Fuel station queue with three dispensers

A fuel station has three dispensers named X, Y and Z. Each starts with a known number of liters. A queue of cars waits at the station, and all of them are in line at time 0. Car i needs demands[i] liters.

Rules:

• Only the car at the front of the queue chooses a dispenser. Cars never overtake one another.
• The car may use a dispenser only if it is free (not serving a car) and holds at least the car's demand.
• If several dispensers qualify, the car takes the first by name: X, then Y, then Z.
• If none qualifies right now, the car waits at the front and blocks everyone behind it.
• Fuel flows at 1 liter per second, so a car with demand d keeps its dispenser busy for d seconds. A dispenser that finishes at time t can serve the next car at time t.

A car's waiting time is the time it starts refuelling. Implement:

def max_wait(demands: list[int], x: int, y: int, z: int) -> int

Return the longest waiting time over all cars. Return -1 if some car can never refuel.

Constraints: 0 <= len(demands) <= 10^5; 0 <= demands[i], x, y, z <= 10^9.

Example 1: demands = [3, 5, 4, 2], x = 10, y = 5, z = 1 returns 7.

• At 0, car 1 takes X (7 liters left, busy until 3). Car 2 takes Y (0 left, busy until 5).
• Car 3 needs 4 liters. Only X has enough, so it waits and starts at 3 (3 left, busy until 7).
• Car 4 needs 2 liters. Only X has enough, so it starts at 7.

Example 2: demands = [5, 1], x = 5, y = 0, z = 0 returns -1. Car 1 empties X, and no dispenser has fuel for car 2.