn) AS s
+        FROM Queue
+    )
+SELECT person_name
+FROM T
+WHERE s <= 1000
+ORDER BY s DESC
+LIMIT 1;
+```
+
 <!-- tabs:end -->

solution/1200-1299/1204.Last Person to Fit in the Bus/README_EN.md

@@ -84,4 +84,20 @@ ORDER BY a.turn DESC
 LIMIT 1;
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            person_name,
+            sum(weight) OVER (ORDER BY turn) AS s
+        FROM Queue
+    )
+SELECT person_name
+FROM T
+WHERE s <= 1000
+ORDER BY s DESC
+LIMIT 1;
+```
+
 <!-- tabs:end -->

solution/1200-1299/1204.Last Person to Fit in the Bus/Solution.sql

@@ -1,10 +1,13 @@
 # Write your MySQL query statement below
-SELECT a.person_name
-FROM
-    Queue AS a,
-    Queue AS b
-WHERE a.turn >= b.turn
-GROUP BY a.person_id
-HAVING SUM(b.weight) <= 1000
-ORDER BY a.turn DESC
+WITH
+    T AS (
+        SELECT
+            person_name,
+            sum(weight) OVER (ORDER BY turn) AS s
+        FROM Queue
+    )
+SELECT person_name
+FROM T
+WHERE s <= 1000
+ORDER BY s DESC
 LIMIT 1;

solution/1200-1299/1205.Monthly Transactions II/README.md

@@ -89,13 +89,13 @@ Chargebacks 表：
 ```sql
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT * FROM Transactions
-        UNION ALL
-        SELECT id, country, 'chargeback' AS state, amount, cb.trans_date
+        UNION
+        SELECT id, country, 'chargeback', amount, c.trans_date
         FROM
-            Chargebacks AS cb
-            LEFT JOIN Transactions AS tx ON cb.trans_id = tx.id
+            Transactions AS t
+            JOIN Chargebacks AS c ON t.id = c.trans_id
     )
 SELECT
     date_format(trans_date, '%Y-%m') AS month,
@@ -104,7 +104,7 @@ SELECT
     sum(if(state = 'approved', amount, 0)) AS approved_amount,
     sum(state = 'chargeback') AS chargeback_count,
     sum(if(state = 'chargeback', amount, 0)) AS chargeback_amount
-FROM t
+FROM T
 GROUP BY 1, 2
 HAVING approved_amount OR chargeback_amount;
 ```

solution/1200-1299/1205.Monthly Transactions II/README_EN.md

@@ -86,13 +86,13 @@ Chargebacks table:
 ```sql
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT * FROM Transactions
-        UNION ALL
-        SELECT id, country, 'chargeback' AS state, amount, cb.trans_date
+        UNION
+        SELECT id, country, 'chargeback', amount, c.trans_date
         FROM
-            Chargebacks AS cb
-            LEFT JOIN Transactions AS tx ON cb.trans_id = tx.id
+            Transactions AS t
+            JOIN Chargebacks AS c ON t.id = c.trans_id
     )
 SELECT
     date_format(trans_date, '%Y-%m') AS month,
@@ -101,7 +101,7 @@ SELECT
     sum(if(state = 'approved', amount, 0)) AS approved_amount,
     sum(state = 'chargeback') AS chargeback_count,
     sum(if(state = 'chargeback', amount, 0)) AS chargeback_amount
-FROM t
+FROM T
 GROUP BY 1, 2
 HAVING approved_amount OR chargeback_amount;
 ```

solution/1200-1299/1205.Monthly Transactions II/Solution.sql

@@ -1,12 +1,12 @@
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT * FROM Transactions
-        UNION ALL
-        SELECT id, country, 'chargeback' AS state, amount, cb.trans_date
+        UNION
+        SELECT id, country, 'chargeback', amount, c.trans_date
         FROM
-            Chargebacks AS cb
-            LEFT JOIN Transactions AS tx ON cb.trans_id = tx.id
+            Transactions AS t
+            JOIN Chargebacks AS c ON t.id = c.trans_id
     )
 SELECT
     date_format(trans_date, '%Y-%m') AS month,
@@ -15,6 +15,6 @@ SELECT
     sum(if(state = 'approved', amount, 0)) AS approved_amount,
     sum(state = 'chargeback') AS chargeback_count,
     sum(if(state = 'chargeback', amount, 0)) AS chargeback_amount
-FROM t
+FROM T
 GROUP BY 1, 2
 HAVING approved_amount OR chargeback_amount;

solution/2700-2799/2772.Apply Operations to Make All Array Elements Equal to Zero/README_EN.md

@@ -48,22 +48,22 @@
 
 **Solution 1: Difference Array + Prefix Sum**
 
-Let's first consider the first element $nums[0]$ of $nums$:
+First, let's consider the first element of $nums$, $nums[0]$:
 
--   If $nums[0] = 0$, then we don't need to do anything;
--   If $nums[0] \gt 0$, then we need to operate $nums[0]$ times on $nums[0..k-1]$ to make the elements in $nums[0..k-1]$ all subtract $nums[0]$, so that $nums[0]$ becomes $0$.
+-   If $nums[0] = 0$, we don't need to do anything.
+-   If $nums[0] > 0$, we need to operate on $nums[0..k-1]$ for $nums[0]$ times, subtracting $nums[0]$ from all elements in $nums[0..k-1]$, so $nums[0]$ becomes $0$.
 
-We can use difference array to maintain the operations on a segment of continuous elements. We use $d[i]$ to represent the difference array, and take the prefix sum of the difference array to get the variation amount of each position.
+To perform the add and subtract operations on a contiguous segment of elements simultaneously, we can use a difference array to manage these operations. We represent the difference array with $d[i]$, and calculating the prefix sum of the difference array gives us the change of the value at each position.
 
-Therefore, we traverse $nums$, for each element $nums[i]$, the current variation amount $s = \sum_{j=0}^{i} d[j]$ and we add $s$ to $nums[i]$ to get the actual value of $nums[i]$.
+Therefore, we iterate over $nums$. For each element $nums[i]$, the current position's change quantity is $s = \sum_{j=0}^{i} d[j]$. We add $s$ to $nums[i]$ to get the actual value of $nums[i]$.
 
--   If $nums[i] = 0$, then we don't need to do anything, just skip.
--   If $nums[i]=0$ or $i + k \gt n$, it means that after the previous operation, $nums[i]$ has become negative, or $nums[i..i+k-1]$ out of bounds, so it is impossible to make all elements in $nums$ equal to $0$, then return `false`. Otherwise, we need to subtract $nums[i]$ from all elements in $[i..i+k-1]$, so we subtract $nums[i]$ from $s$ and add $nums[i]$ to $d[i+k]$.
--   Continue to traverse the next element.
+-   If $nums[i] = 0$, there's no need for any operation, and we can skip directly.
+-   If $nums[i]=0$ or $i + k > n$, it indicates that after the previous operations, $nums[i]$ has become negative, or $nums[i..i+k-1]$ is out of bounds. Therefore, it's impossible to make all elements in $nums$ equal to $0$. We return `false`. Otherwise, we need to subtract $nums[i]$ from all elements in the interval $[i..i+k-1]$. Therefore, we subtract $nums[i]$ from $s$ and add $nums[i]$ to $d[i+k]$.
+-   We continue to iterate over the next element.
 
-If the traversal is over, it means that we can make all elements in $nums$ equal to $0$, return `true`.
+If the iteration ends, it means that all elements in $nums$ can be made equal to $0$, so we return `true`.
 
-Time complexity $O(n)$, space complexity $O(n)$. Where $n$ is the length of the array $nums$.
+The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $nums$.
 
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