feat: add solutions to lc problems: No.3050,3051

solution/3000-3099/3050.Pizza Toppings Cost Analysis/README.md

@@ -67,12 +67,29 @@ Output table is ordered by the total cost in descending order.</pre>
 
 ## 解法
 
-### 方法一
+### 方法一：窗口函数 + 条件连接
+
+我们先使用窗口函数，按照 `topping_name` 字段对表进行排序，并为每一行添加一个 `rk` 字段，表示当前行的排名。
+
+然后我们使用条件连接，连接三次表 `T`，分别为 `t1`, `t2`, `t3`。连接条件是 `t1.rk < t2.rk` 和 `t2.rk < t3.rk`。然后我们计算三个配料的总价，按照总价降序排序，再按照配料名升序排序。
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT *, RANK() OVER (ORDER BY topping_name) AS rk
+        FROM Toppings
+    )
+SELECT
+    CONCAT(t1.topping_name, ',', t2.topping_name, ',', t3.topping_name) AS pizza,
+    t1.cost + t2.cost + t3.cost AS total_cost
+FROM
+    T AS t1
+    JOIN T AS t2 ON t1.rk < t2.rk
+    JOIN T AS t3 ON t2.rk < t3.rk
+ORDER BY 2 DESC, 1 ASC;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3050.Pizza Toppings Cost Analysis/README_EN.md

@@ -65,12 +65,29 @@ Output table is ordered by the total cost in descending order.</pre>
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Window Function + Conditional Join
+
+First, we use a window function to sort the table by the `topping_name` field and add a `rk` field to each row, representing the ranking of the current row.
+
+Then we use conditional join to join the table `T` three times, named as `t1`, `t2`, `t3` respectively. The join conditions are `t1.rk < t2.rk` and `t2.rk < t3.rk`. After that, we calculate the total price of the three toppings, sort by total price in descending order, and then sort by topping name in ascending order.
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT *, RANK() OVER (ORDER BY topping_name) AS rk
+        FROM Toppings
+    )
+SELECT
+    CONCAT(t1.topping_name, ',', t2.topping_name, ',', t3.topping_name) AS pizza,
+    t1.cost + t2.cost + t3.cost AS total_cost
+FROM
+    T AS t1
+    JOIN T AS t2 ON t1.rk < t2.rk
+    JOIN T AS t3 ON t2.rk < t3.rk
+ORDER BY 2 DESC, 1 ASC;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3050.Pizza Toppings Cost Analysis/Solution.sql

@@ -0,0 +1,14 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT *, RANK() OVER (ORDER BY topping_name) AS rk
+        FROM Toppings
+    )
+SELECT
+    CONCAT(t1.topping_name, ',', t2.topping_name, ',', t3.topping_name) AS pizza,
+    t1.cost + t2.cost + t3.cost AS total_cost
+FROM
+    T AS t1
+    JOIN T AS t2 ON t1.rk < t2.rk
+    JOIN T AS t3 ON t2.rk < t3.rk
+ORDER BY 2 DESC, 1 ASC;

solution/3000-3099/3051.Find Candidates for Data Scientist Position/README.md

@@ -65,12 +65,20 @@ The output table is sorted by candidate_id in ascending order.
 
 ## 解法
 
-### 方法一
+### 方法一：条件筛选 + 分组统计
+
+我们首先筛选出具备 `Python`, `Tableau`, `PostgreSQL` 这三个技能的候选人，然后按照 `candidate_id` 进行分组统计，统计每个候选人具备的技能数量，最后筛选出具备这三个技能的候选人，并且按照 `candidate_id` 进行升序排序。
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT candidate_id
+FROM Candidates
+WHERE skill IN ('Python', 'Tableau', 'PostgreSQL')
+GROUP BY 1
+HAVING COUNT(1) = 3
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3051.Find Candidates for Data Scientist Position/README_EN.md

@@ -63,12 +63,20 @@ The output table is sorted by candidate_id in ascending order.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Conditional Filtering + Grouping Statistics
+
+First, we filter out candidates who have the skills `Python`, `Tableau`, and `PostgreSQL`. Then, we group by `candidate_id` and count the number of skills each candidate has. Finally, we filter out candidates who have these three skills and sort them in ascending order by `candidate_id`.
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT candidate_id
+FROM Candidates
+WHERE skill IN ('Python', 'Tableau', 'PostgreSQL')
+GROUP BY 1
+HAVING COUNT(1) = 3
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3051.Find Candidates for Data Scientist Position/Solution.sql

@@ -0,0 +1,7 @@
+# Write your MySQL query statement below
+SELECT candidate_id
+FROM Candidates
+WHERE skill IN ('Python', 'Tableau', 'PostgreSQL')
+GROUP BY 1
+HAVING COUNT(1) = 3
+ORDER BY 1;