What is the time complexity of Display Table of Food Orders in a Restaurant?

The optimal solution has a time complexity of O(m*n*log(n)).

What is the space complexity of Display Table of Food Orders in a Restaurant?

The space complexity is O(m*n).

What data structures are used to solve Display Table of Food Orders in a Restaurant?

This problem uses Array, Hash Table, String, Sorting, Ordered Set. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Display Table of Food Orders in a Restaurant?

Insight 1: We need to aggregate the orders by table number and food item. Insight 2: We need to maintain a set of unique food items and sort them alphabetically to create the header row.

Which companies ask LeetCode 1418. Display Table of Food Orders in a Restaurant in interviews?

This problem is frequently asked at J.P. Morgan, Nordstrom.

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Display Table of Food Orders in a Restaurant - LeetCode 1418 Solution

Display Table of Food Orders in a Restaurant - Complete Solution Guide

Display Table of Food Orders in a Restaurant is LeetCode problem 1418, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given the array orders , which represents the orders that customers have done in a restaurant. More specifically orders[i]=[customerName i ,tableNumber i ,foodItem i ] where customerName i is the name of the customer, tableNumber i is the table customer sit at, and foodItem i is the item customer orders. Return the restaurant's “ display table ” . The “ display table ” is a table whose row entries denote how many of each food item each table ordered. The first column is t

Detailed Explanation

The problem requires creating a display table that summarizes food orders in a restaurant. The input is a list of orders, where each order contains the customer's name, table number, and the food item ordered. The output should be a table represented as a list of lists of strings. The first row of the table is a header row, with "Table" in the first column, followed by the food items in alphabetical order. Subsequent rows represent each table number (sorted numerically), and each cell contains the count of how many times that table ordered the food item corresponding to the column. Customer names are irrelevant to the output. All counts should be represented as strings.

Solution Approach

The solution involves the following steps: First, iterate through the input `orders` to store the order counts for each table and food item using a nested HashMap (or defaultdict of Counters). Simultaneously, keep track of all unique food items in a Set. Then, convert the Set of food items into a List and sort it alphabetically. Create the header row using "Table" followed by the sorted food items. Extract the table numbers from the nested HashMap, convert them to integers, and sort them numerically. Iterate through the sorted table numbers, creating a row for each table. For each row, add the table number as a string, and then iterate through the sorted food items, adding the corresponding order count (or 0 if no order) as a string to the row. Finally, assemble the result table as a list of lists of strings and return it.

Step-by-Step Algorithm

Step 1: Initialize a nested HashMap (table_orders) to store the count of each food item for each table. Also, initialize a Set (food_items) to store unique food items.
Step 2: Iterate through the input orders list. For each order, extract the table number (as a string) and the food item. Convert the table number to an integer.
Step 3: Update the table_orders HashMap to increment the count of the specific food item for the corresponding table. Also, add food item to the food_items Set
Step 4: Convert the food_items Set to a List and sort it alphabetically.
Step 5: Create the header row with "Table" as the first element, followed by the sorted list of food items.
Step 6: Extract the table numbers (keys) from the table_orders HashMap, convert them to integers, and sort them numerically.
Step 7: Iterate through the sorted table numbers.
Step 8: For each table number, create a new row. The first element of the row is the table number (as a string).
Step 9: Iterate through the sorted list of food items. For each food item, check if the table has an order for that food. If yes, add the count (as a string) to the row; otherwise, add "0" to the row.
Step 10: Add the row to the result table.
Step 11: Return the result table.

Key Insights

Insight 1: We need to aggregate the orders by table number and food item.
Insight 2: We need to maintain a set of unique food items and sort them alphabetically to create the header row.
Insight 3: Table numbers must be extracted, converted to integers, and sorted numerically for the row ordering.
Insight 4: A suitable data structure to efficiently store and retrieve counts is a nested HashMap (or defaultdict of Counters in Python).

Complexity Analysis

Time Complexity: O(m*n*log(n))

Space Complexity: O(m*n)

Topics

This problem involves: Array, Hash Table, String, Sorting, Ordered Set.

Companies

Asked at: J.P. Morgan, Nordstrom.

Display Table of Food Orders in a Restaurant | LeetCode 1418

Medium

Array Hash Table String Sorting Ordered Set

Time: O(m*n*log(n))Space: O(m*n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<List<String>> displayTable(List<List<String>> orders) {
        Map<Integer, Map<String, Integer>> tableOrders = new HashMap<>();
        Set<String> foodItems = new HashSet<>();

        for (List<String> order : orders) {
            String tableStr = order.get(1);
            String food = order.get(2);
            int tableNum = Integer.parseInt(tableStr);

            tableOrders.computeIfAbsent(tableNum, k -> new HashMap<>()).put(food, tableOrders.getOrDefault(tableNum, new HashMap<>()).getOrDefault(food, 0) + 1);
            foodItems.add(food);
        }

        List<String> sortedFoods = new ArrayList<>(foodItems);
        Collections.sort(sortedFoods);

        List<String> header = new ArrayList<>();
        header.add("Table");
        header.addAll(sortedFoods);

        List<List<String>> resultTable = new ArrayList<>();
        resultTable.add(header);

        List<Integer> sortedTables = new ArrayList<>(tableOrders.keySet());
        Collections.sort(sortedTables);

        for (int tableNum : sortedTables) {
            List<String> row = new ArrayList<>();
            row.add(String.valueOf(tableNum));
            for (String food : sortedFoods) {
                row.add(String.valueOf(tableOrders.get(tableNum).getOrDefault(food, 0)));
            }
            resultTable.add(row);
        }

        return resultTable;
    }
}

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Problem Info

Description

Problem 1418: Display Table of Food Orders in a Restaurant - Detailed explanation and solution approaches.

Topics

Array Hash Table String Sorting Ordered Set

Companies

J.P. MorganNordstrom

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1418

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize a nested HashMap (table_orders) to store the count of each food item for each table. Also, initialize a Set (food_items) to store unique food items.
2
Step 2: Iterate through the input orders list. For each order, extract the table number (as a string) and the food item. Convert the table number to an integer.
3
Step 3: Update the table_orders HashMap to increment the count of the specific food item for the corresponding table. Also, add food item to the food_items Set
4
Step 4: Convert the food_items Set to a List and sort it alphabetically.
5
Step 5: Create the header row with "Table" as the first element, followed by the sorted list of food items.
6
Step 6: Extract the table numbers (keys) from the table_orders HashMap, convert them to integers, and sort them numerically.
7
Step 7: Iterate through the sorted table numbers.
8
Step 8: For each table number, create a new row. The first element of the row is the table number (as a string).
9
Step 9: Iterate through the sorted list of food items. For each food item, check if the table has an order for that food. If yes, add the count (as a string) to the row; otherwise, add "0" to the row.
10
Step 10: Add the row to the result table.
11
Step 11: Return the result table.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n*log(n))

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity is derived from the sorting operations and the nested loops. The space complexity is determined by the size of the data structures used to store the table orders and the result table. The number of orders does not directly appear in the complexity, as it only affects how quickly the `table_orders` map and `food_items` set are built; after that, it's just about iterating over the *unique* tables and food items.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using TreeMap instead of HashMap + Sorting

TreeMap provides automatic sorting based on keys. Using TreeMap for storing the table orders would eliminate the need to explicitly sort the table numbers.

Time: O(m*n*log(n)), where 'm' is the number of tables and 'n' is the number of unique food items.Space: O(m*n)

Trade-offs: TreeMap adds a slight overhead to insertion and retrieval, but eliminates explicit sorting. It may be more readable but slightly less efficient in some cases, although the asymptotic complexity remains the same. The constant factor may be slightly higher.

✨

Key Insights

1Insight 1: We need to aggregate the orders by table number and food item.
2Insight 2: We need to maintain a set of unique food items and sort them alphabetically to create the header row.
3Insight 3: Table numbers must be extracted, converted to integers, and sorted numerically for the row ordering.
4Insight 4: A suitable data structure to efficiently store and retrieve counts is a nested HashMap (or defaultdict of Counters in Python).

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty orders list. The solution should return a table with only the header row, containing 'Table' and an empty list of food items. The code correctly handles this since the loops are skipped if orders is empty.
Edge case 2: A table ordering no food items. The solution correctly handles this by initializing the count to '0' for food items the table hasn't ordered.
Edge case 3: Orders with duplicate entries (same customer, table, and food). The solution correctly handles this by accumulating the counts in the HashMap.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the food items alphabetically. This would violate the problem requirement.

✗Common mistake 2: Not converting table numbers to integers before sorting. String sorting would lead to incorrect ordering (e.g., '10' < '2' as strings).

✗Common mistake 3: Not handling the case where a table has not ordered a particular food item. This results in a missing entry instead of a zero.

✗Common mistake 4: Not converting the counts to strings when constructing the result table.

✗Common mistake 5: Not properly handling the memory management in C, leading to memory leaks.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Restaurant order management system. This algorithm is useful for generating reports that summarize orders by table and food item, which helps with inventory management and identifying popular items.

🏦Real-world scenario 2: Sales data analysis. This algorithm can be adapted to analyze sales data for different products across different stores, generating a summary table that shows sales volume by product and store.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Display Table of Food Orders in a Restaurant"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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