Rat in a Maze Problem when movement in all possible directions is allowed

Consider a rat placed at (0, 0) in a square matrix m[ ][ ] of order n and has to reach the destination at (n-1, n-1). The task is to find a sorted array of strings denoting all the possible directions which the rat can take to reach the destination at (n-1, n-1). The directions in which the rat can move are ‘U'(up), ‘D'(down), ‘L’ (left), ‘R’ (right).

Examples:

Input : N = 4 
1 0 0 0 
1 1 0 1 
0 1 0 0 
0 1 1 1
Output :
DRDDRR

Input :N = 4 
1 0 0 0 
1 1 0 1 
1 1 0 0 
0 1 1 1
Output :
DDRDRR DRDDRR
Explanation:

Solution:

Brute Force Approach:

We can use simple backtracking without using any extra space .

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
#define MAX 5
using namespace std;


    void getallpath(int matrix[MAX][MAX], int n,int row,int col,vector<string> &ans,string cur)
    {
        if(row>=n or col>=n or row<0 or col<0 or matrix[row][col] == 0)
        return ;

        if(row == n-1 and col == n-1)
        {
            ans.push_back(cur);
            return ;
        }

        //now if its one we have 4 calls
        matrix[row][col] = 0;

        getallpath(matrix,n,row-1,col,ans,cur+"U");
        getallpath(matrix,n,row,col+1,ans,cur+"R");
        getallpath(matrix,n,row,col-1,ans,cur+"L");
        getallpath(matrix,n,row+1,col,ans,cur+"D");

        matrix[row][col] = 1;

        return ;
    }

vector<string> findPath(int matrix[MAX][MAX], int n) {
        // Your code goes here
        vector<string> ans;
        getallpath(matrix,n,0,0,ans,"");
        return ans;
    }
int main()
{
    int m[MAX][MAX] = { { 1, 0, 0, 0, 0 },
                        { 1, 1, 1, 1, 1 },
                        { 1, 1, 1, 0, 1 },
                        { 0, 0, 0, 0, 1 },
                        { 0, 0, 0, 0, 1 } };
    int n = sizeof(m) / sizeof(m[0]);
    vector<string> ans = findPath(m, n);
    for(auto i : ans)
    cout<<i<<" ";

    return 0;
}

Python3

# Python3 implementation of the above approach
def getallpath(m, n, ans, curpath, r, c):
        if(r>=n or c>=n or r<0 or c<0 or m[r] == 0):
            return;

        if(r == n-1 and c == n-1):
            ans.append(curpath)
            return
        m[r] = 0

        # calls
        getallpath(m,n,ans,curpath+"U",r-1,c)
        getallpath(m,n,ans,curpath+"D",r+1,c)
        getallpath(m,n,ans,curpath+"L",r,c-1)
        getallpath(m,n,ans,curpath+"R",r,c+1)

        m[r] = 1;

# Function to store and print
# all the valid paths
def findpath(m ,n):

    # vector to store all the possible paths
    possiblePaths = []
    getallpath(m,n,possiblePaths,"",0,0)
    # Print all possible paths
    for i in range(len(possiblePaths)):
        print(possiblePaths[i], end = " ")

# Driver code
if __name__ == "__main__":

    m = [ [ 1, 0, 0, 0, 0 ],
          [ 1, 1, 1, 1, 1 ],
          [ 1, 1, 1, 0, 1 ],
          [ 0, 0, 0, 0, 1 ],
          [ 0, 0, 0, 0, 1 ] ]
    n = len(m)

    findpath(m, n)

# This code is contributed by Arpit Jain

Output

DRRRRDDD DRDRURRDDD DDRURRRDDD DDRRURRDDD 

Complexity Analysis:

Time Complexity: O(4^(n^2)). 
As we are making 4 calls for every cell in the matrix.
Auxiliary Space: O(1). 
As we are not using any extra space.

Approach 2: 

1. Start from the initial index (i.e. (0,0)) and look for the valid moves through the adjacent cells in the order Down->Left->Right->Up (so as to get the sorted paths) in the grid.
2. If the move is possible, then move to that cell while storing the character corresponding to the move(D,L,R,U) and again start looking for the valid move until the last index (i.e. (n-1,n-1)) is reached.
3. Also, keep on marking the cells as visited and when we traversed all the paths possible from that cell, then unmark that cell for other different paths and remove the character from the path formed.
4. As the last index of the grid(bottom right) is reached, then store the traversed path.

Below is the implementation of the above approach:  

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
#define MAX 5
using namespace std;

// Function returns true if the
// move taken is valid else
// it will return false.
bool isSafe(int row, int col, int m[][MAX],
                 int n, bool visited[][MAX])
{
    if (row == -1 || row == n || col == -1 ||
                  col == n || visited[row][col]
                           || m[row][col] == 0)
        return false;

    return true;
}

// Function to print all the possible
// paths from (0, 0) to (n-1, n-1).
void printPathUtil(int row, int col, int m[][MAX],
              int n, string& path, vector<string>&
               possiblePaths, bool visited[][MAX])
{
    // This will check the initial point
    // (i.e. (0, 0)) to start the paths.
    if (row == -1 || row == n || col == -1
               || col == n || visited[row][col]
                           || m[row][col] == 0)
        return;

    // If reach the last cell (n-1, n-1)
    // then store the path and return
    if (row == n - 1 && col == n - 1) {
        possiblePaths.push_back(path);
        return;
    }

    // Mark the cell as visited
    visited[row][col] = true;

    // Try for all the 4 directions (down, left,
    // right, up) in the given order to get the
    // paths in lexicographical order

    // Check if downward move is valid
    if (isSafe(row + 1, col, m, n, visited))
    {
        path.push_back('D');
        printPathUtil(row + 1, col, m, n,
                 path, possiblePaths, visited);
        path.pop_back();
    }

    // Check if the left move is valid
    if (isSafe(row, col - 1, m, n, visited))
    {
        path.push_back('L');
        printPathUtil(row, col - 1, m, n,
                   path, possiblePaths, visited);
        path.pop_back();
    }

    // Check if the right move is valid
    if (isSafe(row, col + 1, m, n, visited))
    {
        path.push_back('R');
        printPathUtil(row, col + 1, m, n,
                   path, possiblePaths, visited);
        path.pop_back();
    }

     // Check if the upper move is valid
    if (isSafe(row - 1, col, m, n, visited))
    {
        path.push_back('U');
        printPathUtil(row - 1, col, m, n,
               path, possiblePaths, visited);
        path.pop_back();
    }

    // Mark the cell as unvisited for
    // other possible paths
    visited[row][col] = false;
}

// Function to store and print
// all the valid paths
void printPath(int m[MAX][MAX], int n)
{
    // vector to store all the possible paths
    vector<string> possiblePaths;
    string path;
    bool visited[n][MAX];
    memset(visited, false, sizeof(visited));

    // Call the utility function to
    // find the valid paths
    printPathUtil(0, 0, m, n, path,
                      possiblePaths, visited);

    // Print all possible paths
    for (int i = 0; i < possiblePaths.size(); i++)
        cout << possiblePaths[i] << " ";
}

// Driver code
int main()
{
    int m[MAX][MAX] = { { 1, 0, 0, 0, 0 },
                        { 1, 1, 1, 1, 1 },
                        { 1, 1, 1, 0, 1 },
                        { 0, 0, 0, 0, 1 },
                        { 0, 0, 0, 0, 1 } };
    int n = sizeof(m) / sizeof(m[0]);
    printPath(m, n);

    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;

class GFG{

// Vector to store all the possible paths
static Vector<String> possiblePaths = new Vector<>();
static String path = "";
static final int MAX =  5;

// Function returns true if the
// move taken is valid else
// it will return false.
static boolean isSafe(int row, int col, int m[][],
                      int n, boolean visited[][])
{
    if (row == -1 || row == n || col == -1 ||
         col == n || visited[row][col] ||
                     m[row][col] == 0)
        return false;

    return true;
}

// Function to print all the possible
// paths from (0, 0) to (n-1, n-1).
static void printPathUtil(int row, int col, int m[][],
                          int n, boolean visited[][])
{

    // This will check the initial point
    // (i.e. (0, 0)) to start the paths.
    if (row == -1 || row == n || col == -1 ||
         col == n || visited[row][col] ||
                     m[row][col] == 0)
        return;

    // If reach the last cell (n-1, n-1)
    // then store the path and return
    if (row == n - 1 && col == n - 1)
    {
        possiblePaths.add(path);
        return;
    }

    // Mark the cell as visited
    visited[row][col] = true;

    // Try for all the 4 directions (down, left,
    // right, up) in the given order to get the
    // paths in lexicographical order

    // Check if downward move is valid
    if (isSafe(row + 1, col, m, n, visited))
    {
        path += 'D';
        printPathUtil(row + 1, col, m, n,
                      visited);
        path = path.substring(0, path.length() - 1);
    }

    // Check if the left move is valid
    if (isSafe(row, col - 1, m, n, visited))
    {
        path += 'L';
        printPathUtil(row, col - 1, m, n,
                      visited);
        path = path.substring(0, path.length() - 1);
    }

    // Check if the right move is valid
    if (isSafe(row, col + 1, m, n, visited))
    {
        path += 'R';
        printPathUtil(row, col + 1, m, n,
                      visited);
        path = path.substring(0, path.length() - 1);
    }

    // Check if the upper move is valid
    if (isSafe(row - 1, col, m, n, visited))
    {
        path += 'U';
        printPathUtil(row - 1, col, m, n,
                      visited);
        path = path.substring(0, path.length() - 1);
    }

    // Mark the cell as unvisited for
    // other possible paths
    visited[row][col] = false;
}

// Function to store and print
// all the valid paths
static void printPath(int m[][], int n)
{
    boolean [][]visited = new boolean[n][MAX];

    // Call the utility function to
    // find the valid paths
    printPathUtil(0, 0, m, n, visited);

    // Print all possible paths
    for(int i = 0; i < possiblePaths.size(); i++)
        System.out.print(possiblePaths.get(i) + " ");
}

// Driver code
public static void main(String[] args)
{
    int m[][] = { { 1, 0, 0, 0, 0 },
                  { 1, 1, 1, 1, 1 },
                  { 1, 1, 1, 0, 1 },
                  { 0, 0, 0, 0, 1 },
                  { 0, 0, 0, 0, 1 } };
    int n = m.length;

    printPath(m, n);
}
}

// This code is contributed by gauravrajput1

Python3

# Python3 implementation of the above approach
from typing import List

MAX = 5

# Function returns true if the
# move taken is valid else
# it will return false.
def isSafe(row: int, col: int,
           m: List[List[int]], n: int,
           visited: List[List[bool]]) -> bool:

    if (row == -1 or row == n or
        col == -1 or col == n or
        visited[row][col] or m[row][col] == 0):
        return False

    return True

# Function to print all the possible
# paths from (0, 0) to (n-1, n-1).
def printPathUtil(row: int, col: int,
                  m: List[List[int]],
                  n: int, path: str,
                  possiblePaths: List[str],
                  visited: List[List[bool]]) -> None:

    # This will check the initial point
    # (i.e. (0, 0)) to start the paths.
    if (row == -1 or row == n or
        col == -1 or col == n or
        visited[row][col] or m[row][col] == 0):
        return

    # If reach the last cell (n-1, n-1)
    # then store the path and return
    if (row == n - 1 and col == n - 1):
        possiblePaths.append(path)
        return

    # Mark the cell as visited
    visited[row][col] = True

    # Try for all the 4 directions (down, left,
    # right, up) in the given order to get the
    # paths in lexicographical order

    # Check if downward move is valid
    if (isSafe(row + 1, col, m, n, visited)):
        path += 'D'
        printPathUtil(row + 1, col, m, n,
                      path, possiblePaths, visited)
        path = path[:-1]

    # Check if the left move is valid
    if (isSafe(row, col - 1, m, n, visited)):
        path += 'L'
        printPathUtil(row, col - 1, m, n,
                      path, possiblePaths, visited)
        path = path[:-1]

    # Check if the right move is valid
    if (isSafe(row, col + 1, m, n, visited)):
        path += 'R'
        printPathUtil(row, col + 1, m, n,
                      path, possiblePaths, visited)
        path = path[:-1]

    # Check if the upper move is valid
    if (isSafe(row - 1, col, m, n, visited)):
        path += 'U'
        printPathUtil(row - 1, col, m, n,
                      path, possiblePaths, visited)
        path = path[:-1]

    # Mark the cell as unvisited for
    # other possible paths
    visited[row][col] = False

# Function to store and print
# all the valid paths
def printPath(m: List[List[int]], n: int) -> None:

    # vector to store all the possible paths
    possiblePaths = []
    path = ""
    visited = [[False for _ in range(MAX)]
                      for _ in range(n)]

    # Call the utility function to
    # find the valid paths
    printPathUtil(0, 0, m, n, path,
                  possiblePaths, visited)

    # Print all possible paths
    for i in range(len(possiblePaths)):
        print(possiblePaths[i], end = " ")

# Driver code
if __name__ == "__main__":

    m = [ [ 1, 0, 0, 0, 0 ],
          [ 1, 1, 1, 1, 1 ],
          [ 1, 1, 1, 0, 1 ],
          [ 0, 0, 0, 0, 1 ],
          [ 0, 0, 0, 0, 1 ] ]
    n = len(m)

    printPath(m, n)

# This code is contributed by sanjeev2552

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG{

// List to store all the possible paths
static List<String> possiblePaths = new List<String>();
static String path = "";
static readonly int MAX =  5;

// Function returns true if the
// move taken is valid else
// it will return false.
static bool isSafe(int row, int col, int [,]m,
                      int n, bool [,]visited)
{
    if (row == -1 || row == n || col == -1 ||
         col == n || visited[row,col] ||
                     m[row,col] == 0)
        return false;
    return true;
}

// Function to print all the possible
// paths from (0, 0) to (n-1, n-1).
static void printPathUtil(int row, int col, int [,]m,
                          int n, bool [,]visited)
{

    // This will check the initial point
    // (i.e. (0, 0)) to start the paths.
    if (row == -1 || row == n || col == -1 ||
         col == n || visited[row,col] ||
                     m[row,col] == 0)
        return;

    // If reach the last cell (n-1, n-1)
    // then store the path and return
    if (row == n - 1 && col == n - 1)
    {
        possiblePaths.Add(path);
        return;
    }

    // Mark the cell as visited
    visited[row,col] = true;

    // Try for all the 4 directions (down, left,
    // right, up) in the given order to get the
    // paths in lexicographical order

    // Check if downward move is valid
    if (isSafe(row + 1, col, m, n, visited))
    {
        path += 'D';
        printPathUtil(row + 1, col, m, n,
                      visited);
        path = path.Substring(0, path.Length - 1);
    }

    // Check if the left move is valid
    if (isSafe(row, col - 1, m, n, visited))
    {
        path += 'L';
        printPathUtil(row, col - 1, m, n,
                      visited);
        path = path.Substring(0, path.Length - 1);
    }

    // Check if the right move is valid
    if (isSafe(row, col + 1, m, n, visited))
    {
        path += 'R';
        printPathUtil(row, col + 1, m, n,
                      visited);
        path = path.Substring(0, path.Length - 1);
    }

    // Check if the upper move is valid
    if (isSafe(row - 1, col, m, n, visited))
    {
        path += 'U';
        printPathUtil(row - 1, col, m, n,
                      visited);
        path = path.Substring(0, path.Length - 1);
    }

    // Mark the cell as unvisited for
    // other possible paths
    visited[row,col] = false;
}

// Function to store and print
// all the valid paths
static void printPath(int [,]m, int n)
{
    bool [,]visited = new bool[n,MAX];

    // Call the utility function to
    // find the valid paths
    printPathUtil(0, 0, m, n, visited);

    // Print all possible paths
    for(int i = 0; i < possiblePaths.Count; i++)
        Console.Write(possiblePaths[i] + " ");
}

// Driver code
public static void Main(String[] args)
{
    int [,]m = { { 1, 0, 0, 0, 0 },
                  { 1, 1, 1, 1, 1 },
                  { 1, 1, 1, 0, 1 },
                  { 0, 0, 0, 0, 1 },
                  { 0, 0, 0, 0, 1 } };
    int n = m.GetLength(0);
    printPath(m, n);
}
}

// This code is contributed by gauravrajput1

Javascript

<script>

// Javascript implementation of the above approach

// Vector to store all the possible paths
let possiblePaths = [];
let path = "";
let MAX =  5;

// Function returns true if the
// move taken is valid else
// it will return false.
function isSafe(row, col, m, n, visited)
{
    if (row == -1 || row == n || col == -1 ||
         col == n || visited[row][col] ||
                           m[row][col] == 0)
        return false;

    return true;
}

// Function to print all the possible
// paths from (0, 0) to (n-1, n-1).
function printPathUtil(row, col, m, n, visited)
{

    // This will check the initial point
    // (i.e. (0, 0)) to start the paths.
    if (row == -1 || row == n || col == -1 ||
         col == n || visited[row][col] ||
                           m[row][col] == 0)
        return;

    // If reach the last cell (n-1, n-1)
    // then store the path and return
    if (row == n - 1 && col == n - 1)
    {
        possiblePaths.push(path);
        return;
    }

    // Mark the cell as visited
    visited[row][col] = true;

    // Try for all the 4 directions (down, left,
    // right, up) in the given order to get the
    // paths in lexicographical order

    // Check if downward move is valid
    if (isSafe(row + 1, col, m, n, visited))
    {
        path += 'D';
        printPathUtil(row + 1, col, m, n,
                      visited);
        path = path.substring(0, path.length - 1);
    }

    // Check if the left move is valid
    if (isSafe(row, col - 1, m, n, visited))
    {
        path += 'L';
        printPathUtil(row, col - 1, m, n,
                      visited);
        path = path.substring(0, path.length - 1);
    }

    // Check if the right move is valid
    if (isSafe(row, col + 1, m, n, visited))
    {
        path += 'R';
        printPathUtil(row, col + 1, m, n,
                      visited);
        path = path.substring(0, path.length - 1);
    }

    // Check if the upper move is valid
    if (isSafe(row - 1, col, m, n, visited))
    {
        path += 'U';
        printPathUtil(row - 1, col, m, n,
                      visited);
        path = path.substring(0, path.length - 1);
    }

    // Mark the cell as unvisited for
    // other possible paths
    visited[row][col] = false;
}

// Function to store and print
// all the valid paths
function printPath(m, n)
{
    let visited = new Array(n);
    for(let i = 0; i < n; i++)
    {
        visited[i] = new Array(MAX);
        for(let j = 0; j < MAX; j++)
            visited[i][j] = false;
    }

    // Call the utility function to
    // find the valid paths
    printPathUtil(0, 0, m, n, visited);

    // Print all possible paths
    for(let i = 0; i < possiblePaths.length; i++)
        document.write(possiblePaths[i] + " ");
}

// Driver code
let m = [ [ 1, 0, 0, 0, 0 ],
          [ 1, 1, 1, 1, 1 ],
          [ 1, 1, 1, 0, 1 ],
          [ 0, 0, 0, 0, 1 ],
          [ 0, 0, 0, 0, 1 ] ];
let n = m.length;

printPath(m, n);

// This code is contributed by patel2127

</script>

Output

DDRRURRDDD DDRURRRDDD DRDRURRDDD DRRRRDDD 

Complexity Analysis:

• Time Complexity: O(3^(n^2)). 
  As there are N^2 cells from each cell there are 3 unvisited neighboring cells. So the time complexity O(3^(N^2).
• Auxiliary Space: O(3^(n^2)). 
  As there can be at most 3^(n^2) cells in the answer so the space complexity is O(3^(n^2)).

Efficient Approach:

Instead of maintaining visited matrix, we can modify the given matrix to treat it as visited matrix.

Below is the implementation:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
#define MAX 5
using namespace std;

vector<string> res;

bool isValid(int row, int col, int m[][MAX], int n)
{
    if (row >= 0 && row < n && col >= 0 && col < n
        && m[row][col] == 1) {
        return true;
    }
    return false;
}

void findPathHelper(int m[][MAX], int n, int x, int y,
                    int dx[], int dy[], string path)
{
    if (x == n - 1 && y == n - 1) {
        res.push_back(path);
        return;
    }
    string dir = "DLRU";
    for (int i = 0; i < 4; i++) {
        int row = x + dx[i];
        int col = y + dy[i];
        if (isValid(row, col, m, n)) {
            m[row][col] = 2; // used to track visited cells
                             // of matrix
            findPathHelper(m, n, row, col, dx, dy,
                           path + dir[i]);
            m[row][col]
                = 1; // mark it unvisited yet explorable
        }
    }
}

vector<string> findPath(int m[][MAX], int n)
{
    // dx, dy will be used to follow `DLRU` exploring
    // approach which is lexicographically sorted order
    int dx[] = { 1, 0, 0, -1 };
    int dy[] = { 0, -1, 1, 0 };
    if (m[0][0] == 1) {
        m[0][0] = 2;
        findPathHelper(m, n, 0, 0, dx, dy, "");
    }
    return res;
}

int main()
{
    int m[MAX][MAX] = { { 1, 0, 0, 0, 0 },
                        { 1, 1, 1, 1, 1 },
                        { 1, 1, 1, 0, 1 },
                        { 0, 0, 0, 0, 1 },
                        { 0, 0, 0, 0, 1 } };
    int n = sizeof(m) / sizeof(m[0]);

    findPath(m, n);

    for (int i = 0; i < res.size(); ++i)
        cout << res[i] << ' ';
    return 0;
}

// This code is contributed by jainlovely450

Java

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

class GFG {

    static ArrayList<String> res;
    public static ArrayList<String> findPath(int[][] m, int n) {
        res = new ArrayList<>();
          // dx, dy will be used to follow `DLRU` exploring approach
          // which is lexicographically sorted order
        int[] dx = { 1,  0, 0, -1 };
        int[] dy = { 0, -1, 1,  0 };
        if (m[0][0] == 1) {
            m[0][0] = 2;
            findPathHelper(m, n, 0, 0, dx, dy, "");
        }
        return res;
    }

    private static void findPathHelper(int[][] m, int n, int x, int y, int[] dx, int[] dy, String path) {
        if (x == n - 1 && y == n - 1) {
            res.add(path);
            return;
        }
        String dir = "DLRU";
        for (int i = 0; i < 4; i++) {
            int row = x + dx[i];
            int col = y + dy[i];
            if (isValid(row, col, m, n)) {
                m[row][col] = 2;                // used to track visited cells of matrix
                findPathHelper(m, n, row, col, dx, dy, path + dir.charAt(i));
                m[row][col] = 1;                // mark it unvisited yet explorable
            }
        }
    }

    private static boolean isValid(int row, int col, int[][] m, int n) {
        if (row >= 0 && row < n && col >= 0 && col < n && m[row][col] == 1) {
            return true;
        }
        return false;
    }

    public static void main(String[] args) {
        int m[][] = { { 1, 0, 0, 0, 0 },
                      { 1, 1, 1, 1, 1 },
                      { 1, 1, 1, 0, 1 },
                      { 0, 0, 0, 0, 1 },
                      { 0, 0, 0, 0, 1 } };
        int n = m.length;

        ArrayList<String> ans = findPath(m, n);
        System.out.println(ans);
    }
}

// Contributed by: jmamtora99

Python3

# Python code for the approach
res = []

def isValid(row, col, m, n):

    if (row >= 0 and row < n and col >= 0 and col < n and m[row][col] == 1):
        return True

    return False


def findPathHelper(m, n, x, y, dx, dy, path):

    global res

    if (x == n - 1 and y == n - 1):
        res.append(path)
        return

    dir = "DLRU"
    for i in range(4):
        row = x + dx[i]
        col = y + dy[i]
        if (isValid(row, col, m, n)):
            m[row][col] = 2             # used to track visited cells of matrix
            findPathHelper(m, n, row, col, dx, dy, path + dir[i])
            m[row][col] = 1             # mark it unvisited yet explorable

def findPath(m,n):
    global res

    res.clear()

    # dx, dy will be used to follow `DLRU` exploring approach
    # which is lexicographically sorted order
    dx = [ 1, 0, 0, -1 ]
    dy = [ 0, -1, 1, 0 ]
    if (m[0][0] == 1):
        m[0][0] = 2
        findPathHelper(m, n, 0, 0, dx, dy, "")

    return res

# driver code
m = [ [ 1, 0, 0, 0, 0 ],
      [ 1, 1, 1, 1, 1 ],
      [ 1, 1, 1, 0, 1 ],
      [ 0, 0, 0, 0, 1 ],
      [ 0, 0, 0, 0, 1 ] ]
n = len(m)

ans = findPath(m, n)
print(ans)

# This code is contributed by shinjanpatra

Javascript

<script>

// JavaScript implementation of the above approach
const MAX = 5;
let res = [];

function isValid(row, col, m, n)
{
    if (row >= 0 && row < n && col >= 0 && col < n
        && m[row][col] == 1) {
        return true;
    }
    return false;
}

function findPathHelper(m, n, x, y, dx, dy, path)
{
    if (x == n - 1 && y == n - 1) {
        res.push(path);
        return;
    }
    let dir = "DLRU";
    for (let i = 0; i < 4; i++) {
        let row = x + dx[i];
        let col = y + dy[i];
        if (isValid(row, col, m, n)) {
            m[row][col] = 2; // used to track visited cells
                             // of matrix
            findPathHelper(m, n, row, col, dx, dy,
                           path + dir[i]);
            m[row][col] = 1; // mark it unvisited yet explorable
        }
    }
}

function findPath(m, n)
{

    // dx, dy will be used to follow `DLRU` exploring
    // approach which is lexicographically sorted order
    let dx = [ 1, 0, 0, -1 ];
    let dy = [ 0, -1, 1, 0 ];
    if (m[0][0] == 1) {
        m[0][0] = 2;
        findPathHelper(m, n, 0, 0, dx, dy, "");
    }
    return res;
}

// driver code

let m = [ [ 1, 0, 0, 0, 0 ],
          [ 1, 1, 1, 1, 1 ],
          [ 1, 1, 1, 0, 1 ],
          [ 0, 0, 0, 0, 1 ],
          [ 0, 0, 0, 0, 1 ] ];

let n = m.length;
findPath(m, n);

for (let i = 0; i < res.length; ++i)
    document.write(res[i],' ');

// This code is contributed by shinjanpatra

</script>

Output

DDRRURRDDD DDRURRRDDD DRDRURRDDD DRRRRDDD 

Complexity Analysis: 

• Time Complexity: O(3^(n^2)). 
• Space Complexity: O(1); Since we are not using visited matrix.