1847. Closest Room

Problem

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

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

• The room has a size of at least minSizej, and

• abs(id - preferredj) 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 jth 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 <= 10^5

• k == queries.length

• 1 <= k <= 10^4

• 1 <= roomIdi, preferredj <= 10^7

• 1 <= sizei, minSizej <= 10^7

Solution

class Solution {
    public int[] closestRoom(int[][] rooms, int[][] queries) {
        int numRoom = rooms.length;
        int numQuery = queries.length;
        for (int i = 0; i < numQuery; i++) {
            queries[i] = new int[] {queries[i][0], queries[i][1], i};
        }
        Arrays.sort(rooms, (a, b) -> (a[1] != b[1] ? (a[1] - b[1]) : (a[0] - b[0])));
        Arrays.sort(queries, (a, b) -> (a[1] != b[1] ? (a[1] - b[1]) : (a[0] - b[0])));
        TreeSet<Integer> roomIds = new TreeSet<>();
        int[] result = new int[numQuery];
        int j = numRoom - 1;
        for (int i = numQuery - 1; i >= 0; i--) {
            int currRoomId = queries[i][0];
            int currRoomSize = queries[i][1];
            int currQueryIndex = queries[i][2];
            while (j >= 0 && rooms[j][1] >= currRoomSize) {
                roomIds.add(rooms[j--][0]);
            }
            if (roomIds.contains(currRoomId)) {
                result[currQueryIndex] = currRoomId;
                continue;
            }
            Integer nextRoomId = roomIds.higher(currRoomId);
            Integer prevRoomId = roomIds.lower(currRoomId);
            if (nextRoomId == null && prevRoomId == null) {
                result[currQueryIndex] = -1;
            } else if (nextRoomId == null) {
                result[currQueryIndex] = prevRoomId;
            } else if (prevRoomId == null) {
                result[currQueryIndex] = nextRoomId;
            } else {
                if ((currRoomId - prevRoomId) <= (nextRoomId - currRoomId)) {
                    result[currQueryIndex] = prevRoomId;
                } else {
                    result[currQueryIndex] = nextRoomId;
                }
            }
        }
        return result;
    }
}

Explain:

nope.

Complexity:

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