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Kotlin-based LeetCode algorithm problem solutions, regularly updated
1847\. Closest Room
Hard
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 <= 105
* `k == queries.length`
* 1 <= k <= 104
* 1 <= roomIdi, preferredj <= 107
* 1 <= sizei, minSizej <= 107