Hotel Room Allocation

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.

เงื่อนไข

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.

ตัวอย่าง

อินพุต:
rooms = [3, 1, 4]
groups = [2, 3, 2]

ผลลัพธ์ที่คาดหวัง:
expected = [1, 3, 0]

อินพุต:
std::vector<long long> rooms = {3, 1, 4};
std::vector<long long> groups = {2, 3, 2};

ผลลัพธ์ที่คาดหวัง:
std::vector<long long> expected = {1, 3, 0};

อินพุต:
long[] rooms = new long[] {3, 1, 4};
long[] groups = new long[] {2, 3, 2};

ผลลัพธ์ที่คาดหวัง:
long[] expected = new long[] {1, 3, 0};

อินพุต:
rooms = [2, 2]
groups = [2, 2]

ผลลัพธ์ที่คาดหวัง:
expected = [1, 2]

อินพุต:
std::vector<long long> rooms = {2, 2};
std::vector<long long> groups = {2, 2};

ผลลัพธ์ที่คาดหวัง:
std::vector<long long> expected = {1, 2};

อินพุต:
long[] rooms = new long[] {2, 2};
long[] groups = new long[] {2, 2};

ผลลัพธ์ที่คาดหวัง:
long[] expected = new long[] {1, 2};

แหล่งอ้างอิง

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• CSES: Hotel Queries