feat: add solutions to lc problem: No.3482

solution/3400-3499/3482.Analyze Organization Hierarchy/README.md

@@ -20,7 +20,7 @@ tags:
 
 <pre>
 +----------------+---------+
-| Column Name    | Type    | 
+| Column Name    | Type    |
 +----------------+---------+
 | employee_id    | int     |
 | employee_name  | varchar |
@@ -151,7 +151,89 @@ manager_id is null for the top-level manager (CEO).
 #### MySQL
 
 ```sql
+# Write your MySQL query statement below
+WITH RECURSIVE
+    level_cte AS (
+        SELECT employee_id, manager_id, 1 AS level, salary FROM Employees
+        UNION ALL
+        SELECT a.employee_id, b.manager_id, level + 1, a.salary
+        FROM
+            level_cte a
+            JOIN Employees b ON b.employee_id = a.manager_id
+    ),
+    employee_with_level AS (
+        SELECT a.employee_id, a.employee_name, a.salary, b.level
+        FROM
+            Employees a,
+            (SELECT employee_id, level FROM level_cte WHERE manager_id IS NULL) b
+        WHERE a.employee_id = b.employee_id
+    )
+SELECT
+    a.employee_id,
+    a.employee_name,
+    a.level,
+    COALESCE(b.team_size, 0) AS team_size,
+    a.salary + COALESCE(b.budget, 0) AS budget
+FROM
+    employee_with_level a
+    LEFT JOIN (
+        SELECT manager_id AS employee_id, COUNT(*) AS team_size, SUM(salary) AS budget
+        FROM level_cte
+        WHERE manager_id IS NOT NULL
+        GROUP BY manager_id
+    ) b
+        ON a.employee_id = b.employee_id
+ORDER BY level, budget DESC, employee_name;
+```
 
+#### Pandas
+
+```python
+import pandas as pd
+
+
+def analyze_organization_hierarchy(employees: pd.DataFrame) -> pd.DataFrame:
+    # 初始化 CEO (level 1)
+    employees = employees.copy()
+    employees["level"] = None
+    ceo_id = employees.loc[employees["manager_id"].isna(), "employee_id"].values[0]
+    employees.loc[employees["employee_id"] == ceo_id, "level"] = 1
+
+    # 递归计算层级
+    def compute_levels(emp_df, level):
+        next_level_ids = emp_df[emp_df["level"] == level]["employee_id"].tolist()
+        if not next_level_ids:
+            return
+        emp_df.loc[emp_df["manager_id"].isin(next_level_ids), "level"] = level + 1
+        compute_levels(emp_df, level + 1)
+
+    compute_levels(employees, 1)
+
+    # 计算 team_size 和 budget
+    team_size = {eid: 0 for eid in employees["employee_id"]}
+    budget = {
+        eid: salary
+        for eid, salary in zip(employees["employee_id"], employees["salary"])
+    }
+
+    for eid in sorted(employees["employee_id"], reverse=True):
+        manager_id = employees.loc[
+            employees["employee_id"] == eid, "manager_id"
+        ].values[0]
+        if pd.notna(manager_id):
+            team_size[manager_id] += team_size[eid] + 1
+            budget[manager_id] += budget[eid]
+
+    # 生成最终 DataFrame
+    employees["team_size"] = employees["employee_id"].map(team_size)
+    employees["budget"] = employees["employee_id"].map(budget)
+
+    # 按 level 升序，budget 降序，employee_name 升序排序
+    employees = employees.sort_values(
+        by=["level", "budget", "employee_name"], ascending=[True, False, True]
+    )
+
+    return employees[["employee_id", "employee_name", "level", "team_size", "budget"]]
 ```
 
 <!-- tabs:end -->

solution/3400-3499/3482.Analyze Organization Hierarchy/README_EN.md

@@ -20,7 +20,7 @@ tags:
 
 <pre>
 +----------------+---------+
-| Column Name    | Type    | 
+| Column Name    | Type    |
 +----------------+---------+
 | employee_id    | int     |
 | employee_name  | varchar |
@@ -151,7 +151,84 @@ manager_id is null for the top-level manager (CEO).
 #### MySQL
 
 ```sql
+# Write your MySQL query statement below
+WITH RECURSIVE
+    level_cte AS (
+        SELECT employee_id, manager_id, 1 AS level, salary FROM Employees
+        UNION ALL
+        SELECT a.employee_id, b.manager_id, level + 1, a.salary
+        FROM
+            level_cte a
+            JOIN Employees b ON b.employee_id = a.manager_id
+    ),
+    employee_with_level AS (
+        SELECT a.employee_id, a.employee_name, a.salary, b.level
+        FROM
+            Employees a,
+            (SELECT employee_id, level FROM level_cte WHERE manager_id IS NULL) b
+        WHERE a.employee_id = b.employee_id
+    )
+SELECT
+    a.employee_id,
+    a.employee_name,
+    a.level,
+    COALESCE(b.team_size, 0) AS team_size,
+    a.salary + COALESCE(b.budget, 0) AS budget
+FROM
+    employee_with_level a
+    LEFT JOIN (
+        SELECT manager_id AS employee_id, COUNT(*) AS team_size, SUM(salary) AS budget
+        FROM level_cte
+        WHERE manager_id IS NOT NULL
+        GROUP BY manager_id
+    ) b
+        ON a.employee_id = b.employee_id
+ORDER BY level, budget DESC, employee_name;
+```
+
+#### Pandas
+
+```python
+import pandas as pd
+
+def analyze_organization_hierarchy(employees: pd.DataFrame) -> pd.DataFrame:
+    # Copy the input DataFrame to avoid modifying the original
+    employees = employees.copy()
+    employees['level'] = None
+
+    # Identify the CEO (level 1)
+    ceo_id = employees.loc[employees['manager_id'].isna(), 'employee_id'].values[0]
+    employees.loc[employees['employee_id'] == ceo_id, 'level'] = 1
+
+    # Recursively compute employee levels
+    def compute_levels(emp_df, level):
+        next_level_ids = emp_df[emp_df['level'] == level]['employee_id'].tolist()
+        if not next_level_ids:
+            return
+        emp_df.loc[emp_df['manager_id'].isin(next_level_ids), 'level'] = level + 1
+        compute_levels(emp_df, level + 1)
+
+    compute_levels(employees, 1)
+
+    # Initialize team size and budget dictionaries
+    team_size = {eid: 0 for eid in employees['employee_id']}
+    budget = {eid: salary for eid, salary in zip(employees['employee_id'], employees['salary'])}
+
+    # Compute team size and budget for each employee
+    for eid in sorted(employees['employee_id'], reverse=True):
+        manager_id = employees.loc[employees['employee_id'] == eid, 'manager_id'].values[0]
+        if pd.notna(manager_id):
+            team_size[manager_id] += team_size[eid] + 1
+            budget[manager_id] += budget[eid]
+
+    # Map computed team size and budget to employees DataFrame
+    employees['team_size'] = employees['employee_id'].map(team_size)
+    employees['budget'] = employees['employee_id'].map(budget)
+
+    # Sort the final result by level (ascending), budget (descending), and employee name (ascending)
+    employees = employees.sort_values(by=['level', 'budget', 'employee_name'], ascending=[True, False, True])
 
+    return employees[['employee_id', 'employee_name', 'level', 'team_size', 'budget']]
 ```
 
 <!-- tabs:end -->

solution/3400-3499/3482.Analyze Organization Hierarchy/Solution.py

@@ -0,0 +1,48 @@
+import pandas as pd
+
+
+def analyze_organization_hierarchy(employees: pd.DataFrame) -> pd.DataFrame:
+    # Copy the input DataFrame to avoid modifying the original
+    employees = employees.copy()
+    employees["level"] = None
+
+    # Identify the CEO (level 1)
+    ceo_id = employees.loc[employees["manager_id"].isna(), "employee_id"].values[0]
+    employees.loc[employees["employee_id"] == ceo_id, "level"] = 1
+
+    # Recursively compute employee levels
+    def compute_levels(emp_df, level):
+        next_level_ids = emp_df[emp_df["level"] == level]["employee_id"].tolist()
+        if not next_level_ids:
+            return
+        emp_df.loc[emp_df["manager_id"].isin(next_level_ids), "level"] = level + 1
+        compute_levels(emp_df, level + 1)
+
+    compute_levels(employees, 1)
+
+    # Initialize team size and budget dictionaries
+    team_size = {eid: 0 for eid in employees["employee_id"]}
+    budget = {
+        eid: salary
+        for eid, salary in zip(employees["employee_id"], employees["salary"])
+    }
+
+    # Compute team size and budget for each employee
+    for eid in sorted(employees["employee_id"], reverse=True):
+        manager_id = employees.loc[
+            employees["employee_id"] == eid, "manager_id"
+        ].values[0]
+        if pd.notna(manager_id):
+            team_size[manager_id] += team_size[eid] + 1
+            budget[manager_id] += budget[eid]
+
+    # Map computed team size and budget to employees DataFrame
+    employees["team_size"] = employees["employee_id"].map(team_size)
+    employees["budget"] = employees["employee_id"].map(budget)
+
+    # Sort the final result by level (ascending), budget (descending), and employee name (ascending)
+    employees = employees.sort_values(
+        by=["level", "budget", "employee_name"], ascending=[True, False, True]
+    )
+
+    return employees[["employee_id", "employee_name", "level", "team_size", "budget"]]

solution/3400-3499/3482.Analyze Organization Hierarchy/Solution.sql

@@ -0,0 +1,33 @@
+# Write your MySQL query statement below
+WITH RECURSIVE
+    level_cte AS (
+        SELECT employee_id, manager_id, 1 AS level, salary FROM Employees
+        UNION ALL
+        SELECT a.employee_id, b.manager_id, level + 1, a.salary
+        FROM
+            level_cte a
+            JOIN Employees b ON b.employee_id = a.manager_id
+    ),
+    employee_with_level AS (
+        SELECT a.employee_id, a.employee_name, a.salary, b.level
+        FROM
+            Employees a,
+            (SELECT employee_id, level FROM level_cte WHERE manager_id IS NULL) b
+        WHERE a.employee_id = b.employee_id
+    )
+SELECT
+    a.employee_id,
+    a.employee_name,
+    a.level,
+    COALESCE(b.team_size, 0) AS team_size,
+    a.salary + COALESCE(b.budget, 0) AS budget
+FROM
+    employee_with_level a
+    LEFT JOIN (
+        SELECT manager_id AS employee_id, COUNT(*) AS team_size, SUM(salary) AS budget
+        FROM level_cte
+        WHERE manager_id IS NOT NULL
+        GROUP BY manager_id
+    ) b
+        ON a.employee_id = b.employee_id
+ORDER BY level, budget DESC, employee_name;