Условие
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.
Ограничения
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.
Примеры
Ввод:
rooms = [3, 1, 4]
groups = [2, 3, 2]
Ожидаемый вывод:
expected = [1, 3, 0]
Ввод:
std::vector<long long> rooms = {3, 1, 4};
std::vector<long long> groups = {2, 3, 2};
Ожидаемый вывод:
std::vector<long long> expected = {1, 3, 0};
Ввод:
long[] rooms = new long[] {3, 1, 4};
long[] groups = new long[] {2, 3, 2};
Ожидаемый вывод:
long[] expected = new long[] {1, 3, 0};
Пояснение
The first group leaves [1, 1, 4]. The second uses hotel 3, leaving [1, 1, 1]. The last group of 2 cannot fit.
Ввод:
rooms = [2, 2]
groups = [2, 2]
Ожидаемый вывод:
expected = [1, 2]
Ввод:
std::vector<long long> rooms = {2, 2};
std::vector<long long> groups = {2, 2};
Ожидаемый вывод:
std::vector<long long> expected = {1, 2};
Ввод:
long[] rooms = new long[] {2, 2};
long[] groups = new long[] {2, 2};
Ожидаемый вывод:
long[] expected = new long[] {1, 2};
Пояснение
Both hotels fit the first group, so choose the earlier hotel 1. It becomes full, so the next group uses hotel 2.
Ссылки
Это задание адаптировано для CodeUtility. Следуйте условию и ограничениям выше.