What is the brute force solution for Filter Restaurants by Vegan-Friendly, Price and Distance?

The brute force approach for Filter Restaurants by Vegan-Friendly, Price and Distance is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will iterate through all the restaurants and check if each one meets the filtering criteria. If it does, we will add it to our result list. This is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Filter Restaurants by Vegan-Friendly, Price and Distance?

Filter Restaurants by Vegan-Friendly, Price and Distance uses the following concepts: Array, Sorting. The recommended patterns to study are: Sorting, Filtering, Array.

What are the interview tips for Filter Restaurants by Vegan-Friendly, Price and Distance?

Always clarify the requirements before coding. Think about edge cases, such as empty input or all restaurants being filtered out. Discuss your thought process and approach with the interviewer.

What are common mistakes when solving Filter Restaurants by Vegan-Friendly, Price and Distance?

Not considering the vegan-friendly filter correctly. Forgetting to sort the results as specified.

#1333

Filter Restaurants by Vegan-Friendly, Price and Distance

Medium

ArraySortingSortingFilteringArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution uses a single pass to filter the restaurants and then sorts the filtered results. This is more efficient as it reduces the number of times we loop through the data.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty list to store the filtered restaurant IDs.
2Step 2: Loop through each restaurant and check if it meets the filtering criteria.
3Step 3: If it meets the criteria, add a tuple of (rating, id) to the result list.
4Step 4: Sort the result list by rating (descending) and then by ID (descending).
5Step 5: Extract and return only the IDs from the sorted list.

solution.py7 lines

1def filterRestaurants(restaurants, veganFriendly, maxPrice, maxDistance):
2    filtered = []
3    for r in restaurants:
4        if (veganFriendly == 0 or r[2] == 1) and r[3] <= maxPrice and r[4] <= maxDistance:
5            filtered.append((r[1], r[0]))
6    filtered.sort(key=lambda x: (-x[0], -x[1]))
7    return [id for _, id in filtered]

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step after filtering, while the filtering itself is O(n). The space complexity is O(n) because we store the filtered results before sorting.

1Filtering is crucial before sorting to reduce unnecessary computations.
2Sorting can be done efficiently using built-in functions.

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