LeetCode 586: Customer Placing the Largest Number of Orders Solution in Python – A Step-by-Step Guide

What Is LeetCode 586: Customer Placing the Largest Number of Orders?

In LeetCode 586: Customer Placing the Largest Number of Orders, you’re given a list of tuples orders where each tuple (order_id, customer_id) represents an order ID and the customer who placed it. Your task is to return the customer_id of the customer who placed the most orders. If there’s a tie, return any one of the tied customers (typically the smallest ID in practice, but here we’ll assume the first encountered for simplicity). For example, with orders = [(1, 101), (2, 101), (3, 102), (4, 101)], the result is 101 with 3 orders. This problem builds on LeetCode 347: Top K Frequent Elements for frequency counting but focuses on selecting the single most frequent element from a structured list.

Problem Statement

• Input: orders (List[Tuple[int, int]])—list of (order_id, customer_id).
• Output: Integer—customer_id with the most orders.
• Rules: Count orders per customer; return ID with highest count; ties resolved arbitrarily (e.g., first encountered).

Constraints

• 1 <= orders.length <= 5 * 10⁴
• 1 <= order_id, customer_id <= 5 * 10⁴ (order_id unique)

Examples

• Input: orders = [(1,101),(2,101),(3,102),(4,101)]
• Output: 101
• Counts: 101: 3, 102: 1; 101 wins.
• Input: orders = [(1,101),(2,102)]
• Output: 101
• Counts: 101: 1, 102: 1; 101 returned (first encountered).
• Input: orders = [(1,101)]
• Output: 101
• Single order, 101 wins.

Understanding the Problem: Finding the Top Customer

To solve LeetCode 586: Customer Placing the Largest Number of Orders in Python, we need to count the number of orders per customer_id in the orders list and identify the customer with the highest count, returning their ID, handling up to 5 * 10⁴ orders efficiently. A brute-force approach scanning the list repeatedly for each customer (O(n²)) could work but would be slow and unnecessary. Instead, a dictionary-based counting method processes the list in O(n) time, leveraging Python’s efficient key-value storage to track frequencies. We’ll explore:

• Best Solution (Dictionary-Based Counting): O(n) time, O(k) space—fast and optimal (n = number of orders, k = unique customers).
• 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 586 because it identifies the customer with the most orders in O(n) time and O(k) space (k ≤ n, typically small) by using a dictionary to tally orders for each customer_id in a single pass, then finding the ID with the maximum count efficiently. It’s like keeping an order tally sheet, marking each customer’s count as orders come in, and spotlighting the leader—all in one quick, organized sweep!

How It Works

Think of this as an order-tally champion:

• Step 1: Build Order Dictionary:
• Map customer_id to the number of orders from orders.
• Step 2: Count Orders:
• Iterate orders, increment count for each customer_id.
• Step 3: Find Top Customer:
• Scan dictionary for customer_id with highest count.
• If tie, pick first encountered (or smallest ID, adjustable).
• Step 4: Return Result:
• customer_id with most orders.
• Why It Works:
• Single pass builds counts.
• Dictionary enables O(1) updates and lookups.

It’s like an order-counting maestro!

Step-by-Step Example

Example: orders = [(1,101),(2,101),(3,102),(4,101)]

• Step 1: Initialize: order_counts = {}.
• Step 2: Count orders:
• (1, 101): order_counts = {101: 1}.
• (2, 101): order_counts = {101: 2}.
• (3, 102): order_counts = {101: 2, 102: 1}.
• (4, 101): order_counts = {101: 3, 102: 1}.
• Step 3: Find top:
• Max count = 3, customer_id = 101.
• Result: 101.

Example: orders = [(1,101),(2,102)]

• Step 2: Count:
• order_counts = {101: 1, 102: 1}.
• Step 3: Max count = 1, customer_id = 101 (first encountered).
• Result: 101.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

from typing import List, Tuple

class Solution:
    def findCustomer(self, orders: List[Tuple[int, int]]) -> int:
        # Step 1: Initialize dictionary for order counts
        order_counts = {}

        # Step 2: Count orders per customer
        for _, customer_id in orders:
            order_counts[customer_id] = order_counts.get(customer_id, 0) + 1

        # Step 3: Find customer with maximum orders
        max_orders = 0
        top_customer = 0
        for customer_id, count in order_counts.items():
            if count > max_orders:
                max_orders = count
                top_customer = customer_id

        # Step 4: Return top customer's ID
        return top_customer

• Line 7: Initialize empty dictionary for counts.
• Lines 10-11: Iterate orders, increment count using get for default 0.
• Lines 14-19: Find max count and corresponding customer_id.
• Line 22: Return top customer’s ID.
• Time Complexity: O(n)—single pass over orders.
• Space Complexity: O(k)—dictionary size (k = unique customers).

It’s like a top-customer tally wizard!

Alternative Solution: Brute-Force Traversal

Why an Alternative Approach?

The brute-force traversal approach checks each customer ID against the entire list to count their orders, running in O(n²) time and O(1) space (excluding minimal variables). It’s simple but inefficient for large lists, making it a good baseline for small datasets or understanding when avoiding dictionaries.

How It Works

Picture this as an order-scanning seeker:

• Step 1: Get unique customer IDs.
• Step 2: For each ID, count orders by traversing list.
• Step 3: Track max count and top customer.
• Step 4: Return top customer’s ID.

It’s like a manual order counter!

Step-by-Step Example

Example: orders = [(1,101),(2,101),(3,102)]

• Step 1: Unique IDs: [101, 102].
• Step 2: Count:
• 101: Scan → 2 orders.
• 102: Scan → 1 order.
• Step 3: Max = 2, top = 101.
• Result: 101.

Code for Brute-Force Approach

from typing import List, Tuple

class Solution:
    def findCustomer(self, orders: List[Tuple[int, int]]) -> int:
        # Step 1: Get unique customer IDs
        unique_ids = set(customer_id for _, customer_id in orders)

        # Step 2: Find customer with most orders
        max_orders = 0
        top_customer = 0

        for customer_id in unique_ids:
            count = 0
            for _, order_customer_id in orders:
                if order_customer_id == customer_id:
                    count += 1
            if count > max_orders:
                max_orders = count
                top_customer = customer_id

        # Step 3: Return top customer
        return top_customer

• Line 6: Extract unique IDs with set.
• Lines 9-18: For each ID, count orders, update max.
• Line 21: Return top ID.
• Time Complexity: O(n²)—nested loops.
• Space Complexity: O(1)—minimal space (excluding unique IDs).

It’s a brute-force order scanner!

Comparing the Two Solutions

• Dictionary (Best):
• Pros: O(n), O(k), fast and concise.
• Cons: Extra space for dictionary.
• Brute-Force (Alternative):
• Pros: O(n²), O(1), space-efficient.
• Cons: Slower, less elegant.

Dictionary wins for efficiency!

Additional Examples and Edge Cases

• Single order: That customer’s ID.
• All tied: First encountered ID.
• Empty list: Undefined (assume handled).

Dictionary handles them all elegantly!

Complexity Recap

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

Dictionary’s the speed champ!

Key Takeaways

• Dictionary: Count finesse—learn at Python Basics!
• Brute-Force: Scan simplicity.
• Orders: Counting is fun.
• Python: Dict or loops, your pick!

Final Thoughts: Crown That Customer!

LeetCode 586: Customer Placing the Largest Number of Orders in Python is a straightforward 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 586 on LeetCode and identify that top customer today!