Traveling the Graphs

Parse an undirected weighted graph and one route request.
edgeLine contains one or more edge tokens [A,B,5], separated by exactly one space.
routeLine has the form A->D,10: start node, destination node, and maximum allowed travel time.
Node names are single uppercase letters, and weights and the limit are unsigned decimal integers.
Return the unique shortest route as uppercase node names joined by -> when its total distance is at most the limit.
Otherwise return the first applicable error by priority:
E1: input syntax error.
E2: logical input error: a duplicate undirected edge, a self-edge, an undefined endpoint, a disconnected graph, or more than one shortest route.
E3: the unique shortest route exceeds the allowed travel time.

Function
travelingTheGraphs(edgeLine: String, routeLine: String) → String

Examples
Example 1
edgeLine = "[A,B,3] [B,C,5] [C,D,2]"
routeLine = "A->D,10"
return = "A->B->C->D"
The only route from A to D has total distance 10, exactly the limit.
Example 2
edgeLine = "[A,B,3] [A,C,7] [C,D,2] [B,C,5]"
routeLine = "A->D,10"
return = "A->C->D"
A->C->D has distance 9, shorter than the alternative through B.
Example 3
edgeLine = "[A,B,5] [A,C,2] [B,C,4] [B,D,6] [C,B,7]"
routeLine = "A->C,10"
return = "E2"
[C,B,7] duplicates the already defined undirected edge [B,C,4].

Constraints
Every node is one of A through Z.
0 <= edgeWeight, maximumTravelTime <= 10^9
The edge line contains at most 325 tokens.
No leading, trailing, or repeated whitespace is valid.

The pattern is Dijkstra with path counting. Run it from the start node, track the shortest distance to each node and how many shortest paths reach it. If the destination count is above 1, return E2. Zero-weight edges are allowed, so don't assume positive weights, but Dijkstra still works. The pitfall is order. First validate syntax strictly: exact brackets, single spaces, single uppercase letters, unsigned integers, no stray whitespace. Only then check logic: duplicate undirected edges (B,C equals C,B), self-edges, endpoints in the route that aren't defined, and a disconnected graph across all nodes seen. Only after all that compare distance to the limit for E3. Reconstruct the path with parent pointers. Where this breaks naive solutions is checking connectivity only between start and end, or counting ties incorrectly with zero weights. If you freeze on the parsing, StealthCoder is the hedge during the live OA.

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