There is a maze of N x N size, each element will have an alphabet from ‘A’ to ‘Z’. (for ease of calculation N is odd). ‘T’ represents Treasure, X represents the passage. The treasure is placed at the center of the maze (e.g. for a 5 x 5 maze, treasure is at 3rd row & 3rd Column). Only way to reach the treasure is from the top left corner of the maze & the entry is allowed to next layer only if top left corner has passage ( ‘X’ ). Your aim is to find the minimum number of steps at each layer of the maze to be rotated counter clockwise ( +ve) /clockwise ( -ve) so that all ‘X’ come to the left corner & treasure can be reached.
It is guaranteed that only one ‘X’ is present in each of the layer.
Example 1
Input
5
O I M U R
J V U X A
X W T S R
K Z F H D
Q W K V M
Output
-2 2
X J O I M
K X S H U
Q U T F R
W V W Z A
K V M D R
Example 2
Input
7
L E A R K X G
N Z N Z Q B S
O P E B M S I
A Z R T X M U
A W S S O K U
A X B A G M O
D F Q A C U U
Output
5 -4 3
X G S I U U O
K X W Z P Z U
R B X O S N U
A A M T S Z C
E G B E R Q A
L M K M S B Q
N O A A A D F



My solution:


import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args)throws IOException {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
        int n=Integer.parseInt(br.readLine());
        char m[][]=new char[n][n];
        for(int i=0;i<n;i++)
        {   String s[]=br.readLine().split(" ");
            for(int j=0;j<n;j++)
                m[i][j]=s[j].charAt(0);
        }
        for(int k=0;k<(n/2);k++)
        {   int s1=0,r=k,c=k,f=0,i=0;
            int te=2*(n-2*k)+2*(n-2*k-2);//total elements
            char temp[]=new char[te];
            for(;c<(n-k);c++)
            {   temp[i]=m[k][c];
                i++;
            }
            for(r=k+1;r<(n-k);r++)
            {   temp[i]=m[r][c-1];
                i++;
            }
            for(c=c-2;c>k;c--)
            {   temp[i]=m[r-1][c];
                i++;
            }
            for(r=r-1;r>k;r--)
            {   temp[i]=m[r][k];
                i++;
            }
			r=k;c=k;
            for(;c<(n-k);c++)
            {   if(m[k][c]=='X')
                {   f=1;
                    break;
                }
                s1++;
            }
            for(r=k+1;r<(n-k)&&f==0;r++)
            {   if(m[r][c-1]=='X')
                {   f=1;
                    break;
                }
                s1++;
            }
            for(c=c-2;c>k && f==0;c--)
            {   if(m[r-1][c]=='X')
                {   f=1;
                    break;
                }
                s1++;
            }
            for(r=r-1;r>k && f==0;r--)
            {   if(m[r][k]=='X')
                {   f=1;
                    break;
                }
                s1++;
            }
            if(s1<=(te/2))
                bw.write(s1+" ");
            else
                bw.write("-"+(te-s1)+" ");
            char temp2[]=new char[te];
            for(i=0;i<te;i++)
            {   int t=(i-s1+te)%te;
                temp2[t]=temp[i];
            }
            r=k;c=k;i=0;
            for(;c<(n-k);c++)
            {   m[r][c]=temp2[i];
                i++;
            }
            for(r=k+1;r<(n-k);r++)
            {   m[r][c-1]=temp2[i];
                i++;
            }
            for(c=c-2;c>k;c--)
            {   m[r-1][c]=temp2[i];
                i++;
            }
            for(r=r-1;r>k;r--)
            {   m[r][k]=temp2[i];
                i++;
            }
        }
        bw.write("\n");
        for(int i=0;i<n;i++)
        {   for(int j=0;j<n;j++)
                bw.write(m[i][j]+" ");
            bw.write("\n");
        }
        bw.flush();
    }
}