Minimum time required to reach a given cell (x, y) from (0,0). A particular cell may have a wall i.e. you cannot go to that particular cell. (0,0) and (x, y) does not have a wall. Solution: We just have to do BFS from (0,0) and keep storing the time also of every particular cell when it is pushed back in the BFS queue. Given many messages. messages consist of (Type, Port, “message”, message_length), you have to sort (on the basis of decreasing length of messages) in such a way that different messages with the same Type and Port will combine i.e. the message will concatenate and message length will be the length of message 1 + length of message 2. Output final messages.