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.

Filter Restaurants by Vegan-Friendly, Price and Distance — LeetCode #1333 (Medium)

Tags: Array, Sorting

Related patterns: Sorting, Filtering, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

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 time complexity is O(n²) because we sort the filtered list, which can take up to O(n log n) time, and the filtering process is O(n). The space complexity is O(1) since we are not using any additional data structures that grow with input size.

Step 1: Initialize an empty list to store the filtered restaurant IDs.
Step 2: Loop through each restaurant in the input array.
Step 3: For each restaurant, check if it meets the vegan-friendly, price, and distance criteria.
Step 4: If it meets the criteria, add its ID to the result list.
Step 5: Sort the result list first by rating (descending) and then by ID (descending).
Step 6: Return the sorted list of IDs.

1. Start with 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. 2. Check restaurant 1: meets criteria, add ID 1. 3. Check restaurant 2: does not meet vegan-friendly criteria. 4. Check restaurant 3: meets criteria, add ID 3. 5. Check restaurant 4: does not meet vegan-friendly criteria. 6. Check restaurant 5: meets criteria, add ID 5. Result before sorting: [1, 3, 5]. After sorting: [3, 1, 5].

Optimal Solution approach

Time complexity: O(n log n). Space complexity: 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.

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.

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

1. Start with 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]]. 2. Check restaurant 1: meets criteria, add (4, 1). 3. Check restaurant 2: does not meet vegan-friendly criteria. 4. Check restaurant 3: meets criteria, add (8, 3). 5. Check restaurant 4: does not meet vegan-friendly criteria. 6. Check restaurant 5: meets criteria, add (1, 5). Result before sorting: [(4, 1), (8, 3), (1, 5)]. After sorting: [(8, 3), (4, 1), (1, 5)]. Final IDs: [3, 1, 5].

Key Insights

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

Common Mistakes

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

Interview Tips

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.

#1333

Filter Restaurants by Vegan-Friendly, Price and Distance

Medium

Array↗Sorting↗Sorting↗Filtering↗Array↗

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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