George - Implementation

plaintext
function count_available_rooms(n, rooms)
// rooms is array of pairs (occupancy, capacity) count := 0
for each room in rooms
free_space := room.capacity - room.occupancy
if free_space >= 2 then
 count := count + 1
 return count

The loop processes each room once. For each pair, you compute the difference between capacity and occupancy. If that difference is at least $2$, you increment the counter. The time complexity is $O(n)$ where $n$ is the number of rooms. You check each room exactly once. The space complexity is $O(1)$ if you process rooms as you read them.