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 {
    public int[] closestRoom(int[][] rooms, int[][] queries) {
        int n = rooms.length;
        int k = queries.length;
        Arrays.sort(rooms, (a, b) -> a[1] - b[1]);
        Integer[] idx = new Integer[k];
        for (int i = 0; i < k; i++) {
            idx[i] = i;
        }
        Arrays.sort(idx, (i, j) -> queries[i][1] - queries[j][1]);
        int i = 0;
        TreeMap<Integer, Integer> tm = new TreeMap<>();
        for (int[] room : rooms) {
            tm.merge(room[0], 1, Integer::sum);
        }
        int[] ans = new int[k];
        Arrays.fill(ans, -1);
        for (int j : idx) {
            int prefer = queries[j][0], minSize = queries[j][1];
            while (i < n && rooms[i][1] < minSize) {
                if (tm.merge(rooms[i][0], -1, Integer::sum) == 0) {
                    tm.remove(rooms[i][0]);
                }
                ++i;
            }
            if (i == n) {
                break;
            }
            Integer p = tm.ceilingKey(prefer);
            if (p != null) {
                ans[j] = p;
            }
            p = tm.floorKey(prefer);
            if (p != null && (ans[j] == -1 || ans[j] - prefer >= prefer - p)) {
                ans[j] = p;
            }
        }
        return ans;
    }
}