Leetcode #1333: Filter Restaurants by Vegan-Friendly, Price and Distance
In this guide, we solve Leetcode #1333 Filter Restaurants by Vegan-Friendly, Price and Distance 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 restaurants where restaurants[i] = [idi, ratingi, veganFriendlyi, pricei, distancei]. You have to filter the restaurants using three filters.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Sorting
Intuition
Sorting reveals structure that is hard to see in the original order.
Once sorted, a linear scan is usually enough to compute the answer.
Approach
Sort the data and sweep through it while maintaining a small state.
This keeps the logic straightforward and reliable.
Steps:
- Sort the data.
- Scan in order while maintaining state.
- Update the best answer.
Example
Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 1, maxPrice = 50, maxDistance = 10
Output: [3,1,5]
Explanation:
The restaurants are:
Restaurant 1 [id=1, rating=4, veganFriendly=1, price=40, distance=10]
Restaurant 2 [id=2, rating=8, veganFriendly=0, price=50, distance=5]
Restaurant 3 [id=3, rating=8, veganFriendly=1, price=30, distance=4]
Restaurant 4 [id=4, rating=10, veganFriendly=0, price=10, distance=3]
Restaurant 5 [id=5, rating=1, veganFriendly=1, price=15, distance=1]
After filter restaurants with veganFriendly = 1, maxPrice = 50 and maxDistance = 10 we have restaurant 3, restaurant 1 and restaurant 5 (ordered by rating from highest to lowest).
Python Solution
class Solution:
def filterRestaurants(
self,
restaurants: List[List[int]],
veganFriendly: int,
maxPrice: int,
maxDistance: int,
) -> List[int]:
restaurants.sort(key=lambda x: (-x[1], -x[0]))
ans = []
for idx, _, vegan, price, dist in restaurants:
if vegan >= veganFriendly and price <= maxPrice and dist <= maxDistance:
ans.append(idx)
return ans
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
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.