Minimum Bus Routes to a Destination

Each row in routes lists all stops served by one bus route. Boarding a route costs one bus; after boarding, you may travel to any stop on it. Return the minimum buses needed to travel from sourceStop to targetStop, or -1 if impossible.

Function
minimumBusRoutes(routes: int[][], sourceStop: int, targetStop: int) → int

Examples
Example 1
routes = [[1,2,7],[3,6,7]]
sourceStop = 1
targetStop = 6
return = 2
Take the first route to 7, then the second to 6.
Example 2
routes = [[1,5,7],[3,5]]
sourceStop = 5
targetStop = 5
return = 0
The trip is already complete.
Example 3
routes = [[1,2],[3,4]]
sourceStop = 1
targetStop = 4
return = -1
The route groups are disconnected.

Constraints
1 <= routes.length <= 500.
The total number of listed stops is at most 10^5.
Stops on one route are unique.

Build a map from each stop to the list of routes that serve it. Then run BFS over routes, not stops. Start by queuing every route that contains sourceStop with a count of 1. Pop a route, and if it contains targetStop, return its count. Otherwise, for each stop on that route, push every unvisited route from the map and mark it visited with count plus one. Handle the edge case first: if source equals target, return 0. The common pitfall is BFS over stops, where every stop-to-stop move costs one and you overcount, or you revisit routes thousands of times and time out. Mark visited routes AND clear or skip stops you've already expanded, since the total stop count reaches 10^5. If the queue empties, return -1. StealthCoder is the hedge if the live OA wipes your memory of this setup.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.