You wrote a trendy new messaging app, MeshMessage, to get around flaky cell phone coverage.

Instead of routing texts through cell towers, your app sends messages via the phones of nearby users, passing each message along from one phone to the next until it reaches the intended recipient. (Don't worry—the messages are encrypted while they're in transit.)

Some friends have been using your service, and they're complaining that it takes a long time for messages to get delivered. After some preliminary debugging, you suspect messages might not be taking the most direct route from the sender to the recipient.

Given information about active users on the network, find the shortest route for a message from one user (the sender) to another (the recipient). Return a vector of users that make up this route.

There might be a few shortest delivery routes, all with the same length. For now, let's just return any shortest route.

Your network information takes the form of an unordered map mapping username strings to a vector of other users nearby:

For the network above, a message from Jayden to Adam should have this route:

Our solution has two main steps. First, we do a breadth-first search of the user network starting from the sender. Then, we use the results of our search to backtrack and find the shortest path.

How much work is a breadth-first search?

In the worst case, we'll go through the BFS loop once for every node in the graph, since we only ever add each node to nodesToVisit once (we check howWeReachedNodes to see if we've already added a node before). Each loop iteration involves a constant amount of work to dequeue the node and check if it's our end node. If we have n nodes, then this portion of the loop is .

But there's more to each loop iteration: we also look at the current node's neighbors. Over all of the nodes in the graph, checking the neighbors is , since it involves crossing each edge twice: once for each node at either end.

Putting this together, the complexity of the breadth-first search is .

What about backtracking to determine the shortest path? Handling each node in the path is , and we could have at most N nodes in our shortest path. So, that's for building up the path. Then, it's another to reverse it. So, the total time complexity of our backtracking step is .

Putting these together, the time complexity of our entire algorithm is .

What about space complexity? The queue of nodes to visit, the mapping of nodes to previous nodes, and the final path ... they all store a constant amount of information per node. So, each data structure could take up to space if it stored information about all of our nodes. That means our overall space complexity is .

1. In our solution, we assumed that if one user (Min) could transmit a message to another (Jayden), then Jayden would also be able to transmit a message to Min. Suppose this wasn't guaranteed—maybe Min's cell phone transmits over shorter distances than Jayden's. How would our graph change to represent this? Could we still use BFS?
2. What if we wanted to find the shortest path? Assume we're given a GPS location for each user. How could we incorporate the distance between users into our graph? How would our algorithm change?
3. In our solution, we assumed that users never moved around. How could we extend our algorithm to handle the graph changing over time?