Closest Room

Hard

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ArraysBinary Search

There is a hotel with n rooms. The rooms are represented by a 2D integer array rooms where rooms[i] = [roomId_i, size_i] denotes that there is a room with room number roomId_i and size equal to size_i. Each roomId_i is guaranteed to be unique.

You are also given k queries in a 2D array queries where queries[j] = [preferred_j, minSize_j]. The answer to the j^th query is the room number id of a room such that:

The room has a size of at least minSize_j, and
abs(id - preferred_j) is minimized, where abs(x) is the absolute value of x.

If there is a tie in the absolute difference, then use the room with the smallest such id. If there is no such room, the answer is -1.

Return an array answer of length k where answer[j] contains the answer to the j^th query.

Example 1:

Input: rooms = [[2,2],[1,2],[3,2]], queries = [[3,1],[3,3],[5,2]]
Output: [3,-1,3]
Explanation: The answers to the queries are as follows:
Query = [3,1]: Room number 3 is the closest as abs(3 - 3) = 0, and its size of 2 is at least 1. The answer is 3.
Query = [3,3]: There are no rooms with a size of at least 3, so the answer is -1.
Query = [5,2]: Room number 3 is the closest as abs(3 - 5) = 2, and its size of 2 is at least 2. The answer is 3.

Example 2:

Input: rooms = [[1,4],[2,3],[3,5],[4,1],[5,2]], queries = [[2,3],[2,4],[2,5]]
Output: [2,1,3]
Explanation: The answers to the queries are as follows:
Query = [2,3]: Room number 2 is the closest as abs(2 - 2) = 0, and its size of 3 is at least 3. The answer is 2.
Query = [2,4]: Room numbers 1 and 3 both have sizes of at least 4. The answer is 1 since it is smaller.
Query = [2,5]: Room number 3 is the only room with a size of at least 5. The answer is 3.

Constraints:

n == rooms.length
1 <= n <= 10⁵
k == queries.length
1 <= k <= 10⁴
1 <= roomId_i, preferred_j <= 10⁷
1 <= size_i, minSize_j <= 10⁷

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the room IDs and sizes, and for the preferred room ID and minimum size?
If multiple rooms have the same minimum size requirement met and are equidistant to the preferred room, which one should I return?
If no room meets the minimum size requirement, what should the function return?
Can the `rooms` array or the `queries` array be empty or null?
Are room IDs guaranteed to be unique, or is it possible to have multiple rooms with the same ID?

Brute Force Solution

Approach

Imagine searching for a room that best fits your size requirement and is as close as possible to a desired room number. The brute force method is like checking every single room to find the best one. We go through each room individually and compare it to our requirements.

Here's how the algorithm would work step-by-step:

Take the very first room.
Check if the room's size meets our minimum size requirement. If it doesn't, forget about it and move to the next room.
If the room is big enough, calculate how far away its room number is from our desired room number.
Remember this distance and the room's number as the 'best' room so far.
Now, move to the next room and repeat the process of checking size and calculating the distance.
If this new room is big enough and closer to our desired room number than the previous 'best' room, then replace the 'best' room with this new room.
Continue doing this for every single room.
Once we have checked all the rooms, the room we remembered as the 'best' one will be the closest room that also meets our size requirement.

Code Implementation

def find_closest_room_brute_force(rooms, query_room_number, minimum_size):

    closest_room = -1
    min_difference = float('inf')

    for room_number, room_size in rooms:
        # Filter out rooms that don't meet the minimum size requirement
        if room_size < minimum_size:
            continue

        # Calculate the difference between the room number and the target
        difference = abs(room_number - query_room_number)

        # If we find a closer room or an equally close room, update closest_room
        if difference < min_difference:
            min_difference = difference
            closest_room = room_number
        elif difference == min_difference and room_number < closest_room:
            # Prefer smaller room numbers if distances are equal
            closest_room = room_number

    return closest_room

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates through each of the m queries. For each query, the algorithm iterates through all n rooms. Inside the inner loop, it performs constant-time operations to check the room's size and calculate the distance if it meets the criteria. Therefore, the overall time complexity is O(n*m), where n is the number of rooms and m is the number of queries.

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It keeps track of the 'best' room so far, which involves storing the room number and its distance. These are a fixed number of variables that do not depend on the number of rooms, N. Therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

The optimal strategy solves this problem efficiently by first organizing the available rooms and then quickly finding the best room for each query. We can use organization and intelligent searching to avoid checking every room for every query.

Here's how the algorithm would work step-by-step:

First, sort all the rooms based on their size. This will help us quickly find rooms that are large enough to satisfy the size requirement in our queries.
Now, let's also sort the queries so that we can process them in a manner that takes advantage of our sorted room list. We will store the original order of the queries, so we can return the results in the correct order later.
Go through the sorted queries one at a time. For each query, find the rooms that are big enough by walking through our sorted room list. We stop when we encounter rooms that are too small, since the list is sorted.
From the rooms that are big enough, find the room whose ID is closest to the preferred room ID in the query. Keep track of the best room ID found so far.
Once we have the best room ID for the query, store it in the correct position according to the original order of the queries. This will ensure that the results match the input order.
Repeat the process for all queries. By sorting and then only looking at potentially valid rooms, we significantly reduce the search time.

Code Implementation

def closest_room(rooms, queries):
    room_size_index = 1
    query_preferred_id_index = 0
    query_minimum_size_index = 1

    # Sort rooms based on their size, largest to smallest.
    rooms.sort(key=lambda x: x[room_size_index], reverse=True)
    
    indexed_queries = []
    for index, query in enumerate(queries):
        indexed_queries.append((query[query_preferred_id_index], query[query_minimum_size_index], index))

    # Sort queries based on minimum size, largest to smallest.
    indexed_queries.sort(key=lambda x: x[1], reverse=True)

    number_of_rooms = len(rooms)
    number_of_queries = len(queries)
    result = [0] * number_of_queries
    available_rooms = []
    room_id_index = 0
    current_room_index = 0

    # Process each query in order of decreasing minimum size.
    for preferred_id, minimum_size, query_index in indexed_queries:

        # Add rooms that meet the minimum size requirement
        while current_room_index < number_of_rooms and rooms[current_room_index][room_size_index] >= minimum_size:
            available_rooms.append(rooms[current_room_index][room_id_index])
            current_room_index += 1
        
        if not available_rooms:
            result[query_index] = -1
            continue

        # Find the closest room to the preferred ID
        closest_id = -1
        min_difference = float('inf')

        for room_id in available_rooms:
            difference = abs(room_id - preferred_id)

            if difference < min_difference:
                min_difference = difference
                closest_id = room_id
            elif difference == min_difference:
                closest_id = min(closest_id, room_id)

        result[query_index] = closest_id

    return result

Big(O) Analysis

Time Complexity

O(n log n + m log m + m*n) – Sorting the rooms takes O(n log n) time where n is the number of rooms. Sorting the queries takes O(m log m) time, where m is the number of queries. For each query, in the worst case, we iterate through all the rooms to find those that meet the minimum size requirement and then find the closest ID, taking O(n) time. Since we do this for each of the m queries, the overall complexity becomes O(n log n + m log m + m*n). In scenarios where the number of rooms and queries are similar, the complexity simplifies to O(m*n) because it dominates the logarithmic terms.

Space Complexity

O(N) – The algorithm sorts the rooms array, which may require O(N) auxiliary space depending on the sorting algorithm used. A queries array, maintaining the original order of the queries, is also created, requiring O(N) space, where N is the number of queries. Additionally, a result array of size N is needed to store the results corresponding to each query. Therefore, the auxiliary space complexity is dominated by these arrays, resulting in O(N) space.

Edge Cases

Case	How to Handle
Empty rooms array	Return an empty list of results since no rooms exist to query.
Empty queries array	Return an empty list of results since there are no queries to process.
Single room and single query	Check if the room size meets the minimum size and return the room ID if it does, otherwise return -1.
Multiple rooms with the same size.	Iterate through all suitable rooms, keeping track of the closest ID to the preferred ID, returning the best one.
Room IDs or query IDs are negative.	Handle negative IDs correctly by ensuring they don't cause issues with indexing or other operations (e.g., use a data structure that supports negative keys).
Room sizes or query minimum sizes are zero.	Ensure zero size is a valid condition and handled appropriately in the filtering logic.
No room meets the minimum size requirement for a query.	Return -1 as the closest room ID when no room satisfies the minimum size.
Integer overflow when calculating differences between preferred ID and room IDs.	Use a data type with sufficient range (e.g., long) for calculations to prevent overflow.

Empty rooms array

How to Handle:

Return an empty list of results since no rooms exist to query.

Empty queries array

How to Handle:

Return an empty list of results since there are no queries to process.

Single room and single query

How to Handle:

Check if the room size meets the minimum size and return the room ID if it does, otherwise return -1.

Multiple rooms with the same size.

How to Handle:

Iterate through all suitable rooms, keeping track of the closest ID to the preferred ID, returning the best one.

Room IDs or query IDs are negative.

How to Handle:

Handle negative IDs correctly by ensuring they don't cause issues with indexing or other operations (e.g., use a data structure that supports negative keys).

Room sizes or query minimum sizes are zero.

How to Handle:

Ensure zero size is a valid condition and handled appropriately in the filtering logic.

No room meets the minimum size requirement for a query.

How to Handle:

Return -1 as the closest room ID when no room satisfies the minimum size.

Integer overflow when calculating differences between preferred ID and room IDs.

How to Handle:

Use a data type with sufficient range (e.g., long) for calculations to prevent overflow.