跳转至

3482. Analyze Organization Hierarchy

题目描述

Table: Employees

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| employee_id    | int     |
| employee_name  | varchar |
| manager_id     | int     |
| salary         | int     |
| department     | varchar |
+----------------+----------+
employee_id is the unique key for this table.
Each row contains information about an employee, including their ID, name, their manager's ID, salary, and department.
manager_id is null for the top-level manager (CEO).

Write a solution to analyze the organizational hierarchy and answer the following:

  1. Hierarchy Levels: For each employee, determine their level in the organization (CEO is level 1, employees reporting directly to the CEO are level 2, and so on).
  2. Team Size: For each employee who is a manager, count the total number of employees under them (direct and indirect reports).
  3. Salary Budget: For each manager, calculate the total salary budget they control (sum of salaries of all employees under them, including indirect reports, plus their own salary).

Return the result table ordered by the result ordered by level in ascending order, then by budget in descending order, and finally by employee_name in ascending order.

The result format is in the following example.

 

Example:

Input:

Employees table:

+-------------+---------------+------------+--------+-------------+
| employee_id | employee_name | manager_id | salary | department  |
+-------------+---------------+------------+--------+-------------+
| 1           | Alice         | null       | 12000  | Executive   |
| 2           | Bob           | 1          | 10000  | Sales       |
| 3           | Charlie       | 1          | 10000  | Engineering |
| 4           | David         | 2          | 7500   | Sales       |
| 5           | Eva           | 2          | 7500   | Sales       |
| 6           | Frank         | 3          | 9000   | Engineering |
| 7           | Grace         | 3          | 8500   | Engineering |
| 8           | Hank          | 4          | 6000   | Sales       |
| 9           | Ivy           | 6          | 7000   | Engineering |
| 10          | Judy          | 6          | 7000   | Engineering |
+-------------+---------------+------------+--------+-------------+

Output:

+-------------+---------------+-------+-----------+--------+
| employee_id | employee_name | level | team_size | budget |
+-------------+---------------+-------+-----------+--------+
| 1           | Alice         | 1     | 9         | 84500  |
| 3           | Charlie       | 2     | 4         | 41500  |
| 2           | Bob           | 2     | 3         | 31000  |
| 6           | Frank         | 3     | 2         | 23000  |
| 4           | David         | 3     | 1         | 13500  |
| 7           | Grace         | 3     | 0         | 8500   |
| 5           | Eva           | 3     | 0         | 7500   |
| 9           | Ivy           | 4     | 0         | 7000   |
| 10          | Judy          | 4     | 0         | 7000   |
| 8           | Hank          | 4     | 0         | 6000   |
+-------------+---------------+-------+-----------+--------+

Explanation:

  • Organization Structure:
    • Alice (ID: 1) is the CEO (level 1) with no manager
    • Bob (ID: 2) and Charlie (ID: 3) report directly to Alice (level 2)
    • David (ID: 4), Eva (ID: 5) report to Bob, while Frank (ID: 6) and Grace (ID: 7) report to Charlie (level 3)
    • Hank (ID: 8) reports to David, and Ivy (ID: 9) and Judy (ID: 10) report to Frank (level 4)
  • Level Calculation:
    • The CEO (Alice) is at level 1
    • Each subsequent level of management adds 1 to the level
  • Team Size Calculation:
    • Alice has 9 employees under her (the entire company except herself)
    • Bob has 3 employees (David, Eva, and Hank)
    • Charlie has 4 employees (Frank, Grace, Ivy, and Judy)
    • David has 1 employee (Hank)
    • Frank has 2 employees (Ivy and Judy)
    • Eva, Grace, Hank, Ivy, and Judy have no direct reports (team_size = 0)
  • Budget Calculation:
    • Alice's budget: Her salary (12000) + all employees' salaries (72500) = 84500
    • Charlie's budget: His salary (10000) + Frank's budget (23000) + Grace's salary (8500) = 41500
    • Bob's budget: His salary (10000) + David's budget (13500) + Eva's salary (7500) = 31000
    • Frank's budget: His salary (9000) + Ivy's salary (7000) + Judy's salary (7000) = 23000
    • David's budget: His salary (7500) + Hank's salary (6000) = 13500
    • Employees with no direct reports have budgets equal to their own salary

Note:

  • The result is ordered first by level in ascending order
  • Within the same level, employees are ordered by budget in descending order then by name in ascending order

解法

方法一

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
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;

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
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"]]

评论