Rat in a Maze with multiple steps or jump allowed

This is the variation of Rat in Maze

A Maze is given as N*N binary matrix of blocks where source block is the upper left most block i.e., maze[0][0] and destination block is lower rightmost block i.e., maze[N-1][N-1]. A rat starts from source and has to reach destination. The rat can move only in two directions: forward and down.

In the maze matrix, 0 means the block is dead end and non-zero number means the block can be used in the path from source to destination. The non-zero value of mat[i][j] indicates number of maximum jumps rat can make from cell mat[i][j].

In this variation, Rat is allowed to jump multiple steps at a time instead of 1.

Examples:

Input : { {2, 1, 0, 0},
         {3, 0, 0, 1},
         {0, 1, 0, 1},
          {0, 0, 0, 1}
        }
Output : { {1, 0, 0, 0},
           {1, 0, 0, 1},
           {0, 0, 0, 1},
           {0, 0, 0, 1}
         }

Explanation 
Rat started with M[0][0] and can jump upto 2 steps right/down. 
Let's try in horizontal direction - 
M[0][1] won't lead to solution and M[0][2] is 0 which is dead end. 
So, backtrack and try in down direction. 
Rat jump down to M[1][0] which eventually leads to solution.  

Input : { 
      {2, 1, 0, 0},
      {2, 0, 0, 1},
      {0, 1, 0, 1},
      {0, 0, 0, 1}
    }
Output : Solution doesn't exist

Naive Algorithm:

The Naive Algorithm is to generate all paths from source to destination and one by one check if the generated path satisfies the constraints.

while there are untried paths
{
   generate the next path
   if this path has all blocks as non-zero
   {
      print this path;
   }
}

Backtracking Algorithm:

If destination is reached
    print the solution matrix
Else
   a) Mark current cell in solution matrix as 1. 
   b) Move forward/jump (for each valid steps) in horizontal direction 
      and recursively check if this move leads to a solution. 
   c) If the move chosen in the above step doesn't lead to a solution
       then move down and check if this move leads to a solution. 
   d) If none of the above solutions work then unmark this cell as 0 
       (BACKTRACK) and return false.

Implementation of Backtracking solution

C++

/* C/C++ program to solve Rat in a Maze problem
using backtracking */
#include <stdio.h>
 
// Maze size
#define N 4
 
bool solveMazeUtil(int maze[N][N], int x, int y,
                                int sol[N][N]);
 
/* A utility function to print solution matrix
sol[N][N] */
void printSolution(int sol[N][N])
{
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++)
            printf(" %d ", sol[i][j]);
        printf("\n");
    }
}
 
/* A utility function to check if x, y is valid
index for N*N maze */
bool isSafe(int maze[N][N], int x, int y)
{
    // if (x, y outside maze) return false
    if (x >= 0 && x < N && y >= 0 &&
    y < N && maze[x][y] != 0)
        return true;
 
    return false;
}
 
/* This function solves the Maze problem using
Backtracking. It mainly uses solveMazeUtil() to
solve the problem. It returns false if no path
is possible, otherwise return true and prints
the path in the form of 1s. Please note that
there may be more than one solutions,
this function prints one of the feasible solutions.*/
bool solveMaze(int maze[N][N])
{
    int sol[N][N] = { { 0, 0, 0, 0 },
                    { 0, 0, 0, 0 },
                    { 0, 0, 0, 0 },
                    { 0, 0, 0, 0 } };
 
    if (solveMazeUtil(maze, 0, 0, sol) == false) {
        printf("Solution doesn't exist");
        return false;
    }
 
    printSolution(sol);
    return true;
}
 
/* A recursive utility function to solve Maze problem */
bool solveMazeUtil(int maze[N][N], int x, int y,
                                int sol[N][N])
{
    // if (x, y is goal) return true
    if (x == N - 1 && y == N - 1) {
        sol[x][y] = 1;
        return true;
    }
 
    // Check if maze[x][y] is valid
    if (isSafe(maze, x, y) == true) {
 
        // mark x, y as part of solution path
        sol[x][y] = 1;
 
        /* Move forward in x direction */
        for (int i = 1; i <= maze[x][y] && i < N; i++) {
 
            /* Move forward in x direction */
            if (solveMazeUtil(maze, x + i, y, sol) == true)
                return true;
 
            /* If moving in x direction doesn't give
            solution then Move down in y direction */
            if (solveMazeUtil(maze, x, y + i, sol) == true)
                return true;
        }
 
        /* If none of the above movements work then
        BACKTRACK: unmark x, y as part of solution
        path */
        sol[x][y] = 0;
        return false;
    }
 
    return false;
}
 
// driver program to test above function
int main()
{
    int maze[N][N] = { { 2, 1, 0, 0 },
                    { 3, 0, 0, 1 },
                    { 0, 1, 0, 1 },
                    { 0, 0, 0, 1 } };
 
    solveMaze(maze);
    return 0;
}

Java

// Java program to solve Rat in a Maze problem
// using backtracking
class GFG
{
 
    // Maze size
    static int N = 4;
 
    /* A utility function to print solution matrix
    sol[N][N] */
    static void printSolution(int sol[][])
    {
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
            {
                System.out.printf(" %d ", sol[i][j]);
            }
            System.out.printf("\n");
        }
    }
 
    /* A utility function to check if x, y is valid
    index for N*N maze */
    static boolean isSafe(int maze[][], int x, int y)
    {
         
        // if (x, y outside maze) return false
        if (x >= 0 && x < N && y >= 0 &&
             y < N && maze[x][y] != 0)
        {
            return true;
        }
 
        return false;
    }
 
    /* This function solves the Maze problem using
    Backtracking. It mainly uses solveMazeUtil() to
    solve the problem. It returns false if no path
    is possible, otherwise return true and prints
    the path in the form of 1s. Please note that
    there may be more than one solutions,
    this function prints one of the feasible solutions.*/
    static boolean solveMaze(int maze[][])
    {
        int sol[][] = {{0, 0, 0, 0},
                       {0, 0, 0, 0},
                       {0, 0, 0, 0},
                       {0, 0, 0, 0}};
 
        if (solveMazeUtil(maze, 0, 0, sol) == false)
        {
            System.out.printf("Solution doesn't exist");
            return false;
        }
 
        printSolution(sol);
        return true;
    }
 
    /* A recursive utility function to solve Maze problem */
    static boolean solveMazeUtil(int maze[][], int x,
                                 int y, int sol[][])
    {
        // if (x, y is goal) return true
        if (x == N - 1 && y == N - 1)
        {
            sol[x][y] = 1;
            return true;
        }
 
        // Check if maze[x][y] is valid
        if (isSafe(maze, x, y) == true)
        {
 
            // mark x, y as part of solution path
            sol[x][y] = 1;
 
            /* Move forward in x direction */
            for (int i = 1; i <= maze[x][y] && i < N; i++)
            {
 
                /* Move forward in x direction */
                if (solveMazeUtil(maze, x + i, y, sol) == true)
                {
                    return true;
                }
 
                /* If moving in x direction doesn't give
                solution then Move down in y direction */
                if (solveMazeUtil(maze, x, y + i, sol) == true)
                {
                    return true;
                }
            }
 
            /* If none of the above movements work then
            BACKTRACK: unmark x, y as part of solution
            path */
            sol[x][y] = 0;
            return false;
        }
 
        return false;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int maze[][] = {{2, 1, 0, 0},
                        {3, 0, 0, 1},
                        {0, 1, 0, 1},
                        {0, 0, 0, 1}};
 
        solveMaze(maze);
    }
}
 
// This code is contributed by Princi Singh

Python3

""" Python3 program to solve Rat in a
Maze problem using backtracking """
 
# Maze size
N = 4
 
""" A utility function to print solution matrix
sol """
def printSolution(sol):
    for i in range(N):
        for j in range(N):
            print(sol[i][j], end = " ")
        print()
         
""" A utility function to check if
x, y is valid index for N*N maze """
def isSafe(maze, x, y):
     
    # if (x, y outside maze) return false
    if (x >= 0 and x < N and y >= 0 and
         y < N and maze[x][y] != 0):
        return True
    return False
 
""" This function solves the Maze problem using
Backtracking. It mainly uses solveMazeUtil() to
solve the problem. It returns false if no path
is possible, otherwise return True and prints
the path in the form of 1s. Please note that
there may be more than one solutions,
this function prints one of the feasible solutions."""
def solveMaze(maze):
    sol = [[0, 0, 0, 0],
           [0, 0, 0, 0],
           [0, 0, 0, 0],
           [0, 0, 0, 0]]
    if (solveMazeUtil(maze, 0, 0, sol) == False):
        print("Solution doesn't exist")
        return False
    printSolution(sol)
    return True
     
""" A recursive utility function
to solve Maze problem """
def solveMazeUtil(maze, x, y, sol):
     
    # if (x, y is goal) return True
    if (x == N - 1 and y == N - 1) :
        sol[x][y] = 1
        return True
         
    # Check if maze[x][y] is valid
    if (isSafe(maze, x, y) == True):
         
        # mark x, y as part of solution path
        sol[x][y] = 1
         
        """ Move forward in x direction """
        for i in range(1, N):
            if (i <= maze[x][y]):
                 
                """ Move forward in x direction """
                if (solveMazeUtil(maze, x + i,
                                  y, sol) == True):
                    return True
                     
                """ If moving in x direction doesn't give
                solution then Move down in y direction """
                if (solveMazeUtil(maze, x,
                                  y + i, sol) == True):
                    return True
                     
        """ If none of the above movements work then
        BACKTRACK: unmark x, y as part of solution
        path """
        sol[x][y] = 0
        return False
    return False
 
# Driver Code
maze = [[2, 1, 0, 0],
        [3, 0, 0, 1],
        [0, 1, 0, 1],
        [0, 0, 0, 1]]
solveMaze(maze)
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# program to solve Rat in a Maze problem
// using backtracking
using System;
     
class GFG
{
 
    // Maze size
    static int N = 4;
 
    /* A utility function to print
    solution matrix sol[N, N] */
    static void printSolution(int [,]sol)
    {
        for (int i = 0; i < N; i++)
        {
            for (int j = 0; j < N; j++)
            {
                Console.Write(" {0} ", sol[i, j]);
            }
            Console.Write("\n");
        }
    }
 
    /* A utility function to check if
    x, y is valid index for N*N maze */
    static Boolean isSafe(int [,]maze,
                          int x, int y)
    {
         
        // if (x, y outside maze) return false
        if (x >= 0 && x < N && y >= 0 &&
            y < N && maze[x, y] != 0)
        {
            return true;
        }
 
        return false;
    }
 
    /* This function solves the Maze problem using
    Backtracking. It mainly uses solveMazeUtil() to
    solve the problem. It returns false if no path
    is possible, otherwise return true and prints
    the path in the form of 1s. Please note that
    there may be more than one solutions,
    this function prints one of the feasible solutions.*/
    static Boolean solveMaze(int [,]maze)
    {
        int [,]sol = {{0, 0, 0, 0},
                      {0, 0, 0, 0},
                      {0, 0, 0, 0},
                      {0, 0, 0, 0}};
 
        if (solveMazeUtil(maze, 0, 0, sol) == false)
        {
            Console.Write("Solution doesn't exist");
            return false;
        }
 
        printSolution(sol);
        return true;
    }
 
    /* A recursive utility function to solve Maze problem */
    static Boolean solveMazeUtil(int [,]maze, int x,
                                 int y, int [,]sol)
    {
        // if (x, y is goal) return true
        if (x == N - 1 && y == N - 1)
        {
            sol[x, y] = 1;
            return true;
        }
 
        // Check if maze[x,y] is valid
        if (isSafe(maze, x, y) == true)
        {
 
            // mark x, y as part of solution path
            sol[x, y] = 1;
 
            /* Move forward in x direction */
            for (int i = 1;
                     i <= maze[x, y] && i < N; i++)
            {
 
                /* Move forward in x direction */
                if (solveMazeUtil(maze, x + i,
                                  y, sol) == true)
                {
                    return true;
                }
 
                /* If moving in x direction doesn't give
                solution then Move down in y direction */
                if (solveMazeUtil(maze, x,
                                  y + i, sol) == true)
                {
                    return true;
                }
            }
 
            /* If none of the above movements work then
            BACKTRACK: unmark x, y as part of solution
            path */
            sol[x, y] = 0;
            return false;
        }
 
        return false;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int [,]maze = {{2, 1, 0, 0},
                       {3, 0, 0, 1},
                       {0, 1, 0, 1},
                       {0, 0, 0, 1}};
 
        solveMaze(maze);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// JavaScript program to solve Rat in a Maze problem
// using backtracking
 
    // Maze size
    let N = 4;
   
    /* A utility function to print solution matrix
    sol[N][N] */
    function printSolution(sol)
    {
        for (let i = 0; i < N; i++)
        {
            for (let j = 0; j < N; j++)
            {
                 document.write(sol[i][j] + " ");
            }
             document.write("<br/>");
        }
    }
   
    /* A utility function to check if x, y is valid
    index for N*N maze */
    function isSafe(maze, x, y)
    {
           
        // if (x, y outside maze) return false
        if (x >= 0 && x < N && y >= 0 &&
             y < N && maze[x][y] != 0)
        {
            return true;
        }
   
        return false;
    }
   
    /* This function solves the Maze problem using
    Backtracking. It mainly uses solveMazeUtil() to
    solve the problem. It returns false if no path
    is possible, otherwise return true and print
    the path in the form of 1s. Please note that
    there may be more than one solutions,
    this function prints one of the feasible solutions.*/
    function solveMaze(maze)
    {
        let sol = [[0, 0, 0, 0],
                       [0, 0, 0, 0],
                       [0, 0, 0, 0],
                       [0, 0, 0, 0]];
   
        if (solveMazeUtil(maze, 0, 0, sol) == false)
        {
             document.write("Solution doesn't exist");
            return false;
        }
   
        printSolution(sol);
        return true;
    }
   
    /* A recursive utility function to solve Maze problem */
    function solveMazeUtil(maze,  x,
                                 y, sol)
    {
        // if (x, y is goal) return true
        if (x == N - 1 && y == N - 1)
        {
            sol[x][y] = 1;
            return true;
        }
   
        // Check if maze[x][y] is valid
        if (isSafe(maze, x, y) == true)
        {
   
            // mark x, y as part of solution path
            sol[x][y] = 1;
   
            /* Move forward in x direction */
            for (let i = 1; i <= maze[x][y] && i < N; i++)
            {
   
                /* Move forward in x direction */
                if (solveMazeUtil(maze, x + i, y, sol) == true)
                {
                    return true;
                }
   
                /* If moving in x direction doesn't give
                solution then Move down in y direction */
                if (solveMazeUtil(maze, x, y + i, sol) == true)
                {
                    return true;
                }
            }
   
            /* If none of the above movements work then
            BACKTRACK: unmark x, y as part of solution
            path */
            sol[x][y] = 0;
            return false;
        }
   
        return false;
    }
 
// Driver code
 
        let maze = [[2, 1, 0, 0],
                        [3, 0, 0, 1],
                        [0, 1, 0, 1],
                        [0, 0, 0, 1]];
   
        solveMaze(maze);
 
// This code is contributed by splevel62.
</script>

Output:

Time Complexity: The time complexity of the backtracking algorithm used in the code is exponential, which means O(2^(n^2)) in the worst case scenario where n is the size of the maze.

Space Complexity: The space complexity of the code is O(n^2) because it requires a 2D array of size n x n to store the solution matrix. Additionally, the recursive calls to the function also require additional space on the stack, which can go up to O(n^2) in the worst case.

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Last Updated : 30 Apr, 2023