Java Program for Rat in a Maze | Backtracking-2

08 Jun, 2022
We have discussed Backtracking and Knight’s tour problem in Set 1. Let us discuss Rat in a Maze as another example problem that can be solved using Backtracking.
A Maze is given as N*N binary matrix of blocks where source block is the upper left most block i.e., maze[0][0], and destination block is lower rightmost block i.e., maze[N-1][N-1]. A rat starts from the source and has to reach its destination. The rat can move only in two directions: forward and down.
In the maze matrix, 0 means the block is a dead end, and 1 means the block can be used in the path from source to destination. Note that this is a simple version of the typical Maze problem. For example, a more complex version can be that the rat can move in 4 directions and a more complex version can be with a limited number of moves.
Following is an example maze.
 Gray blocks are dead ends (value = 0). 
Following is binary matrix representation of the above maze.
                {1, 0, 0, 0}
                {1, 1, 0, 1}
                {0, 1, 0, 0}
                {1, 1, 1, 1}
Following is a maze with highlighted solution path.
Following is the solution matrix (output of program) for the above input matrix.
                {1, 0, 0, 0}
                {1, 1, 0, 0}
                {0, 1, 0, 0}
                {0, 1, 1, 1}
 All entries in solution path are marked as 1.
Java
/* Java program to solve Rat in a Maze problem using   backtracking */ public class RatMaze {    final int N = 4;     /* A utility function to print solution matrix       sol[N][N] */    void printSolution(int sol[][])    {        for (int i = 0; i < N; i++) {            for (int j = 0; j < N; j++)                System.out.print(" " + sol[i][j] + " ");            System.out.println();        }    }     /* A utility function to check if x, y is valid        index for N*N maze */    boolean isSafe(int maze[][], int x, int y)    {        // if (x, y outside maze) return false        return (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1);    }     /* This function solves the Maze problem using       Backtracking. It mainly uses solveMazeUtil()       to solve the problem. It returns false if no       path is possible, otherwise return true and       prints the path in the form of 1s. Please note       that there may be more than one solutions, this       function prints one of the feasible solutions.*/    boolean solveMaze(int maze[][])    {        int sol[][] = { { 0, 0, 0, 0 },                        { 0, 0, 0, 0 },                        { 0, 0, 0, 0 },                        { 0, 0, 0, 0 } };         if (solveMazeUtil(maze, 0, 0, sol) == false) {            System.out.print("Solution doesn't exist");            return false;        }         printSolution(sol);        return true;    }     /* A recursive utility function to solve Maze       problem */    boolean solveMazeUtil(int maze[][], int x, int y,                          int sol[][])    {        // if (x, y is goal) return true        if (x == N - 1 && y == N - 1) {            sol[x][y] = 1;            return true;        }         // Check if maze[x][y] is valid        if (isSafe(maze, x, y) == true) {            // mark x, y as part of solution path            sol[x][y] = 1;             /* Move forward in x direction */            if (solveMazeUtil(maze, x + 1, y, sol))                return true;             /* If moving in x direction doesn't give               solution then  Move down in y direction */            if (solveMazeUtil(maze, x, y + 1, sol))                return true;             /* If none of the above movements works then               BACKTRACK: unmark x, y as part of solution               path */            sol[x][y] = 0;            return false;        }         return false;    }     public static void main(String args[])    {        RatMaze rat = new RatMaze();        int maze[][] = { { 1, 0, 0, 0 },                         { 1, 1, 0, 1 },                         { 0, 1, 0, 0 },                         { 1, 1, 1, 1 } };        rat.solveMaze(maze);    }}// This code is contributed by Abhishek Shankhadhar
1  0  0  0 
1  1  0  0 
0  1  0  0 
0  1  1  1
 
Time Complexity: O(2^(n^2))
Auxiliary Space: O(n^2)
Please refer complete article on Rat in a Maze | Backtracking-2 for more details! 
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