Garage occupancy at every timestamp

A parking garage records occupancy over n discrete timestamps, numbered 0 to n - 1. You get a list of parking tickets. Ticket [entry, exit] means one car arrived at timestamp entry and left at timestamp exit.

A car counts as present at every timestamp t with entry <= t <= exit. It still counts at its exit timestamp and stops counting at exit + 1.

Implement:

def garage_occupancy(n: int, tickets: list[list[int]]) -> list[int]:
    ...

Return a list of length n where position t is the number of cars in the garage at timestamp t.

Constraints

• 0 <= n <= 10^5, 0 <= len(tickets) <= 10^5
• 0 <= entry <= exit <= n - 1
• Several cars can share the same entry or exit timestamp.

Example 1

n = 5
tickets = [[0, 2], [1, 1], [2, 4]]
output  = [1, 2, 2, 1, 1]

At timestamp 2, the first car is leaving and the third car is arriving. Both count, so the total is 2.

Example 2

n = 4
tickets = [[3, 3]]
output  = [0, 0, 0, 1]

Aim for better than O(n * len(tickets)) time.

Reconstructed from the reported question; details may differ from what interviewers ask.