Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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, employees reporting directly to the CEO are level 2, and so on).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:
Note:
Problem summary: 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: 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). Team Size: For each employee who is a manager, count the total number of employees under them (direct and indirect reports). Salary Budget: For each manager, calculate the total salary budget they
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"Employees":["employee_id","employee_name","manager_id","salary","department"]},"rows":{"Employees":[[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"]]}}Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3482: Analyze Organization Hierarchy
// Auto-generated Java example from py.
class Solution {
public void exampleSolution() {
}
}
// Reference (py):
// # Accepted solution for LeetCode #3482: Analyze Organization Hierarchy
// 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"]]
// Accepted solution for LeetCode #3482: Analyze Organization Hierarchy
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #3482: Analyze Organization Hierarchy
// 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"]]
# Accepted solution for LeetCode #3482: Analyze Organization Hierarchy
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"]]
// Accepted solution for LeetCode #3482: Analyze Organization Hierarchy
// Rust example auto-generated from py reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (py):
// # Accepted solution for LeetCode #3482: Analyze Organization Hierarchy
// 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"]]
// Accepted solution for LeetCode #3482: Analyze Organization Hierarchy
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #3482: Analyze Organization Hierarchy
// 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"]]
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.