LeetCode 584: Find Customer Referee Solution in Python – A Step-by-Step Guide

What Is LeetCode 584: Find Customer Referee?

In LeetCode 584: Find Customer Referee, you’re given a list of tuples customers where each tuple (id, name, referee_id) represents a customer’s ID, name, and the ID of the customer who referred them (or None if no referee). Your task is to return a list of customer names where the referee_id is either not equal to 2 or is None. For example, with customers = [(1, "Alice", None), (2, "Bob", 2), (3, "Charlie", 1)], the result is ["Alice", "Charlie"] because Alice has no referee and Charlie’s referee isn’t 2. This problem builds on LeetCode 349: Intersection of Two Arrays for filtering but focuses on simple condition-based selection from a structured list.

Problem Statement

• Input: customers (List[Tuple[int, str, Optional[int]]])—list of (id, name, referee_id).
• Output: List[str]—names of customers where referee_id ≠ 2 or is None.
• Rules: Filter based on referee_id; include null referees; return names only.

Constraints

• 1 <= customers.length <= 5 * 10⁴
• 1 <= id <= 5 * 10⁴ (unique)
• 1 <= name.length <= 10
• referee_id is either an integer in [1, 5 * 10⁴] or None

Examples

• Input: customers = [(1,"Alice",None),(2,"Bob",2),(3,"Charlie",1)]
• Output: ["Alice","Charlie"]
• Alice: referee None, Charlie: referee 1 ≠ 2.
• Input: customers = [(1,"Alice",2),(2,"Bob",2)]
• Output: []
• All referees are 2, no matches.
• Input: customers = [(1,"Alice",None)]
• Output: ["Alice"]
• Referee None, included.

Understanding the Problem: Filtering Customers

To solve LeetCode 584: Find Customer Referee in Python, we need to filter a list of customer records to extract names where the referee_id is either not 2 or None, returning them as a list, handling up to 5 * 10⁴ records efficiently. A brute-force approach scanning the list repeatedly for each condition could work but is unnecessary since a single pass suffices. Instead, a dictionary-based approach (or even a simple list comprehension) processes the list in O(n) time, leveraging Python’s efficient data handling. We’ll explore:

• Best Solution (Dictionary-Based Filtering): O(n) time, O(n) space—fast and optimal (n = number of customers).
• Alternative Solution (Brute-Force Traversal): O(n) time, O(1) space—thorough but redundant for this simplicity.

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

Best Solution: Dictionary-Based Filtering

Why Dictionary Wins

The dictionary-based filtering solution is the best for LeetCode 584 because it efficiently identifies qualifying customers in O(n) time and O(n) space by using a dictionary to map customer IDs to names and referee IDs, then filtering in one pass with a clear condition check, ensuring all matches are collected without unnecessary overhead. It’s like keeping a customer roster and quickly flagging those not tied to referee 2—all in a neat, streamlined process!

How It Works

Think of this as a customer-sifting organizer:

• Step 1: Build Customer Dictionary:
• Map id to (name, referee_id) from customers.
• Step 2: Filter Customers:
• Iterate dictionary, check referee_id ≠ 2 or None.
• Step 3: Collect Names:
• Add qualifying names to result list.
• Step 4: Return Result:
• List of customer names.
• Why It Works:
• Dictionary organizes data for O(1) lookups (though not needed for filtering here).
• Single pass ensures efficiency.

It’s like a referee-checking maestro!

Step-by-Step Example

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

• Step 1: Build dictionary:
• customer_data = {1: ("Alice", None), 2: ("Bob", 2), 3: ("Charlie", 1)}.
• Step 2: Filter:
• ID 1: referee_id = None, include "Alice".
• ID 2: referee_id = 2, skip "Bob".
• ID 3: referee_id = 1 ≠ 2, include "Charlie".
• Step 3: Result: ["Alice", "Charlie"].
• Result: ["Alice", "Charlie"].

Example: customers = [(1,"Alice",2),(2,"Bob",2)]

• Step 1: customer_data = {1: ("Alice", 2), 2: ("Bob", 2)}.
• Step 2: Filter:
• ID 1: referee_id = 2, skip.
• ID 2: referee_id = 2, skip.
• Step 3: Result: [].
• Result: [].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners (simplified to a list comprehension for optimal clarity):

Copy codefrom typing import List, Tuple, Optional

class Solution:
    def findCustomers(self, customers: List[Tuple[int, str, Optional[int]]]) -> List[str]:
        # Step 1: Filter customers in one pass using list comprehension
        result = [name for _, name, referee_id in customers 
                  if referee_id != 2 and referee_id is not None or referee_id is None]

        # Step 2: Return result
        return result

• Line 6-7: List comprehension:
• Iterate customers, unpack tuple.
• Condition: referee_id ≠ 2 or referee_id is None.
• Collect name if condition met.
• Line 10: Return list of names.
• Time Complexity: O(n)—single pass over customers.
• Space Complexity: O(n)—result list storage.

Note: While a dictionary could be used (as initially described), it’s overkill here since we don’t need ID lookups post-filtering. The list comprehension is more concise and equally efficient, aligning with Python’s simplicity. For consistency with the "Best Solution" title, I’ll keep the dictionary approach as an option:

Copy codefrom typing import List, Tuple, Optional

class Solution:
    def findCustomers(self, customers: List[Tuple[int, str, Optional[int]]]) -> List[str]:
        # Step 1: Build dictionary
        customer_data = {id: (name, referee_id) for id, name, referee_id in customers}

        # Step 2: Filter and collect names
        result = []
        for id, (name, referee_id) in customer_data.items():
            if referee_id != 2 or referee_id is None:
                result.append(name)

        # Step 3: Return result
        return result

This dictionary version mirrors the detailed explanation, though the list comprehension is preferred for brevity.

It’s like a customer-filtering wizard!

Alternative Solution: Brute-Force Traversal

Why an Alternative Approach?

The brute-force traversal approach scans the customer list once but mimics a less optimized mindset by explicitly iterating and checking conditions without leveraging Python’s concise features, running in O(n) time and O(1) space (excluding output). It’s thorough but redundant compared to the list comprehension, making it a good baseline for understanding or when avoiding advanced constructs.

How It Works

Picture this as a customer-scanning seeker:

• Step 1: Initialize empty result list.
• Step 2: Iterate customers, check referee condition.
• Step 3: Add qualifying names to result.
• Step 4: Return result list.

It’s like a manual referee checker!

Step-by-Step Example

Example: customers = [(1,"Alice",None),(2,"Bob",2)]

• Step 1: result = [].
• Step 2: Iterate:
• (1, "Alice", None): None ≠ 2 or None, add "Alice".
• (2, "Bob", 2): 2 = 2, skip.
• Step 3: Result: ["Alice"].
• Result: ["Alice"].

Code for Brute-Force Approach

Copy codefrom typing import List, Tuple, Optional

class Solution:
    def findCustomers(self, customers: List[Tuple[int, str, Optional[int]]]) -> List[str]:
        # Step 1: Initialize result list
        result = []

        # Step 2: Traverse and filter
        for _, name, referee_id in customers:
            if referee_id != 2 or referee_id is None:
                result.append(name)

        # Step 3: Return result
        return result

• Line 7: Empty list for results.
• Lines 10-12: Check condition, append name if satisfied.
• Line 15: Return list.
• Time Complexity: O(n)—single pass.
• Space Complexity: O(1)—minimal space (excluding output).

It’s a brute-force customer sifter!

Comparing the Two Solutions

• Dictionary/List Comprehension (Best):
• Pros: O(n), O(n), concise and fast.
• Cons: Slight space overhead (dictionary version).
• Brute-Force (Alternative):
• Pros: O(n), O(1), simple explicit logic.
• Cons: Less Pythonic, redundant for this task.

List comprehension wins for elegance and efficiency!

Additional Examples and Edge Cases

• All referee 2: Empty list.
• All None: All names included.
• Empty list: Empty list.

Both handle them all, but list comprehension shines!

Complexity Recap

• Dictionary/List: Time O(n), Space O(n).
• Brute-Force: Time O(n), Space O(1).

List comprehension’s the clarity champ!

Key Takeaways

• Dictionary/List: Filter finesse—learn at Python Basics!
• Brute-Force: Scan simplicity.
• Customers: Filtering is fun.
• Python: List comp or loops, your pick!

Final Thoughts: Find Those Customers!

LeetCode 584: Find Customer Referee in Python is a straightforward data challenge. Dictionary-based filtering (or list comprehension) is your fast track, while brute-force offers a thorough dive. Want more? Try LeetCode 349: Intersection of Two Arrays or LeetCode 217: Contains Duplicate. Ready to sift? Head to Solve LeetCode 584 on LeetCode and identify those customers today!