How do you solve Display Table of Food Orders in a Restaurant on LeetCode?

Display Table of Food Orders in a Restaurant (LeetCode #1418) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient counting of food items per table.

What is the brute force solution for Display Table of Food Orders in a Restaurant?

The brute force approach for Display Table of Food Orders in a Restaurant is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will iterate through each order and manually count the occurrences of each food item for every table. This is straightforward but inefficient, as it requires multiple passes through the data. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Display Table of Food Orders in a Restaurant?

Display Table of Food Orders in a Restaurant uses the following concepts: Array, Hash Table, String, Sorting, Ordered Set. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Display Table of Food Orders in a Restaurant?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches. Be prepared to optimize your solution if asked.

What are common mistakes when solving Display Table of Food Orders in a Restaurant?

Not using a hash map to count occurrences efficiently. Forgetting to sort the food items and table numbers before constructing the display.

#1418

Display Table of Food Orders in a Restaurant

Medium

ArrayHash TableStringSortingOrdered SetHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

In the optimal solution, we utilize a hash map to efficiently count the occurrences of each food item for each table in a single pass through the orders. This reduces the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Use a hash map to count the frequency of each food item for each table.
2Step 2: Extract the unique table numbers and food items from the hash map.
3Step 3: Sort the table numbers and food items, then construct the display table.

solution.py17 lines

1from collections import defaultdict
2
3orders = [["David","3","Ceviche"],["Corina","10","Beef Burrito"],["David","3","Fried Chicken"],["Carla","5","Water"],["Carla","5","Ceviche"],["Rous","3","Ceviche"]]
4
5def displayTable(orders):
6    table_food_count = defaultdict(lambda: defaultdict(int))
7    for order in orders:
8        table_food_count[order[1]][order[2]] += 1
9    table_numbers = sorted(table_food_count.keys(), key=int)
10    food_items = sorted(set(item for table in table_food_count.values() for item in table.keys()))
11    display = [["Table"] + food_items]
12    for table in table_numbers:
13        row = [table] + [str(table_food_count[table][food]) for food in food_items]
14        display.append(row)
15    return display
16
17print(displayTable(orders))

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of table numbers and food items. The space complexity is O(n) because we use a hash map to store counts, which can grow with the number of unique food items and tables.

1Using a hash map allows for efficient counting of food items per table.
2Sorting is essential for organizing the final output.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.