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This is a problem of directed graph. Say there n (=6 here) tickets
- sumitpandey.cse September 25, 2016T1 (A to B)
T2 (B to C)
T3 (C to A)
T4 (B to D)
T5 (A to B)
T6 (X to Y)
1. Draw a graph with START and DESTINATION as vertices and Ticket as edge. Vertex having in-degree 0 (connects only out edge and no in edge) will be the starting point. Vertex having out-degree 0 (no out edge) will be the end point.
2. There is a possibility that graph contains one or multiple cycles (ticket T1, T2 and T3, A->B->C->A) in that case order doesn't matter. Traveler can pick any of the ticket which form cycle to start the journey. Here any of T1/T2/T3.
3. There is also a possibility that graph contains an isolated edge (ticket T6). This can be flagged as - traveler needs at least one more connecting ticket.