Closest Room
<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>
Array; Binary Search; Ordered Set; Sorting
class Solution:
    def closestRoom(
        self, rooms: List[List[int]], queries: List[List[int]]
    ) -> List[int]:
        rooms.sort(key=lambda x: x[1])
        k = len(queries)
        idx = sorted(range(k), key=lambda i: queries[i][1])
        ans = [-1] * k
        i, n = 0, len(rooms)
        sl = SortedList(x[0] for x in rooms)
        for j in idx:
            prefer, minSize = queries[j]
            while i < n and rooms[i][1] < minSize:
                sl.remove(rooms[i][0])
                i += 1
            if i == n:
                break
            p = sl.bisect_left(prefer)
            if p < len(sl):
                ans[j] = sl[p]
            if p and (ans[j] == -1 or ans[j] - prefer >= prefer - sl[p - 1]):
                ans[j] = sl[p - 1]
        return ans
