Problem
Assign arriving groups in order. A group must stay entirely in the earliest hotel with enough free rooms. Subtract the group size from that hotel’s free rooms. If no hotel fits, return 0 for that group and leave all capacities unchanged. Hotel positions start at 1.
Implement hotel_room_allocation(rooms, groups).
Input
rooms: Initial free rooms per hotel, in priority order.
groups: Group sizes in arrival order.
Output
Return the selected hotel position for each group, or 0 when it cannot be placed.
Constraints
1…200,000 hotels and 0…200,000 groups. Initial room counts are 0…1,000,000,000. Group sizes are 1…1,000,000,000. Empty groups input returns an empty list.
Examples
Input:
rooms = [3, 1, 4]
groups = [2, 3, 2]
Expected output:
expected = [1, 3, 0]Input:
std::vector<long long> rooms = {3, 1, 4};
std::vector<long long> groups = {2, 3, 2};
Expected output:
std::vector<long long> expected = {1, 3, 0};Input:
long[] rooms = new long[] {3, 1, 4};
long[] groups = new long[] {2, 3, 2};
Expected output:
long[] expected = new long[] {1, 3, 0};Explanation
The first group leaves [1, 1, 4]. The second uses hotel 3, leaving [1, 1, 1]. The last group of 2 cannot fit.
Input:
rooms = [2, 2]
groups = [2, 2]
Expected output:
expected = [1, 2]Input:
std::vector<long long> rooms = {2, 2};
std::vector<long long> groups = {2, 2};
Expected output:
std::vector<long long> expected = {1, 2};Input:
long[] rooms = new long[] {2, 2};
long[] groups = new long[] {2, 2};
Expected output:
long[] expected = new long[] {1, 2};Explanation
Both hotels fit the first group, so choose the earlier hotel 1. It becomes full, so the next group uses hotel 2.
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.