Last Truck to Leave the Lane

A lane is represented by the line segment from 0 to laneLength. Several trucks begin at distinct coordinates on the lane.
initial[i] is the starting coordinate of truck i, and velocity[i] is its signed velocity. A positive velocity moves a truck to the right, while a negative velocity moves it to the left.
When a truck crosses either boundary, it leaves the lane. When two trucks collide, they exchange their velocities and directions.
Return the time at which the last truck leaves the lane.

Function
lastTruckExitTime(laneLength: int, initial: int[], velocity: int[]) → double

Examples
Example 1
laneLength = 10
initial = [2,6]
velocity = [2,-1]
return = 6.0
The trucks collide after 4/3 time units and exchange velocities. One truck then leaves through the right boundary at time 4, while the other leaves through the left boundary at time 6. Therefore, the last truck leaves at time 6.0.

Treat each truck as a ghost that passes straight through every other truck. When two trucks swap velocities, the picture looks identical to two ghosts crossing. So the lane at any moment contains the same positions and velocities as the no-collision version. The last exit time is the max over all ghosts of their individual exit times. For a positive velocity, that's (laneLength - initial[i]) / velocity[i]. For a negative velocity, it's initial[i] / -velocity[i]. Check it on the example: the first truck gives (10-2)/2 = 4, the second gives 6/1 = 6, max is 6.0. That matches. The pitfall is building an event-queue simulation with floating point meeting times, which is slow, buggy, and unnecessary. Also watch your division and return a double, not an integer. Handle a zero velocity explicitly, since it never reaches a boundary. If the trick slips away mid-assessment, StealthCoder is the hedge that gets you to the one-liner fast. Complexity is O(n) time, O(1) space.

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