How to find connected friends in a social media using java steam - graph

Question

In a social media, suppose each user is represented like below:
public class User {
private int userId;
private List<User> friendsList;
User(int id) {
this.userID = id;
this.friendsList = new LinkedList<>();
}
void addFriend(User a) {
this.friendsList.add(a);
}
int getUserId() {
return this.userID;
}
List<User> getFriendsList() {
return this.friendsList;
}
}
if A and B are friends(A->B), then A will be added to B's friend list and B will be added to A's friend list.
A.addFriend(B);
B.addFriend(A);
How i can efficiently find two users X and Y are connected by using Java 8 steam methods ?
ie if X->p->q->r->s->Y, then
bool isConnected(User X, User Y) should return true
Here X and Y are connected through common friends p, q, r and s.

Answer 1

You would need a way to keep track of the object that was visited or else you will go in loops.
public bool NodeVisited { get; private set; }
boolean isConnected(User a, User b){
for(final User u : a.getFriendsList()){
if(NodeVisited == false){
NodeVisited = true;
if(u.getUserId() == b.getUserId() || isConnected(u, b)){
return true;
}
}
else continue;
}
return false;
}

Answer 2

The close problem to "Find the shortest way (if any) between 2 points in a labyrinth."
Main idea:
Spread 2 waves from 2 points (or 2 friends) which cover near points (friends)
Remember only(!) previous layers of points (friends) to restrict the future spreading out of the wave in that directions
After each spreading out step check wawes intersection for common points (friends)
Repeat until getting intersections of waves or 1 wave will be stopped:
there are no new directions to spread except in backward directions from the previous step.
P.S.:
The most optimal algorithm involves "several" (depends on data) spots of wave spreading processes.
https://github.com/Dimoniada/FriendsProblem

Answer 3

You don't really have to use streams, but just a simple for-loop and recursion:
boolean isConnected(User a, User b){
for(final User u : a.getFriendsList()){
if(u.getUserId() == b.getUserId() || isConnected(u, b)){
return true;
}
}
return false;
}
If you still want to use streams, you can use this snippet (still uses recursion though):
boolean isConnected(User a, User b){
return a.getFriendsList()
.stream()
.anyMatch(u -> u.getUserId() == b.getUserId() || isConnected(u, b));
}