# Rat in a Maze

Last Updated : 08 Apr, 2024

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.

Consider a rat placed atÂ (0, 0)Â in a square matrixÂ of orderÂ N * N. It has to reach the destination atÂ (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 areÂ ‘U'(up),Â ‘D'(down),Â ‘L’ (left),Â ‘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 thatÂ ratÂ can be travelÂ through it. Return the list ofÂ paths in lexicographically increasing order.
Note: In a path, no cell can be visited more than one time.Â If the source cell isÂ 0, the rat cannot move to any other cell.

Example:

Input:

Output: DRDDRR
Explanation:

## Rat in a Maze using Backtracking:

We use a backtracking algorithm to explore all possible paths. While exploring the paths we keep track of the directions we have moved so far and when we reach to the bottom right cell, we record the path in a vector of strings.

Step-by-step approach:

• Create isValid() function to check if a cell at position (row, col) is inside the maze and unblocked.
• Create findPath() to get all valid paths:
• Base case: If the current position is the bottom-right cell, add the current path to the result and return.
• Mark the current cell as blocked.
• Iterate through all possible directions.
• Calculate the next position based on the current direction.
• If the next position is valid (i.e, if isValid() return true), append the direction to the current path and recursively call the findPath() function for the next cell.
• Backtrack by removing the last direction from the current path.
• Mark the current cell as unblocked before returning.

Below is the implementation of the above approach:

C++ ```#include <bits/stdc++.h> using namespace std; // Initialize a string direction which represents all the // directions. string direction = "DLRU"; // Arrays to represent change in rows and columns int dr[4] = { 1, 0, 0, -1 }; int dc[4] = { 0, -1, 1, 0 }; // Function to check if cell(row, col) is inside the maze // and unblocked bool isValid(int row, int col, int n, vector<vector<int> >& maze) { return row >= 0 && col >= 0 && row < n && col < n && maze[row][col]; } // Function to get all valid paths void findPath(int row, int col, vector<vector<int> >& maze, int n, vector<string>& ans, string& currentPath) { // If we reach the bottom right cell of the matrix, add // the current path to ans and return if (row == n - 1 && col == n - 1) { ans.push_back(currentPath); return; } // Mark the current cell as blocked maze[row][col] = 0; for (int i = 0; i < 4; i++) { // Find the next row based on the current row (row) // and the dr[] array int nextrow = row + dr[i]; // Find the next column based on the current column // (col) and the dc[] array int nextcol = col + dc[i]; // Check if the next cell is valid or not if (isValid(nextrow, nextcol, n, maze)) { currentPath += direction[i]; // Recursively call the FindPath function for // the next cell findPath(nextrow, nextcol, maze, n, ans, currentPath); // Remove the last direction when backtracking currentPath.pop_back(); } } // Mark the current cell as unblocked maze[row][col] = 1; } int main() { vector<vector<int> > maze = { { 1, 0, 0, 0 }, { 1, 1, 0, 1 }, { 1, 1, 0, 0 }, { 0, 1, 1, 1 } }; int n = maze.size(); // vector to store all the valid paths vector<string> result; // Store current path string currentPath = ""; if (maze[0][0] != 0 && maze[n - 1][n - 1] != 0) { // Function call to get all valid paths findPath(0, 0, maze, n, result, currentPath); } if (result.size() == 0) cout << -1; else for (int i = 0; i < result.size(); i++) cout << result[i] << " "; cout << endl; return 0; } ``` Java ```import java.util.ArrayList; import java.util.List; public class MazePaths { // Initialize a string direction which represents all // the directions. static String direction = "DLRU"; // Arrays to represent change in rows and columns static int[] dr = { 1, 0, 0, -1 }; static int[] dc = { 0, -1, 1, 0 }; // Function to check if cell(row, col) is inside the // maze and unblocked static boolean isValid(int row, int col, int n, int[][] maze) { return row >= 0 && col >= 0 && row < n && col < n && maze[row][col] == 1; } // Function to get all valid paths static void findPath(int row, int col, int[][] maze, int n, ArrayList<String> ans, StringBuilder currentPath) { // If we reach the bottom right cell of the matrix, // add the current path to ans and return if (row == n - 1 && col == n - 1) { ans.add(currentPath.toString()); return; } // Mark the current cell as blocked maze[row][col] = 0; for (int i = 0; i < 4; i++) { // Find the next row based on the current row // (row) and the dr[] array int nextrow = row + dr[i]; // Find the next column based on the current // column (col) and the dc[] array int nextcol = col + dc[i]; // Check if the next cell is valid or not if (isValid(nextrow, nextcol, n, maze)) { currentPath.append(direction.charAt(i)); // Recursively call the FindPath function // for the next cell findPath(nextrow, nextcol, maze, n, ans, currentPath); // Remove the last direction when // backtracking currentPath.deleteCharAt( currentPath.length() - 1); } } // Mark the current cell as unblocked maze[row][col] = 1; } public static void main(String[] args) { int[][] maze = { { 1, 0, 0, 0 }, { 1, 1, 0, 1 }, { 1, 1, 0, 0 }, { 0, 1, 1, 1 } }; int n = maze.length; // List to store all the valid paths ArrayList<String> result = new ArrayList<>(); // Store current path StringBuilder currentPath = new StringBuilder(); if (maze[0][0] != 0 && maze[n - 1][n - 1] != 0) { // Function call to get all valid paths findPath(0, 0, maze, n, result, currentPath); } if (result.size() == 0) System.out.println(-1); else for (String path : result) System.out.print(path + " "); System.out.println(); } } ``` Python3 ```# Initialize a string direction which represents all the directions. direction = "DLRU" # Arrays to represent change in rows and columns dr = [1, 0, 0, -1] dc = [0, -1, 1, 0] # Function to check if cell(row, col) is inside the maze # and unblocked def is_valid(row, col, n, maze): return 0 <= row < n and 0 <= col < n and maze[row][col] == 1 # Function to get all valid paths def find_path(row, col, maze, n, ans, current_path): # If we reach the bottom right cell of the matrix, add # the current path to ans and return if row == n - 1 and col == n - 1: ans.append(current_path) return # Mark the current cell as blocked maze[row][col] = 0 for i in range(4): # Find the next row based on the current row (row) # and the dr[] array next_row = row + dr[i] # Find the next column based on the current column # (col) and the dc[] array next_col = col + dc[i] # Check if the next cell is valid or not if is_valid(next_row, next_col, n, maze): current_path += direction[i] # Recursively call the find_path function for # the next cell find_path(next_row, next_col, maze, n, ans, current_path) # Remove the last direction when backtracking current_path = current_path[:-1] # Mark the current cell as unblocked maze[row][col] = 1 # Driver code maze = [ [1, 0, 0, 0], [1, 1, 0, 1], [1, 1, 0, 0], [0, 1, 1, 1] ] n = len(maze) # List to store all the valid paths result = [] # Store current path current_path = "" if maze[0][0] != 0 and maze[n - 1][n - 1] != 0: # Function call to get all valid paths find_path(0, 0, maze, n, result, current_path) if not result: print(-1) else: print(" ".join(result)) ``` C# ```using System; using System.Collections.Generic; public class Program { // Initialize an array to represent the change in rows and columns static int[] dr = { 1, 0, 0, -1 }; static int[] dc = { 0, -1, 1, 0 }; static string direction = "DLRU"; // Function to check if cell (row, col) is inside the maze and unblocked static bool IsValid(int row, int col, int n, int[,] maze) { return row >= 0 && col >= 0 && row < n && col < n && maze[row, col] == 1; } // Function to get all valid paths static void FindPath(int row, int col, int[,] maze, int n, List<string> ans, ref string currentPath) { // If we reach the bottom right cell of the matrix, add the current path to ans and return if (row == n - 1 && col == n - 1) { ans.Add(currentPath); return; } // Mark the current cell as blocked maze[row, col] = 0; for (int i = 0; i < 4; i++) { // Find the next row based on the current row (row) and the dr[] array int nextrow = row + dr[i]; // Find the next column based on the current column (col) and the dc[] array int nextcol = col + dc[i]; // Check if the next cell is valid or not if (IsValid(nextrow, nextcol, n, maze)) { currentPath += direction[i]; // Recursively call the FindPath function for the next cell FindPath(nextrow, nextcol, maze, n, ans, ref currentPath); // Remove the last direction when backtracking currentPath = currentPath.Remove(currentPath.Length - 1); } } // Mark the current cell as unblocked maze[row, col] = 1; } public static void Main() { int[,] maze = { { 1, 0, 0, 0 }, { 1, 1, 0, 1 }, { 1, 1, 0, 0 }, { 0, 1, 1, 1 } }; int n = maze.GetLength(0); // List to store all the valid paths List<string> result = new List<string>(); // Store current path string currentPath = ""; if (maze[0, 0] != 0 && maze[n - 1, n - 1] != 0) { // Function call to get all valid paths FindPath(0, 0, maze, n, result, ref currentPath); } if (result.Count == 0) Console.WriteLine(-1); else foreach (string path in result) Console.Write(path + " "); Console.WriteLine(); } } ``` JavaScript ```// Initialize a string direction which represents all the directions. const direction = "DLRU"; // Arrays to represent change in rows and columns const dr = [1, 0, 0, -1]; const dc = [0, -1, 1, 0]; // Function to check if cell(row, col) is inside the maze // and unblocked function isValid(row, col, n, maze) { return row >= 0 && col >= 0 && row < n && col < n && maze[row][col] === 1; } // Function to get all valid paths function findPath(row, col, maze, n, ans, currentPath) { // If we reach the bottom right cell of the matrix, add // the current path to ans and return if (row === n - 1 && col === n - 1) { ans.push(currentPath); return; } // Mark the current cell as blocked maze[row][col] = 0; for (let i = 0; i < 4; i++) { // Find the next row based on the current row (row) // and the dr[] array const nextrow = row + dr[i]; // Find the next column based on the current column // (col) and the dc[] array const nextcol = col + dc[i]; // Check if the next cell is valid or not if (isValid(nextrow, nextcol, n, maze)) { currentPath += direction[i]; // Recursively call the findPath function for // the next cell findPath(nextrow, nextcol, maze, n, ans, currentPath); // Remove the last direction when backtracking currentPath = currentPath.slice(0, -1); } } // Mark the current cell as unblocked maze[row][col] = 1; } // Driver code const maze = [ [1, 0, 0, 0], [1, 1, 0, 1], [1, 1, 0, 0], [0, 1, 1, 1] ]; const n = maze.length; // Array to store all the valid paths const result = []; // Store current path let currentPath = ""; if (maze[0][0] !== 0 && maze[n - 1][n - 1] !== 0) { // Function call to get all valid paths findPath(0, 0, maze, n, result, currentPath); } if (result.length === 0) console.log(-1); else console.log(result.join(" ")); ```

Output
```DDRDRR DRDDRR
```

Time Complexity: O(3^(m*n)), because on every cell we have to try 3 different directions.
Auxiliary Space: O(m*n), Maximum Depth of the recursion tree(auxiliary space).

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