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

Last Updated : 15 Jul, 2024

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; }

Java

import java.util.*;  public class Main {     static final int MAX = 5;      static void getallpath(int[][] matrix, int n, int row,                            int col, List<String> ans,                            String cur)     {         if (row >= n || col >= n || row < 0 || col < 0             || matrix[row][col] == 0) {             return;         }          if (row == n - 1 && col == n - 1) {             ans.add(cur);             return;         }          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;     }      static List<String> findPath(int[][] matrix, int n)     {         List<String> ans = new ArrayList<>();         getallpath(matrix, n, 0, 0, ans, "");         return ans;     }      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;         List<String> ans = findPath(m, n);         for (String i : ans) {             System.out.println(i + " ");         }     } }  // This code is contributed by abn95knd1.

Python

# Python3 implementation of the above approach def getallpath(mat, n, ans, curpath, r, c):         if(r>=n or c>=n or r<0 or c<0 or mat[r][c] == 0):             return;                  if(r == n-1 and c == n-1):             ans.append(curpath)             return         mat[r][c] = 0                  # calls         getallpath(mat,n,ans,curpath+"U",r-1,c)         getallpath(mat,n,ans,curpath+"D",r+1,c)         getallpath(mat,n,ans,curpath+"L",r,c-1)         getallpath(mat,n,ans,curpath+"R",r,c+1)                  mat[r][c] = 1;  # Function to store and print # all the valid paths def findpath(mat ,n):      # vector to store all the possible paths     possiblePaths = []     getallpath(mat,n,possiblePaths,"",0,0)     # Print all possible paths     for i in range(len(possiblePaths)):         print(possiblePaths[i], end = " ")  # Driver code if __name__ == "__main__":          mat = [ [ 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(mat)          findpath(mat, n)  # This code is contributed by Arpit Jain

using System; using System.Collections.Generic;  namespace Algorithm {     class MainClass     {         static readonly int MAX = 5;          static void getallpath(int[,] matrix, int n, int row,                                int col, List<string> ans,                                string cur)         {             if (row >= n || col >= n || row < 0 || col < 0                 || matrix[row, col] == 0)             {                 return;             }              if (row == n - 1 && col == n - 1)             {                 ans.Add(cur);                 return;             }              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;         }          static List<string> findPath(int[,] matrix, int n)         {             List<string> ans = new List<string>();             getallpath(matrix, n, 0, 0, ans, "");             return ans;         }          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);             List<string> ans = findPath(m, n);             foreach (string i in ans)             {                 Console.Write(i + " ");             }         }     } }

JavaScript

// The equivalent JavaScript code is as follows:  const MAX = 5;  function getallpath(matrix, n, row, col, ans, cur) {   if (row >= n || col >= n || row < 0 || col < 0 || matrix[row][col] === 0) {     return;   }    if (row === n - 1 && col === n - 1) {     ans.push(cur);     return;   }    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; }  function findPath(matrix, n) {   let ans = [];   getallpath(matrix, n, 0, 0, ans, "");   return ans; }  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; let ans = findPath(m, n); for (let i of ans) {   console.log(i + " "); }

Output

DRRRRDDD DRDRURRDDD DDRURRRDDD DDRRURRDDD

Complexity Analysis:

Time Complexity: O(3^(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.

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

Rishav Saxena

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

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