Leetcode #1418: Display Table of Food Orders in a Restaurant
In this guide, we solve Leetcode #1418 Display Table of Food Orders in a Restaurant in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given the array orders, which represents the orders that customers have done in a restaurant. More specifically orders[i]=[customerNamei,tableNumberi,foodItemi] where customerNamei is the name of the customer, tableNumberi is the table customer sit at, and foodItemi is the item customer orders.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, String, Ordered Set, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: orders = [["David","3","Ceviche"],["Corina","10","Beef Burrito"],["David","3","Fried Chicken"],["Carla","5","Water"],["Carla","5","Ceviche"],["Rous","3","Ceviche"]]
Output: [["Table","Beef Burrito","Ceviche","Fried Chicken","Water"],["3","0","2","1","0"],["5","0","1","0","1"],["10","1","0","0","0"]]
Explanation:
The displaying table looks like:
Table,Beef Burrito,Ceviche,Fried Chicken,Water
3 ,0 ,2 ,1 ,0
5 ,0 ,1 ,0 ,1
10 ,1 ,0 ,0 ,0
For the table 3: David orders "Ceviche" and "Fried Chicken", and Rous orders "Ceviche".
For the table 5: Carla orders "Water" and "Ceviche".
For the table 10: Corina orders "Beef Burrito".
Python Solution
class Solution:
def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
tables = defaultdict(list)
items = set()
for _, table, foodItem in orders:
tables[int(table)].append(foodItem)
items.add(foodItem)
sorted_items = sorted(items)
ans = [["Table"] + sorted_items]
for table in sorted(tables):
cnt = Counter(tables[table])
row = [str(table)] + [str(cnt[item]) for item in sorted_items]
ans.append(row)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.