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feat: add solutions to lc problem: No.3482 (#4151)
No.3482.Analyze Organization Hierarchy
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‎solution/3400-3499/3482.Analyze Organization Hierarchy/README.md

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2020

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

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

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@@ -20,7 +20,7 @@ tags:
2020

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<pre>
2222
+----------------+---------+
23-
| Column Name | Type |
23+
| Column Name | Type |
2424
+----------------+---------+
2525
| employee_id | int |
2626
| employee_name | varchar |
@@ -151,7 +151,84 @@ manager_id is null for the top-level manager (CEO).
151151
#### MySQL
152152

153153
```sql
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# Write your MySQL query statement below
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WITH RECURSIVE
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level_cte AS (
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SELECT employee_id, manager_id, 1 AS level, salary FROM Employees
158+
UNION ALL
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SELECT a.employee_id, b.manager_id, level + 1, a.salary
160+
FROM
161+
level_cte a
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JOIN Employees b ON b.employee_id = a.manager_id
163+
),
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employee_with_level AS (
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SELECT a.employee_id, a.employee_name, a.salary, b.level
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FROM
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Employees a,
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(SELECT employee_id, level FROM level_cte WHERE manager_id IS NULL) b
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WHERE a.employee_id = b.employee_id
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)
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SELECT
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a.employee_id,
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a.employee_name,
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a.level,
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COALESCE(b.team_size, 0) AS team_size,
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a.salary + COALESCE(b.budget, 0) AS budget
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FROM
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employee_with_level a
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LEFT JOIN (
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SELECT manager_id AS employee_id, COUNT(*) AS team_size, SUM(salary) AS budget
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FROM level_cte
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WHERE manager_id IS NOT NULL
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GROUP BY manager_id
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) b
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ON a.employee_id = b.employee_id
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ORDER BY level, budget DESC, employee_name;
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```
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#### Pandas
190+
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```python
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import pandas as pd
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def analyze_organization_hierarchy(employees: pd.DataFrame) -> pd.DataFrame:
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# Copy the input DataFrame to avoid modifying the original
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employees = employees.copy()
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employees['level'] = None
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# Identify the CEO (level 1)
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ceo_id = employees.loc[employees['manager_id'].isna(), 'employee_id'].values[0]
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employees.loc[employees['employee_id'] == ceo_id, 'level'] = 1
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# Recursively compute employee levels
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def compute_levels(emp_df, level):
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next_level_ids = emp_df[emp_df['level'] == level]['employee_id'].tolist()
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if not next_level_ids:
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return
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emp_df.loc[emp_df['manager_id'].isin(next_level_ids), 'level'] = level + 1
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compute_levels(emp_df, level + 1)
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compute_levels(employees, 1)
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# Initialize team size and budget dictionaries
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team_size = {eid: 0 for eid in employees['employee_id']}
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budget = {eid: salary for eid, salary in zip(employees['employee_id'], employees['salary'])}
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# Compute team size and budget for each employee
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for eid in sorted(employees['employee_id'], reverse=True):
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manager_id = employees.loc[employees['employee_id'] == eid, 'manager_id'].values[0]
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if pd.notna(manager_id):
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team_size[manager_id] += team_size[eid] + 1
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budget[manager_id] += budget[eid]
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# Map computed team size and budget to employees DataFrame
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employees['team_size'] = employees['employee_id'].map(team_size)
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employees['budget'] = employees['employee_id'].map(budget)
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# Sort the final result by level (ascending), budget (descending), and employee name (ascending)
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employees = employees.sort_values(by=['level', 'budget', 'employee_name'], ascending=[True, False, True])
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return employees[['employee_id', 'employee_name', 'level', 'team_size', 'budget']]
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```
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<!-- tabs:end -->

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

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@@ -0,0 +1,48 @@
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import pandas as pd
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3+
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def analyze_organization_hierarchy(employees: pd.DataFrame) -> pd.DataFrame:
5+
# Copy the input DataFrame to avoid modifying the original
6+
employees = employees.copy()
7+
employees["level"] = None
8+
9+
# Identify the CEO (level 1)
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ceo_id = employees.loc[employees["manager_id"].isna(), "employee_id"].values[0]
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employees.loc[employees["employee_id"] == ceo_id, "level"] = 1
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# Recursively compute employee levels
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def compute_levels(emp_df, level):
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next_level_ids = emp_df[emp_df["level"] == level]["employee_id"].tolist()
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if not next_level_ids:
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return
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emp_df.loc[emp_df["manager_id"].isin(next_level_ids), "level"] = level + 1
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compute_levels(emp_df, level + 1)
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compute_levels(employees, 1)
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# Initialize team size and budget dictionaries
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team_size = {eid: 0 for eid in employees["employee_id"]}
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budget = {
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eid: salary
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for eid, salary in zip(employees["employee_id"], employees["salary"])
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}
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# Compute team size and budget for each employee
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for eid in sorted(employees["employee_id"], reverse=True):
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manager_id = employees.loc[
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employees["employee_id"] == eid, "manager_id"
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].values[0]
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if pd.notna(manager_id):
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team_size[manager_id] += team_size[eid] + 1
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budget[manager_id] += budget[eid]
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# Map computed team size and budget to employees DataFrame
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employees["team_size"] = employees["employee_id"].map(team_size)
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employees["budget"] = employees["employee_id"].map(budget)
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# Sort the final result by level (ascending), budget (descending), and employee name (ascending)
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employees = employees.sort_values(
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by=["level", "budget", "employee_name"], ascending=[True, False, True]
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)
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return employees[["employee_id", "employee_name", "level", "team_size", "budget"]]

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

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# Write your MySQL query statement below
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WITH RECURSIVE
3+
level_cte AS (
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SELECT employee_id, manager_id, 1 AS level, salary FROM Employees
5+
UNION ALL
6+
SELECT a.employee_id, b.manager_id, level + 1, a.salary
7+
FROM
8+
level_cte a
9+
JOIN Employees b ON b.employee_id = a.manager_id
10+
),
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employee_with_level AS (
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SELECT a.employee_id, a.employee_name, a.salary, b.level
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FROM
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Employees a,
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(SELECT employee_id, level FROM level_cte WHERE manager_id IS NULL) b
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WHERE a.employee_id = b.employee_id
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)
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SELECT
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a.employee_id,
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a.employee_name,
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a.level,
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COALESCE(b.team_size, 0) AS team_size,
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a.salary + COALESCE(b.budget, 0) AS budget
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FROM
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employee_with_level a
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LEFT JOIN (
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SELECT manager_id AS employee_id, COUNT(*) AS team_size, SUM(salary) AS budget
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FROM level_cte
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WHERE manager_id IS NOT NULL
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GROUP BY manager_id
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) b
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ON a.employee_id = b.employee_id
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ORDER BY level, budget DESC, employee_name;

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