1847

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

Companies: Amazon

Related Topics:
Array, Binary Search, Sorting

Similar Questions:

• Most Beautiful Item for Each Query (Medium)
• Minimum Time to Kill All Monsters (Hard)

Hints:

• Is there a way to sort the queries so it's easier to search the closest room larger than the size?
• Use binary search to speed up the search time.

Solution 1. Offline Query

// OJ: https://leetcode.com/problems/closest-room/
// Author: github.com/lzl124631x
// Time: O(RlogR + QlogQ + QlogR)
// Space: O(R)
class Solution {
public:
    vector<int> closestRoom(vector<vector<int>>& R, vector<vector<int>>& Q) {
        for (int i = 0; i < Q.size(); ++i) Q[i].push_back(i);
        sort(begin(Q), end(Q), [](auto &a, auto &b) { return a[1] > b[1]; });
        sort(begin(R), end(R), [](auto &a, auto &b) { return a[1] > b[1]; });
        set<int> ids;
        vector<int> ans(Q.size());
        int i = 0, N = R.size();
        for (auto &q : Q) {
            int pref = q[0], mn = q[1], idx = q[2], id = -1, diff = INT_MAX;
            while (i < N && R[i][1] >= mn) ids.insert(R[i++][0]);
            auto it = ids.upper_bound(pref);
            if (it != ids.end()) id = *it, diff = id - pref;
            if (it != ids.begin() && pref - *prev(it) <= diff) id = *prev(it);
            ans[idx] = id;
        }
        return ans;
    }
};