<p>There is a hotel with <code>n</code> rooms. The rooms are represented by a 2D integer array <code>rooms</code> where <code>rooms[i] = [roomId<sub>i</sub>, size<sub>i</sub>]</code> denotes that there is a room with room number <code>roomId<sub>i</sub></code> and size equal to <code>size<sub>i</sub></code>. Each <code>roomId<sub>i</sub></code> is guaranteed to be <strong>unique</strong>.</p>

<p>You are also given <code>k</code> queries in a 2D array <code>queries</code> where <code>queries[j] = [preferred<sub>j</sub>, minSize<sub>j</sub>]</code>. The answer to the <code>j<sup>th</sup></code> query is the room number <code>id</code> of a room such that:</p>

<ul>
	<li>The room has a size of <strong>at least</strong> <code>minSize<sub>j</sub></code>, and</li>
	<li><code>abs(id - preferred<sub>j</sub>)</code> is <strong>minimized</strong>, where <code>abs(x)</code> is the absolute value of <code>x</code>.</li>
</ul>

<p>If there is a <strong>tie</strong> in the absolute difference, then use the room with the <strong>smallest</strong> such <code>id</code>. If there is <strong>no such room</strong>, the answer is <code>-1</code>.</p>

<p>Return <em>an array </em><code>answer</code><em> of length </em><code>k</code><em> where </em><code>answer[j]</code><em> contains the answer to the </em><code>j<sup>th</sup></code><em> query</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> rooms = [[2,2],[1,2],[3,2]], queries = [[3,1],[3,3],[5,2]]
<strong>Output:</strong> [3,-1,3]
<strong>Explanation: </strong>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.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> rooms = [[1,4],[2,3],[3,5],[4,1],[5,2]], queries = [[2,3],[2,4],[2,5]]
<strong>Output:</strong> [2,1,3]
<strong>Explanation: </strong>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.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == rooms.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>k == queries.length</code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= roomId<sub>i</sub>, preferred<sub>j</sub> &lt;= 10<sup>7</sup></code></li>
	<li><code>1 &lt;= size<sub>i</sub>, minSize<sub>j</sub> &lt;= 10<sup>7</sup></code></li>
</ul>

from typing import List
import bisect

class Solution:
    def closestRoom(self, rooms: List[List[int]], queries: List[List[int]]) -> List[int]:
        # Sort rooms by size in descending order
        rooms.sort(key=lambda x: (-x[1], x[0]))
        
        # Store the original indices of the queries
        for i, query in enumerate(queries):
            query.append(i)
        
        # Sort queries by minSize in descending order
        queries.sort(key=lambda x: (-x[1], x[0]))
        
        # Initialize variables
        available_rooms = []
        result = [-1] * len(queries)
        room_index = 0
        
        # Process each query
        for preferred, min_size, original_index in queries:
            # Add rooms that meet the minimum size requirement for the current query
            while room_index < len(rooms) and rooms[room_index][1] >= min_size:
                bisect.insort(available_rooms, rooms[room_index][0])
                room_index += 1
            
            # If there are available rooms, find the closest one
            if available_rooms:
                # Find the insertion point for the preferred room ID
                insertion_point = bisect.bisect_left(available_rooms, preferred)
                
                # Check the closest room on the left side (if it exists)
                closest_left = available_rooms[max(insertion_point - 1, 0)]
                
                # Check the closest room on the right side (if it exists)
                closest_right = available_rooms[min(insertion_point, len(available_rooms) - 1)]
                
                # Determine which room is closer
                if abs(closest_left - preferred) <= abs(closest_right - preferred):
                    result[original_index] = closest_left
                else:
                    result[original_index] = closest_right

        return result

# Example usage:
solution = Solution()
rooms = [[2,2],[1,2],[3,2]]
queries = [[3,1],[3,3],[5,2]]
print(solution.closestRoom(rooms, queries))  # Output: [3, -1, 3]

rooms = [[1,4],[2,3],[3,5],[4,1],[5,2]]
queries = [[2,3],[2,4],[2,5]]
print(solution.closestRoom(rooms, queries))  # Output: [2, 1, 3]
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