Time-Ordered Elevator Dispatch

You are given the starting floors of several elevators and a time-ordered sequence of passenger events. Every elevator starts idle. Passenger i appears at floor requestFloors[i], requests direction requestDirections[i], and is processed at absolute time requestTimes[i]. Direction 1 means up and -1 means down.
Process passengers in input order. Before processing each passenger, advance every moving elevator by one integer floor per unit of elapsed time in its current direction. Idle elevators stay at their current floors. The floor line has no endpoints.
An elevator is eligible under these rules:
An idle elevator is always eligible.
An elevator moving up is eligible only when the passenger also requests up and the passenger floor is at or above the elevator's current floor.
An elevator moving down is eligible only when the passenger also requests down and the passenger floor is at or below the elevator's current floor.
Assign the nearest eligible elevator by absolute floor distance. If several eligible elevators are equally near, choose the smaller original index. If no elevator is eligible, the passenger is unserved and the ordinary time advance is the only state change.
After an assignment, reset the selected elevator's modeled floor to the passenger's origin and set its direction to the passenger's requested direction. Travel to the pickup is outside this simulation and consumes no simulated time. The elevator continues in that direction until a later assignment changes it. Passengers with equal times are processed in input order.
Return the index of the elevator assigned to the final passenger, or -1 if the final passenger is unserved.

Function
dispatchFinalPassenger(elevatorFloors: int[], requestFloors: int[], requestDirections: int[], requestTimes: int[]) → int

Examples
Example 1
elevatorFloors = [0,10]
requestFloors = [2,8,6]
requestDirections = [1,-1,-1]
requestTimes = [0,2,4]
return = 1
At time 0, elevator 0 is closest to floor 2, so it is assigned and begins moving up from floor 2. At time 2 it has reached floor 4, but it is ineligible for a down request. Idle elevator 1 is assigned at floor 8 and begins moving down. At time 4 both elevators are at floor 6, but only elevator 1 matches the final down request, so the result is 1.
Example 2
elevatorFloors = [0,10,20]
requestFloors = [5,15]
requestDirections = [1,1]
requestTimes = [0,0]
return = 1
The first request is equally distant from elevators 0 and 1, so index 0 wins. No time elapses before the second request. Elevators 1 and 2 are both five floors away and idle, so the smaller index 1 serves the final passenger.
Example 3
elevatorFloors = [0]
requestFloors = [0,-1]
requestDirections = [1,-1]
requestTimes = [0,1]
return = -1
Elevator 0 accepts the first up request. By time 1 it is at floor 1 and still moving up, so it is ineligible for the final down request at floor -1. The final passenger is unserved.

Constraints
1 <= elevatorFloors.length <= 200
1 <= requestFloors.length <= 2000
requestDirections.length == requestFloors.length
requestTimes.length == requestFloors.length
-10^6 <= elevatorFloors[i], requestFloors[i] <= 10^6
requestDirections[i] is either -1 or 1.
0 <= requestTimes[i] <= 10^9
requestTimes is nondecreasing.

The trick is that there is no trick. Constraints are 200 elevators and 2000 requests, so an O(requests x elevators) simulation is fine, about 400k steps. Keep two arrays: floor and direction (0 for idle). For each passenger, compute dt as the current time minus the previous time, then move each non-idle elevator by direction times dt. Don't advance before the first passenger, since elevators start idle anyway. Then scan for eligible elevators. Idle is always eligible. Up-moving needs an up request with passenger floor >= elevator floor. Down-moving needs a down request with passenger floor <= elevator floor. Pick the smallest distance, and break ties with the smaller index by scanning in order and using strict less-than. The common pitfalls are forgetting that an assigned elevator keeps moving afterward, and mishandling equal times where dt is zero. Track the last result and return it for the final passenger. StealthCoder is your hedge if the eligibility conditions get tangled live.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.