HackerRank: Count Luck

HackerRank: Count Luck
Hermione Granger is lost in the Forbidden Forest while collecting some herbs for her magical potion. The forest is magical and has only 1 exit point which magically transports her back to the Hogwarts School of Wizardy and Witch Craft.
Forest can be considered as a grid of NxM size. Each cell in the forest is either empty (represented by '.') or has a tree (represented by 'X'). Hermione can move through empty cell, but not through cells with tree on it. She can only travel LEFT, RIGHT, UP, DOWN. Her position in the forest is indicated by the marker 'M' and the location of the exit point is indicated by '*'. Top-left corner is indexed (0, 0).

.X.X......X
.X*.X.XXX.X
.XX.X.XM...
......XXXX.
In the above forest, Hermione is located at index (2, 7) and exit is at (1, 2). Each cell is indexed according to Matrix Convention

She starts her commute back to the exit and every time she encounters more than one option to move, she waves her wand and the correct path is illuminated and she proceeds in that way. It is guaranteed that there is only one path to each reachable cell from the starting cell. Can you tell us if she waved her wand exactly K times or not? Ron will be impressed if she is able to do so.

This is a good exercise for the graph theory, as the problem statement states we are dealing with a tree. This can be realized out of the sentence “It is guaranteed that there is only one path to each reachable cell from the starting cell.” So simply you have to count branches on the path from “M” to “*” and check if it is equal to K. This can be done by DFS or BFS in linear time.
Counting the branches is the tricky part. One way is to:

As we traverse through the maze, on path from "M" to "*", lets count number of “.” which leaves current node with more depth and then if that number is more than or equal to 2 then we are at a branch and increase answer by one.

http://blueocean-penn.blogspot.com/2014/12/count-luck-maz.html
Most of the maze question are related to DFS graph search, can be solved by recursive way.

public class Solution {

    public static class Node{

        int r;int c; Node pre;

        public Node(int r, int c){this.r = r; this.c = c;}

    }

    public static void visit(char[][] maz, int startR, int startC, boolean[][] seen,  Node pre, Node tail){

        if(startR<0 || startC<0 || startR>=maz.length || startC>=maz[0].length || seen[startR][startC])

            return;

        if(maz[startR][startC]=='X')

            return;

        if(maz[startR][startC]=='*'){

            tail.pre = pre;

            return;       

        }

        seen[startR][startC] = true;

        Node n = new Node(startR, startC);

        n.pre = pre;

        visit(maz, startR-1, startC, seen,  n, tail);

        visit(maz, startR+1, startC, seen,  n, tail);

        visit(maz, startR, startC-1, seen,  n, tail);

        visit(maz, startR, startC+1, seen,  n, tail);

    }

    public static void main(String[] args) {

        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */

        try(Scanner conin = new Scanner(System.in)){

            int t = conin.nextInt();

            while(t-->0){

                int row = conin.nextInt();

                int col = conin.nextInt();

                int startR=0, startC=0;

                int endR=0, endC=0;

                char[][] maz = new char[row][col];

                for(int r=0; r<row; r++){

                    String str = conin.next();

                    int index = str.indexOf("M");

                    if(index>=0){

                        startR=r;startC=index;

                    }

                    index = str.indexOf("*");

                    if(index>=0){

                        endR=r;endC=index;

                    }

                    maz[r] = str.toCharArray();

                }

                int k=conin.nextInt();

                boolean[][] seen = new boolean[row][col];

                Node tail = new Node(endR, endC);

                Node head = null;

                visit(maz, startR, startC, seen, head, tail);                 

                int counter = 0;

                Node n = tail.pre;

                while(n!=null){

                    //System.out.println(n.r + "-" + n.c);

                    int option = 0;

                    if(n.r-1>=0 && n.r-1<row && maz[n.r-1][n.c]=='.')

                        option++;

                    if(n.r+1>=0 && n.r+1<row && maz[n.r+1][n.c]=='.')

                        option++;

                    if(n.c-1>=0 && n.c-1<col && maz[n.r][n.c-1]=='.')

                        option++;

                    if(n.c+1>=0 && n.c+1<col && maz[n.r][n.c+1]=='.')

                        option++;

                    if(option>2 || (n == tail.pre&&option>1) || (n.r==startR&&n.c==startC&&option>1) || (n.pre!=null && n.pre.r==startR&&n.pre.c==startC&&option>1))

                        counter++; // exit early if counter >k

                    n = n.pre;                    

                }

                

                System.out.println(counter==k?"Impressed":"Oops!");

            }

        }

    }

}