I have a question about this code and time complexity of it.
I know that BFS on matrix has nm time complexity, but here we have to travel for every cell 4 cell more, so it will be 4^(nm)?
We cant go back in our travel.

        public int modDistanceSearch(Para positionZero,char dest){// miejsce na Pare czyli dwie wspolrzede dla Artura
            Trojka start = new Trojka(positionZero.left,positionZero.right,0);
            Queue<Trojka> queue = new LinkedList<>();
            queue.add(start);
            boolean[][] visited = new boolean[map.length][map[0].length];
            visited[start.left][start.right] = true;
            while (!queue.isEmpty()){
                //   System.out.println("sdfg");
                Trojka cur = queue.remove();
                if(map[cur.left][cur.right]==dest){
                    // System.out.println(cur.left+" "+cur.right);
                    tutajKonczyArtur.right = cur.right;
                    tutajKonczyArtur.left = cur.left;
                    return cur.distance;
                }
                //up, down, left, right
                if(correctPosition(cur.left-1,cur.right,visited)){
                    queue.add(new Trojka(cur.left-1,cur.right,cur.distance+1));
                    visited[cur.left-1][cur.right]=true;
                    startToS.left =cur.left-1;
                    startToS.right =cur.right;
                    dir.a +=1;
                }
                if(correctPosition(cur.left+1,cur.right,visited)){
                    queue.add(new Trojka(cur.left+1,cur.right,cur.distance+1));
                    visited[cur.left+1][cur.right]=true;
                    startToS.left =cur.left+1;
                    startToS.right =cur.right;
                    dir.b +=1;
                }
                if(correctPosition(cur.left,cur.right-1,visited)){
                    queue.add(new Trojka(cur.left,cur.right-1,cur.distance+1));
                    visited[cur.left][cur.right-1]=true;
                    startToS.left =cur.left;
                    startToS.right =cur.right-1;
                    dir.c+=1;
                }
                if(correctPosition(cur.left,cur.right+1,visited)){
                    queue.add(new Trojka(cur.left,cur.right+1,cur.distance+1));
                    visited[cur.left][cur.right+1]=true;
                    startToS.left =cur.left;
                    startToS.right =cur.right+1;
                    dir.d+=1;
                }


            }
            return -1;
        }