You are given a N*N maze with a rat placed at maze[0][0]. Find whether any path exist that rat can follow to reach its destination i.e. maze[N-1][N-1]. Rat can move in any direc­tion ( left, right, up and down). Value of every cell in the maze can either be 0 or 1. Cells with value 0 are blocked means rat can­not enter into those cells and those with value 1 are open. Input Format Line 1: Integer N Next N Lines: Each line will contain ith row elements (separated by space) Output Format : The output line contains true if any path exists for the rat to reach its destination otherwise print false. Sample Input 1 : 3 1 0 1 1 0 1 1 1 1 Sample Output 1 : true Sample Input 2 : 3 1 0 1 1 0 1 0 1 1 Sample Output 2 :  false Solution: //// public class Solution {     public static boolean ratInAMaze(int maze[][]){         int n = maze.length;         int path[][] = new int[n][n];          return solveMaze(maze, 0, 0, path);     }        public static boolean solveMaze(int[][] maze, int i, int j, int[][] path) { //        Checking if the current cell which we are going to traverse is a valid cell or not?         if (i < 0 || i >= maze.length || j < 0 || j >= maze.length || maze[i][j] == 0 || path[i][j] == 1) {             return false;         } //        Including the current cell to be in path[i][j]         path[i][j] = 1; //        Checking if the current cell is destination cell or not         if (i == maze.length - 1 && j == maze.length - 1) { //            printing the path before returning             return true;         } //        Explore further the maze in all direction //        Top Direction         if (solveMaze(maze, i - 1, j, path)) {             return true;         } //        Right Direction         if (solveMaze(maze, i, j + 1, path)) {             return true;         } //        Bottom Direction         if (solveMaze(maze, i + 1, j, path)) {             return true;         } //        Left Direction

C++ Programming: From Problem Analysis to Program Design
8th Edition
ISBN:9781337102087
Author:D. S. Malik
Publisher:D. S. Malik
Chapter8: Arrays And Strings
Section: Chapter Questions
Problem 24PE
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You are given a N*N maze with a rat placed at maze[0][0]. Find whether any path exist that rat can follow to reach its destination i.e. maze[N-1][N-1]. Rat can move in any direc­tion ( left, right, up and down).
Value of every cell in the maze can either be 0 or 1. Cells with value 0 are blocked means rat can­not enter into those cells and those with value 1 are open.
Input Format
Line 1: Integer N
Next N Lines: Each line will contain ith row elements (separated by space)
Output Format :
The output line contains true if any path exists for the rat to reach its destination otherwise print false.
Sample Input 1 :
3
1 0 1
1 0 1
1 1 1
Sample Output 1 :
true
Sample Input 2 :
3
1 0 1
1 0 1
0 1 1
Sample Output 2 :
 false

Solution: ////

public class Solution {

    public static boolean ratInAMaze(int maze[][]){

        int n = maze.length;
        int path[][] = new int[n][n];
         return solveMaze(maze, 0, 0, path);
    }
       public static boolean solveMaze(int[][] maze, int i, int j, int[][] path) {
//        Checking if the current cell which we are going to traverse is a valid cell or not?
        if (i < 0 || i >= maze.length || j < 0 || j >= maze.length || maze[i][j] == 0 || path[i][j] == 1) {
            return false;
        }
//        Including the current cell to be in path[i][j]
        path[i][j] = 1;
//        Checking if the current cell is destination cell or not
        if (i == maze.length - 1 && j == maze.length - 1) {
//            printing the path before returning
            return true;
        }
//        Explore further the maze in all direction
//        Top Direction
        if (solveMaze(maze, i - 1, j, path)) {
            return true;
        }
//        Right Direction
        if (solveMaze(maze, i, j + 1, path)) {
            return true;
        }
//        Bottom Direction
        if (solveMaze(maze, i + 1, j, path)) {
            return true;
        }
//        Left Direction

 

 

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