Rat in a Maze

Last Updated : 18 May, 2026

Given a square binary matrix mat[][] representing a maze. Find all possible paths from (0, 0) to (n-1, n-1).

  • A rat starts at the top-left corner (0,0) and needs to reach the bottom-right corner (n-1, n-1).
  • The rat can move in four directions: Up (U), Down (D), Left (L), Right (R).
  • A rat cannot visit the same cell more than once in a path.
  • 1 represents an open cell (rat can visit), and 0 represents a blocked cell (rat cannot visit).
  • If multiple paths exist, return them in lexicographically sorted order otherwise If no path exists, return empty list.

Example:

Input: mat[][] = [[1, 0, 0, 0], [1, 1, 0, 1], [1, 1, 0, 0], [0, 1, 1, 1]]

1

Output: ["DDRDRR", "DRDDRR"]
Explanation: The possible paths are: DDRDRR, DRDDRR

2
Try It Yourself
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[Approach] Using Recursion and Backtracking

The idea is to explore all possible paths from the source to the destination in the maze. We recursively build each path, store it when the destination is reached, and backtrack to explore alternative routes. To avoid revisiting cells in the same path, we mark a cell as visited during exploration and unmark it during backtracking.

C++ 
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

// Directions: Down, Left, Right, Up
string dir = "DLRU";
int dr[4] = {1, 0, 0, -1};
int dc[4] = {0, -1, 1, 0};

// Check if a cell is valid (inside the maze and open)
bool isValid(int r, int c, int n, vector<vector<int>>& maze) {
    return r >= 0 && c >= 0 && r < n && c < n && maze[r][c];
}

// Function to find all valid paths
void findPath(int r, int c, vector<vector<int>>& maze, string& path,
                                vector<string>& res) {
    int n = maze.size(); 

    // If destination is reached, store the path
    if (r == n - 1 && c == n - 1) {
        res.push_back(path);
        return;
    }
    
    // Mark current cell as visited
    maze[r][c] = 0; 

    for (int i = 0; i < 4; i++) {
        int nr = r + dr[i], nc = c + dc[i];
        if (isValid(nr, nc, n, maze)) {
            path.push_back(dir[i]);
            
            // Move to the next cell recursively
            findPath(nr, nc, maze, path, res); 
            
            // Backtrack
            path.pop_back();
        }
    }
    
    // Unmark current cell
    maze[r][c] = 1;  
}

// Function to find all paths and return them
vector<string> ratInMaze(vector<vector<int>>& maze) {
    vector<string> result;
    int n = maze.size();
    string path = "";

    if (maze[0][0] == 1 && maze[n - 1][n - 1] == 1) {
        
        // Start from (0,0)
        findPath(0, 0, maze, path, result);  
    }

    // Sort results lexicographically
    sort(result.begin(), result.end());
    
    return result;
}

int main() {
    vector<vector<int>> maze = {
        {1, 0, 0, 0},
        {1, 1, 0, 1},
        {1, 1, 0, 0},
        {0, 1, 1, 1}
    };

    vector<string> result = ratInMaze(maze);

    for (auto p : result) {
        cout << p << " ";
    }
    
    return 0;
}
Java 
import java.util.ArrayList;
import java.util.Collections;

class GFG {

    // Directions: Down, Left, Right, Up
    static String dir = "DLRU";
    static int[] dr = {1, 0, 0, -1};
    static int[] dc = {0, -1, 1, 0};

    // Check if a cell is valid (inside the maze and open)
    static boolean isValid(int r, int c, int n, int[][] maze) {
        return r >= 0 && c >= 0 && r < n && c < n && maze[r][c] == 1;
    }

    // Function to find all valid paths
    static void findPath(int r, int c, int[][] maze, StringBuilder path,
                                    ArrayList<String> res) {
        int n = maze.length; 

        // If destination is reached, store the path
        if (r == n - 1 && c == n - 1) {
            res.add(path.toString());
            return;
        }
        
        // Mark current cell as visited
        maze[r][c] = 0; 

        for (int i = 0; i < 4; i++) {
            int nr = r + dr[i], nc = c + dc[i];
            if (isValid(nr, nc, n, maze)) {
                path.append(dir.charAt(i));
                
                // Move to the next cell recursively
                findPath(nr, nc, maze, path, res); 
                
                // Backtrack
                path.deleteCharAt(path.length() - 1);
            }
        }
        
        // Unmark current cell
        maze[r][c] = 1;  
    }

    // Function to find all paths and return them
    static ArrayList<String> ratInMaze(int[][] maze) {
        ArrayList<String> result = new ArrayList<>();
        int n = maze.length;
        StringBuilder path = new StringBuilder();

        if (maze[0][0] == 1 && maze[n - 1][n - 1] == 1) {
            
            // Start from (0,0)
            findPath(0, 0, maze, path, result);  
        }

        // Sort results lexicographically
        Collections.sort(result);
        
        return result;
    }

    public static void main(String[] args) {
        int[][] maze = {
            {1, 0, 0, 0},
            {1, 1, 0, 1},
            {1, 1, 0, 0},
            {0, 1, 1, 1}
        };

        ArrayList<String> result = ratInMaze(maze);

        for (String p : result) {
            System.out.print(p + " ");
        }
    }
}
Python 
# Directions: Down, Left, Right, Up
dir = "DLRU"
dr = [1, 0, 0, -1]
dc = [0, -1, 1, 0]

# Check if a cell is valid (inside the maze and open)
def isValid(r, c, n, maze):
    return r >= 0 and c >= 0 and r < n and c < n and maze[r][c] == 1

# Function to find all valid paths
def findPath(r, c, maze, path, res):
    n = len(maze)

    # If destination is reached, store the path
    if r == n - 1 and c == n - 1:
        res.append("".join(path))
        return
    
    # Mark current cell as visited
    maze[r][c] = 0

    for i in range(4):
        nr, nc = r + dr[i], c + dc[i]
        if isValid(nr, nc, n, maze):
            path.append(dir[i])
            
            # Move to the next cell recursively
            findPath(nr, nc, maze, path, res)
            
            # Backtrack
            path.pop()
    
    # Unmark current cell
    maze[r][c] = 1

# Function to find all paths and return them
def ratInMaze(maze):
    result = []
    n = len(maze)
    path = []

    if maze[0][0] == 1 and maze[n - 1][n - 1] == 1:
        
        # Start from (0,0)
        findPath(0, 0, maze, path, result)

    # Sort results lexicographically
    result.sort()
        
    return result

if __name__ == "__main__":
    maze = [
        [1, 0, 0, 0],
        [1, 1, 0, 1],
        [1, 1, 0, 0],
        [0, 1, 1, 1]
    ]

    result = ratInMaze(maze)

    for p in result:
        print(p, end=" ")
C# 
using System;
using System.Collections.Generic;

class GFG {
    
    // Directions: Down, Left, Right, Up
    static string dir = "DLRU";
    static int[] dr = { 1, 0, 0, -1 };
    static int[] dc = { 0, -1, 1, 0 };

    // Check if a cell is valid (inside the maze and open)
    static bool isValid(int r, int c, int n, int[,] maze) {
        return r >= 0 && c >= 0 && r < n && c < n && maze[r, c] == 1;
    }

    // Function to find all valid paths
    static void findPath(int r, int c, int[,] maze, List<char> path, List<string> res) {
        int n = maze.GetLength(0);

        // If destination is reached, store the path
        if (r == n - 1 && c == n - 1) {
            res.Add(new string(path.ToArray()));
            return;
        }

        // Mark current cell as visited
        maze[r, c] = 0;

        for (int i = 0; i < 4; i++) {
            int nr = r + dr[i], nc = c + dc[i];
            if (isValid(nr, nc, n, maze)) {
                path.Add(dir[i]);

                // Move to the next cell recursively
                findPath(nr, nc, maze, path, res);

                // Backtrack
                path.RemoveAt(path.Count - 1);
            }
        }

        // Unmark current cell
        maze[r, c] = 1;
    }

    // Function to find all paths and return them
    static List<string> ratInMaze(int[,] maze) {
        List<string> result = new List<string>();
        int n = maze.GetLength(0);
        List<char> path = new List<char>();

        if (maze[0, 0] == 1 && maze[n - 1, n - 1] == 1)
        {
            // Start from (0,0)
            findPath(0, 0, maze, path, result);
        }

        // Sort results lexicographically
        result.Sort();
    
        return result;
    }

    public static void Main(string[] args) {
        int[,] maze = {
            {1, 0, 0, 0},
            {1, 1, 0, 1},
            {1, 1, 0, 0},
            {0, 1, 1, 1}
        };

        List<string> result = ratInMaze(maze);

        foreach (string p in result) {
            Console.Write(p + " ");
        }
    }
}
JavaScript 
// Directions: Down, Left, Right, Up
const dir = "DLRU";
const dr = [1, 0, 0, -1];
const dc = [0, -1, 1, 0];

// Check if a cell is valid (inside the maze and open)
function isValid(r, c, n, maze) {
    return r >= 0 && c >= 0 && r < n && c < n && maze[r][c] === 1;
}

// Function to find all valid paths
function findPath(r, c, maze, path, res) {
    const n = maze.length;

    // If destination is reached, store the path
    if (r === n - 1 && c === n - 1) {
        res.push(path);
        return;
    }

    // Mark current cell as visited
    maze[r][c] = 0;

    for (let i = 0; i < 4; i++) {
        const nr = r + dr[i], nc = c + dc[i];
        if (isValid(nr, nc, n, maze)) {
            path += dir[i];

            // Move to the next cell recursively
            findPath(nr, nc, maze, path, res);

            // Backtrack
            path = path.slice(0, -1);
        }
    }

    // Unmark current cell
    maze[r][c] = 1;
}

// Function to find all paths and return them
function ratInMaze(maze) {
    let result = [];
    const n = maze.length;
    let path = "";

    if (maze[0][0] === 1 && maze[n - 1][n - 1] === 1) {

        // Start from (0,0)
        findPath(0, 0, maze, path, result);
    }

    // Sort results lexicographically
    result.sort();
    
    return result;
}

// Driver Code
const maze = [
    [1, 0, 0, 0],
    [1, 1, 0, 1],
    [1, 1, 0, 0],
    [0, 1, 1, 1]
];

const result = ratInMaze(maze);

console.log(result.join(" "));

Output
DDRDRR DRDDRR

Time Complexity: O(4n*n), because on every cell we have to try 4 different directions.
Auxiliary Space: O(n2), Maximum Depth of the recursion tree.

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