feat: add solutions to lc problems: No.3053~3056 (doocs#2383)

Author: yanglbme

solution/3000-3099/3053.Classifying Triangles by Lengths/README.md

@@ -66,12 +66,33 @@ Triangles table:
 
 ## 解法
 
-### 方法一
+### 方法一：使用 CASE WHEN 语句
+
+我们可以使用 `CASE WHEN` 语句来判断三角形的类型。
+
+首先，我们需要判断三个边是否能够构成一个三角形。如果不能，我们返回 `Not A Triangle`。
+
+然后，我们判断三个边的长度是否相等。如果相等，我们返回 `Equilateral`。
+
+接着，我们判断是否有两个边的长度相等。如果有，我们返回 `Isosceles`。
+
+否则，说明三个边的长度都不相等，我们返回 `Scalene`。
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    CASE
+        WHEN A + B <= C
+        OR A + C <= B
+        OR B + C <= A THEN 'Not A Triangle'
+        WHEN A = B
+        AND B = c THEN 'Equilateral'
+        WHEN (A = B) + (B = C) + (A = C) = 1 THEN 'Isosceles'
+        ELSE 'Scalene'
+    END AS triangle_type
+FROM Triangles;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3053.Classifying Triangles by Lengths/README_EN.md

@@ -64,12 +64,33 @@ Triangles table:
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Using CASE WHEN Statement
+
+We can use the `CASE WHEN` statement to determine the type of the triangle.
+
+First, we need to determine whether the three sides can form a triangle. If not, we return `Not A Triangle`.
+
+Then, we check if the lengths of the three sides are equal. If they are, we return `Equilateral`.
+
+Next, we check if there are two sides with equal length. If there are, we return `Isosceles`.
+
+Otherwise, it means that the lengths of the three sides are all different, so we return `Scalene`.
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    CASE
+        WHEN A + B <= C
+        OR A + C <= B
+        OR B + C <= A THEN 'Not A Triangle'
+        WHEN A = B
+        AND B = c THEN 'Equilateral'
+        WHEN (A = B) + (B = C) + (A = C) = 1 THEN 'Isosceles'
+        ELSE 'Scalene'
+    END AS triangle_type
+FROM Triangles;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3053.Classifying Triangles by Lengths/Solution.sql

@@ -0,0 +1,12 @@
+# Write your MySQL query statement below
+SELECT
+    CASE
+        WHEN A + B <= C
+        OR A + C <= B
+        OR B + C <= A THEN 'Not A Triangle'
+        WHEN A = B
+        AND B = c THEN 'Equilateral'
+        WHEN (A = B) + (B = C) + (A = C) = 1 THEN 'Isosceles'
+        ELSE 'Scalene'
+    END AS triangle_type
+FROM Triangles;

solution/3000-3099/3054.Binary Tree Nodes/README.md

@@ -70,12 +70,23 @@ Tree table:
 
 ## 解法
 
-### 方法一
+### 方法一：左连接
+
+如果一个节点的父节点为空，则它是根节点；如果一个节点不是任何节点的父节点，则它是叶子节点；否则它是内部节点。
+
+因此，我们使用左连接来连接两次 `Tree` 表，连接条件是 `t1.N = t2.P`。那么如果 `t1.P` 为空，则 `t1.N` 是根节点；如果 `t2.P` 为空，则 `t1.N` 是叶子节点；否则 `t1.N` 是内部节点。
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT DISTINCT
+    t1.N AS N,
+    IF(t1.P IS NULL, 'Root', IF(t2.P IS NULL, 'Leaf', 'Inner')) AS Type
+FROM
+    Tree AS t1
+    LEFT JOIN Tree AS t2 ON t1.N = t2.p
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3054.Binary Tree Nodes/README_EN.md

@@ -68,12 +68,23 @@ Tree table:
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Left Join
+
+If a node's parent is null, then it is a root node; if a node is not the parent of any node, then it is a leaf node; otherwise, it is an internal node.
+
+Therefore, we use left join to join the `Tree` table twice, with the join condition being `t1.N = t2.P`. If `t1.P` is null, then `t1.N` is a root node; if `t2.P` is null, then `t1.N` is a leaf node; otherwise, `t1.N` is an internal node.
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT DISTINCT
+    t1.N AS N,
+    IF(t1.P IS NULL, 'Root', IF(t2.P IS NULL, 'Leaf', 'Inner')) AS Type
+FROM
+    Tree AS t1
+    LEFT JOIN Tree AS t2 ON t1.N = t2.p
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3054.Binary Tree Nodes/Solution.sql

@@ -0,0 +1,8 @@
+# Write your MySQL query statement below
+SELECT DISTINCT
+    t1.N AS N,
+    IF(t1.P IS NULL, 'Root', IF(t2.P IS NULL, 'Leaf', 'Inner')) AS Type
+FROM
+    Tree AS t1
+    LEFT JOIN Tree AS t2 ON t1.N = t2.p
+ORDER BY 1;

solution/3000-3099/3055.Top Percentile Fraud/README.md

@@ -73,12 +73,28 @@ Output table is ordered by state in ascending order, fraud score in descending o
 
 ## 解法
 
-### 方法一
+### 方法一：使用窗口函数
+
+我们可以使用 `RANK()` 窗口函数来计算每个州的欺诈分数的排名，然后筛选出排名为 1 的记录，并且按照题目要求排序。
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            RANK() OVER (
+                PARTITION BY state
+                ORDER BY fraud_score DESC
+            ) AS rk
+        FROM Fraud
+    )
+SELECT policy_id, state, fraud_score
+FROM T
+WHERE rk = 1
+ORDER BY 2, 3 DESC, 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3055.Top Percentile Fraud/README_EN.md

@@ -71,12 +71,28 @@ Output table is ordered by state in ascending order, fraud score in descending o
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Using Window Function
+
+We can use the `RANK()` window function to calculate the ranking of fraud scores for each state, then filter out the records with a rank of 1, and sort them as required by the problem.
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            RANK() OVER (
+                PARTITION BY state
+                ORDER BY fraud_score DESC
+            ) AS rk
+        FROM Fraud
+    )
+SELECT policy_id, state, fraud_score
+FROM T
+WHERE rk = 1
+ORDER BY 2, 3 DESC, 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3055.Top Percentile Fraud/Solution.sql

@@ -0,0 +1,15 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            RANK() OVER (
+                PARTITION BY state
+                ORDER BY fraud_score DESC
+            ) AS rk
+        FROM Fraud
+    )
+SELECT policy_id, state, fraud_score
+FROM T
+WHERE rk = 1
+ORDER BY 2, 3 DESC, 1;

solution/3000-3099/3056.Snaps Analysis/README.md

@@ -98,12 +98,22 @@ All percentages in output table rounded to the two decimal places.
 
 ## 解法
 
-### 方法一
+### 方法一：等值连接 + 分组求和
+
+我们可以通过等值连接，将 `Activities` 表和 `Age` 表按照 `user_id` 进行连接，然后再按照 `age_bucket` 进行分组，最后计算每个年龄段的发送和打开的百分比。
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    age_bucket,
+    ROUND(100 * SUM(IF(activity_type = 'send', time_spent, 0)) / SUM(time_spent), 2) AS send_perc,
+    ROUND(100 * SUM(IF(activity_type = 'open', time_spent, 0)) / SUM(time_spent), 2) AS open_perc
+FROM
+    Activities
+    JOIN Age USING (user_id)
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3056.Snaps Analysis/README_EN.md

@@ -96,12 +96,22 @@ All percentages in output table rounded to the two decimal places.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Equi-Join + Group By Summation
+
+We can perform an equi-join to connect the `Activities` table and the `Age` table based on `user_id`. Then, group by `age_bucket` and finally calculate the percentage of sends and opens for each age group.
 
 <!-- tabs:start -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    age_bucket,
+    ROUND(100 * SUM(IF(activity_type = 'send', time_spent, 0)) / SUM(time_spent), 2) AS send_perc,
+    ROUND(100 * SUM(IF(activity_type = 'open', time_spent, 0)) / SUM(time_spent), 2) AS open_perc
+FROM
+    Activities
+    JOIN Age USING (user_id)
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/3000-3099/3056.Snaps Analysis/Solution.sql

@@ -0,0 +1,9 @@
+# Write your MySQL query statement below
+SELECT
+    age_bucket,
+    ROUND(100 * SUM(IF(activity_type = 'send', time_spent, 0)) / SUM(time_spent), 2) AS send_perc,
+    ROUND(100 * SUM(IF(activity_type = 'open', time_spent, 0)) / SUM(time_spent), 2) AS open_perc
+FROM
+    Activities
+    JOIN Age USING (user_id)
+GROUP BY 1;