Problem Statement

Consider a rat placed at (0, 0) in a square matrix ****of order N * N. It has to reach the destination at (N - 1, N - 1). Find all possible paths that the rat can take to reach from source to destination. The directions in which the rat can move are 'U'(up), 'D'(down), 'L' (left), 'R' (right). Value 0 at a cell in the matrix represents that it is blocked and rat cannot move to it while value 1 at a cell in the matrix represents that rat can be travel through it.Note: In a path, no cell can be visited more than one time.

Example 1:

Input:
N = 4
m[][] = {{1, 0, 0, 0},
         {1, 1, 0, 1},
         {1, 1, 0, 0},
         {0, 1, 1, 1}}
Output:
DDRDRR DRDDRR
Explanation:
The rat can reach the destination at
(3, 3) from (0, 0) by two paths - DRDDRR
and DDRDRR, when printed in sorted order
we get DDRDRR DRDDRR.

Example 2:

Input:
N = 2
m[][] = {{1, 0},
         {1, 0}}
Output:
-1
Explanation:
No path exists and destination cell is
blocked.

Problem Link

Rat in a Maze Problem - I | Practice | GeeksforGeeks

Code (Using Backtracking)

class Solution{
    public:
    vector<string> findPath(vector<vector<int>> &maze, int n) {
        
      vector<string> res;
      string temp = "";
      
        
       if(maze[0][0]==0 or maze[n-1][n-1]==0)
       return res;
       
       vector<vector<bool>> visited(n,vector<bool>(n,false));    
        
       dfs(maze,0,0,n,res,temp,visited);
            
        
        return res;
    }
    void dfs(vector<vector<int>> &maze, int i, int j, int n,vector<string>& res,string &temp,vector<vector<bool>> &visited )
    {
        if(i<0 or j<0 or i>n-1 or j>n-1)
          return;
          
        if(visited[i][j] or !maze[i][j])
           return;
          
        if(i==n-1 and j==n-1)
        {
            res.push_back(temp);
            return;
        }
        
        visited[i][j] = true;
        
         if(i+1<=n-1 and !visited[i+1][j] and maze[i+1][j])
         {
            temp.push_back('D');
            dfs(maze,i+1,j,n,res,temp,visited);
            temp.pop_back();
         }
        
        if(j-1>=0 and !visited[i][j-1] and maze[i][j-1])
        {
            temp.push_back('L');
            dfs(maze,i,j-1,n,res,temp,visited);
            temp.pop_back();
        }
        
        if(j+1<=n-1 and !visited[i][j+1] and maze[i][j+1])
        {
            temp.push_back('R');
            dfs(maze,i,j+1,n,res,temp,visited);
            temp.pop_back();
        }
        
        
        if(i-1>=0 and !visited[i-1][j] and maze[i-1][j])
        {
            temp.push_back('U');
            dfs(maze,i-1,j,n,res,temp,visited);
            temp.pop_back();
        }
        
       
        visited[i][j] = false;
    }
};