LeetCode 570: Managers with at Least 5 Direct Reports Solution in Python – A Step-by-Step Guide

Imagine you’re an HR analyst handed a list of employee records—like [(1, "Alice", None), (2, "Bob", 1), (3, "Charlie", 1)]—where each entry lists an employee’s ID, name, and manager’s ID, and your task is to identify managers who oversee at least 5 direct reports, such as finding "Alice" with enough subordinates. That’s the intriguing challenge of LeetCode 570: Managers with at Least 5 Direct Reports, a medium-level problem that’s a fantastic way to practice data processing in Python. We’ll explore two solutions: the Best Solution, a dictionary-based counting approach that’s efficient and clear, and an Alternative Solution, a brute-force traversal method that’s thorough but less optimized. With detailed examples, clear code, and a friendly tone—especially for the dictionary approach—this guide will help you spot those busy managers, whether you’re new to coding or leveling up. Let’s tally those reports and start counting!

What Is LeetCode 570: Managers with at Least 5 Direct Reports?

In LeetCode 570: Managers with at Least 5 Direct Reports, you’re given a list of tuples employees where each tuple (id, name, manager_id) represents an employee’s ID, name, and the ID of their manager (or None if they have no manager, typically the top boss). Your task is to return a list of names of managers who have at least 5 direct reports—employees who report directly to them. For example, with employees = [(1, "Alice", None), (2, "Bob", 1), (3, "Charlie", 1), (4, "David", 1), (5, "Eve", 1), (6, "Frank", 1)], the result is ["Alice"] because Alice manages 5 employees. This problem builds on LeetCode 347: Top K Frequent Elements for frequency counting but applies it to hierarchical data.

Problem Statement

Input: employees (List[Tuple[int, str, Optional[int]]])—list of (id, name, manager_id) tuples.
Output: List[str]—names of managers with 5 or more direct reports.
Rules: Direct reports have the manager’s ID in their manager_id; count only direct subordinates; manager_id = None for top-level employees.

Constraints

1 <= employees.length <= 10⁵
1 <= id <= 10⁵ (unique)
1 <= name.length <= 10
manager_id is either an integer in range [1, 10⁵] or None

Examples

Input: employees = [(1, "Alice", None), (2, "Bob", 1), (3, "Charlie", 1), (4, "David", 1), (5, "Eve", 1), (6, "Frank", 1)]

Output: ["Alice"]
Alice (ID 1) has 5 direct reports: Bob, Charlie, David, Eve, Frank.

Input: employees = [(1, "Alice", None), (2, "Bob", 1), (3, "Charlie", 2)]

Output: []
Alice has 1 report (Bob), Bob has 1 (Charlie), none reach 5.

Input: employees = [(1, "Alice", None), (2, "Bob", 1), (3, "Charlie", 1), (4, "David", 1), (5, "Eve", 1), (6, "Frank", 1), (7, "Grace", 2), (8, "Hank", 2), (9, "Ivy", 2), (10, "Jack", 2), (11, "Kim", 2)]

Output: ["Alice", "Bob"]
Alice (5), Bob (5).

Understanding the Problem: Identifying Busy Managers

To solve LeetCode 570: Managers with at Least 5 Direct Reports in Python, we need to count the number of direct reports for each manager by grouping employees based on their manager_id, then identify those with 5 or more subordinates and return their names. With up to 10⁵ employees, a brute-force approach traversing the list multiple times for each manager could work but would be inefficient (O(n²)). Instead, a dictionary-based counting method can process the data in one pass, making it both fast and scalable. We’ll explore:

Best Solution (Dictionary-Based Counting): O(n) time, O(n) space—fast and optimal (n = number of employees).
Alternative Solution (Brute-Force Traversal): O(n²) time, O(1) space—simple but slow.

Let’s dive into the dictionary solution with a friendly breakdown!

Best Solution: Dictionary-Based Counting

Why Dictionary Wins

The dictionary-based counting solution is the best for LeetCode 570 because it efficiently identifies managers with 5+ direct reports in O(n) time and O(n) space by using two dictionaries: one to map manager IDs to report counts and another to map IDs to names, processing the employee list in a single pass. It’s like keeping a manager roster, tallying their team sizes as you go, and spotlighting the busiest ones—all in one smooth sweep!

How It Works

Think of this as a manager-tally organizer:

Step 1: Build Dictionaries:

manager_counts: Map manager_id to count of direct reports.
id_to_name: Map id to employee name.

Step 2: Count Reports:

Iterate employees, increment count for each manager_id (skip None).

Step 3: Identify Busy Managers:

Check manager_counts for entries with 5 or more reports.

Step 4: Map to Names:

Use id_to_name to get names of qualifying managers.

Step 5: Return Result:

List of manager names.

Why It Works:

Single pass counts all reports efficiently.
Dictionaries enable O(1) lookups for names.

It’s like a direct-report counter!

Step-by-Step Example

Example: employees = [(1, "Alice", None), (2, "Bob", 1), (3, "Charlie", 1), (4, "David", 1), (5, "Eve", 1), (6, "Frank", 1)]

Step 1: Initialize:

manager_counts = {}.
id_to_name = {1: "Alice", 2: "Bob", 3: "Charlie", 4: "David", 5: "Eve", 6: "Frank"}.

Step 2: Count reports:

(1, "Alice", None): Skip.
(2, "Bob", 1): manager_counts[1] = 1.
(3, "Charlie", 1): manager_counts[1] = 2.
(4, "David", 1): manager_counts[1] = 3.
(5, "Eve", 1): manager_counts[1] = 4.
(6, "Frank", 1): manager_counts[1] = 5.

Step 3: Check counts:

manager_counts = {1: 5}, 5 ≥ 5.

Step 4: Names:

ID 1 → "Alice".

Result: ["Alice"].

Example: employees = [(1, "Alice", None), (2, "Bob", 1), (3, "Charlie", 2)]

Step 1:

manager_counts = {}.
id_to_name = {1: "Alice", 2: "Bob", 3: "Charlie"}.

Step 2:

(1, "Alice", None): Skip.
(2, "Bob", 1): manager_counts[1] = 1.
(3, "Charlie", 2): manager_counts[2] = 1.

Step 3: Check:

manager_counts = {1: 1, 2: 1}, none ≥ 5.

Result: [].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

class Solution:
    def findManagers(self, employees: List[Tuple[int, str, Optional[int]]]) -> List[str]:
        # Step 1: Initialize dictionaries
        manager_counts = {}
        id_to_name = {}

        # Step 2: Populate dictionaries
        for emp_id, name, mgr_id in employees:
            id_to_name[emp_id] = name
            if mgr_id is not None:
                manager_counts[mgr_id] = manager_counts.get(mgr_id, 0) + 1

        # Step 3: Identify managers with 5+ reports
        result = []
        for mgr_id, count in manager_counts.items():
            if count >= 5:
                result.append(id_to_name[mgr_id])

        # Step 4: Return list of manager names
        return result

Lines 4-5: Initialize dictionaries for counts and names.
Lines 8-11: Iterate employees:

Map ID to name.
Increment manager’s report count (skip None).

Lines 14-17: Check counts, add names if ≥ 5.
Line 20: Return result list.
Time Complexity: O(n)—single pass over employees.
Space Complexity: O(n)—dictionary storage.

It’s like a manager-report tally master!

Alternative Solution: Brute-Force Traversal

Why an Alternative Approach?

The brute-force traversal approach checks each employee as a potential manager, counting their direct reports by scanning the list repeatedly, running in O(n²) time and O(1) space (excluding output). It’s simple but inefficient for large lists, making it a good baseline for small datasets or understanding.

How It Works

Picture this as a report-counting wanderer:

Step 1: Build ID-to-name map.
Step 2: For each employee, count direct reports by scanning list.
Step 3: If count ≥ 5, add name to result.
Step 4: Return result list.

It’s like a report-scanning seeker!

Step-by-Step Example

Example: employees = [(1, "Alice", None), (2, "Bob", 1), (3, "Charlie", 1)]

Step 1: id_to_name = {1: "Alice", 2: "Bob", 3: "Charlie"}.
Step 2: Check each:

ID 1: Scan → 2 reports (Bob, Charlie), < 5.
ID 2: Scan → 0 reports, < 5.
ID 3: Scan → 0 reports, < 5.

Result: [].

Code for Brute-Force Approach

class Solution:
    def findManagers(self, employees: List[Tuple[int, str, Optional[int]]]) -> List[str]:
        # Step 1: Build ID-to-name map
        id_to_name = {emp_id: name for emp_id, name, _ in employees}

        # Step 2: Check each employee as manager
        result = []
        for mgr_id, _, _ in employees:
            count = 0
            for _, _, emp_mgr_id in employees:
                if emp_mgr_id == mgr_id:
                    count += 1
            if count >= 5:
                result.append(id_to_name[mgr_id])

        # Step 3: Return unique manager names
        return list(set(result))  # Remove duplicates if any

Line 4: Map IDs to names.
Lines 7-13: For each employee, count reports by scanning.
Line 16: Return unique names.
Time Complexity: O(n²)—nested loops.
Space Complexity: O(1)—minimal extra space.

It’s a brute-force report counter!

Comparing the Two Solutions

Dictionary (Best):

Pros: O(n), O(n), fast.
Cons: Extra space.

Brute-Force (Alternative):

Pros: O(n²), O(1), simple.
Cons: Slower.

Dictionary wins for efficiency!

Additional Examples and Edge Cases

[(1, "Alice", None)]: [].
5+ reports per manager: Handled by count check.
Empty list: [].

Dictionary handles them all!

Complexity Recap

Dictionary: Time O(n), Space O(n).
Brute-Force: Time O(n²), Space O(1).

Dictionary’s the speed champ!

Key Takeaways

Dictionary: Count finesse—learn at Python Basics!
Brute-Force: Full scan.
Reports: Managers are fun.
Python: Dict or brute, your pick!

Final Thoughts: Spot Those Managers!

LeetCode 570: Managers with at Least 5 Direct Reports in Python is a clever data challenge. Dictionary-based counting is your fast track, while brute-force offers a thorough dive. Want more? Try LeetCode 347: Top K Frequent Elements or LeetCode 692: Top K Frequent Words. Ready to tally? Head to Solve LeetCode 570 on LeetCode and find those busy managers today!