LeetCode #1333 — MEDIUM

Filter Restaurants by Vegan-Friendly, Price and Distance

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given the array restaurants where restaurants[i] = [id_i, rating_i, veganFriendly_i, price_i, distance_i]. You have to filter the restaurants using three filters.

The veganFriendly filter will be either true (meaning you should only include restaurants with veganFriendly_i set to true) or false (meaning you can include any restaurant). In addition, you have the filters maxPrice and maxDistance which are the maximum value for price and distance of restaurants you should consider respectively.

Return the array of restaurant IDs after filtering, ordered by rating from highest to lowest. For restaurants with the same rating, order them by id from highest to lowest. For simplicity veganFriendly_i and veganFriendly take value 1 when it is true, and 0 when it is false.

Example 1:

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).

Example 2:

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 = 0, maxPrice = 50, maxDistance = 10
Output: [4,3,2,1,5]
Explanation: The restaurants are the same as in example 1, but in this case the filter veganFriendly = 0, therefore all restaurants are considered.

Example 3:

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 = 0, maxPrice = 30, maxDistance = 3
Output: [4,5]

Constraints:

1 <= restaurants.length <= 10^4
restaurants[i].length == 5
1 <= id_i, rating_i, price_i, distance_i<= 10^5
1 <= maxPrice, maxDistance <= 10^5
veganFriendly_i and veganFriendly are 0 or 1.
All id_i are distinct.

Step 01

Brute Force Baseline

Problem summary: Given the array restaurants where restaurants[i] = [idi, ratingi, veganFriendlyi, pricei, distancei]. You have to filter the restaurants using three filters. The veganFriendly filter will be either true (meaning you should only include restaurants with veganFriendlyi set to true) or false (meaning you can include any restaurant). In addition, you have the filters maxPrice and maxDistance which are the maximum value for price and distance of restaurants you should consider respectively. Return the array of restaurant IDs after filtering, ordered by rating from highest to lowest. For restaurants with the same rating, order them by id from highest to lowest. For simplicity veganFriendlyi and veganFriendly take value 1 when it is true, and 0 when it is false.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[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]]
1
50
10

Example 2

[[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]]
0
50
10

Example 3

[[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]]
0
30
3

Step 02

Core Insight

What unlocks the optimal approach

Do the filtering and sort as said. Note that the id may not be the index in the array.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1333: Filter Restaurants by Vegan-Friendly, Price and Distance
class Solution {
    public List<Integer> filterRestaurants(
        int[][] restaurants, int veganFriendly, int maxPrice, int maxDistance) {
        Arrays.sort(restaurants, (a, b) -> a[1] == b[1] ? b[0] - a[0] : b[1] - a[1]);
        List<Integer> ans = new ArrayList<>();
        for (int[] r : restaurants) {
            if (r[2] >= veganFriendly && r[3] <= maxPrice && r[4] <= maxDistance) {
                ans.add(r[0]);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1333: Filter Restaurants by Vegan-Friendly, Price and Distance
func filterRestaurants(restaurants [][]int, veganFriendly int, maxPrice int, maxDistance int) (ans []int) {
	sort.Slice(restaurants, func(i, j int) bool {
		a, b := restaurants[i], restaurants[j]
		if a[1] != b[1] {
			return a[1] > b[1]
		}
		return a[0] > b[0]
	})
	for _, r := range restaurants {
		if r[2] >= veganFriendly && r[3] <= maxPrice && r[4] <= maxDistance {
			ans = append(ans, r[0])
		}
	}
	return
}

# Accepted solution for LeetCode #1333: Filter Restaurants by Vegan-Friendly, Price and Distance
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

// Accepted solution for LeetCode #1333: Filter Restaurants by Vegan-Friendly, Price and Distance
struct Solution;
use std::cmp::Reverse;

type Pair = (Reverse<i32>, Reverse<i32>);
impl Solution {
    fn filter_restaurants(
        restaurants: Vec<Vec<i32>>,
        vegan_friendly: i32,
        max_price: i32,
        max_distance: i32,
    ) -> Vec<i32> {
        let mut pairs: Vec<Pair> = restaurants
            .into_iter()
            .filter_map(|r| {
                if (vegan_friendly == 0 || r[2] == 1) && r[3] <= max_price && r[4] <= max_distance {
                    Some((Reverse(r[1]), Reverse(r[0])))
                } else {
                    None
                }
            })
            .collect();
        pairs.sort_unstable();
        pairs.into_iter().map(|p| (p.1).0).collect()
    }
}

#[test]
fn test() {
    let restaurants = vec_vec_i32![
        [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]
    ];
    let vegan_friendly = 1;
    let max_price = 50;
    let max_distance = 10;
    let res = vec![3, 1, 5];
    assert_eq!(
        Solution::filter_restaurants(restaurants, vegan_friendly, max_price, max_distance),
        res
    );
    let restaurants = vec_vec_i32![
        [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]
    ];
    let vegan_friendly = 0;
    let max_price = 50;
    let max_distance = 10;
    let res = vec![4, 3, 2, 1, 5];
    assert_eq!(
        Solution::filter_restaurants(restaurants, vegan_friendly, max_price, max_distance),
        res
    );
    let restaurants = vec_vec_i32![
        [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]
    ];
    let vegan_friendly = 0;
    let max_price = 30;
    let max_distance = 3;
    let res = vec![4, 5];
    assert_eq!(
        Solution::filter_restaurants(restaurants, vegan_friendly, max_price, max_distance),
        res
    );
}

// Accepted solution for LeetCode #1333: Filter Restaurants by Vegan-Friendly, Price and Distance
function filterRestaurants(
    restaurants: number[][],
    veganFriendly: number,
    maxPrice: number,
    maxDistance: number,
): number[] {
    restaurants.sort((a, b) => (a[1] === b[1] ? b[0] - a[0] : b[1] - a[1]));
    const ans: number[] = [];
    for (const [id, _, vegan, price, distance] of restaurants) {
        if (vegan >= veganFriendly && price <= maxPrice && distance <= maxDistance) {
            ans.push(id);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.