Count number of ways to reach destination in a Maze

Given a maze with obstacles, count number of paths to reach rightmost-bottommost cell from topmost-leftmost cell. A cell in given maze has value -1 if it is a blockage or dead end, else 0.

From a given cell, we are allowed to move to cells (i+1, j) and (i, j+1) only.

Examples:

Input: maze[R][C] =  {{0,  0, 0, 0},
                      {0, -1, 0, 0},
                      {-1, 0, 0, 0},
                      {0,  0, 0, 0}};
Output: 4
There are four possible paths as shown in
below diagram.
blockage

This problem is an extension of below problem.

Backtracking | Set 2 (Rat in a Maze)



In this post a different solution is discussed that can be used to solve the above Rat in a Maze problem also.

The idea is to modify the given grid[][] so that grid[i][j] contains count of paths to reach (i, j) from (0, 0) if (i, j) is not a blockage, else grid[i][j] remains -1. 

We can recursively compute grid[i][j] using below 
formula and finally return grid[R-1][C-1]

  // If current cell is a blockage
  if (maze[i][j] == -1)
      maze[i][j] = -1; //  Do not change

  // If we can reach maze[i][j] from maze[i-1][j]
  // then increment count.
  else if (maze[i-1][j] > 0)
      maze[i][j] = (maze[i][j] + maze[i-1][j]);

  // If we can reach maze[i][j] from maze[i][j-1]
  // then increment count.
  else if (maze[i][j-1] > 0)
      maze[i][j] = (maze[i][j] + maze[i][j-1]);

Below is the implementation of above idea.

C++

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// C++ program to count number of paths in a maze
// with obstacles.
#include<bits/stdc++.h>
using namespace std;
#define R 4
#define C 4
  
// Returns count of possible paths in a maze[R][C]
// from (0,0) to (R-1,C-1)
int countPaths(int maze[][C])
{
    // If the initial cell is blocked, there is no
    // way of moving anywhere
    if (maze[0][0]==-1)
        return 0;
  
    // Initializing the leftmost column
    for (int i=0; i<R; i++)
    {
        if (maze[i][0] == 0)
            maze[i][0] = 1;
  
        // If we encounter a blocked cell in leftmost
        // row, there is no way of visiting any cell
        // directly below it.
        else
            break;
    }
  
    // Similarly initialize the topmost row
    for (int i=1; i<C; i++)
    {
        if (maze[0][i] == 0)
            maze[0][i] = 1;
  
        // If we encounter a blocked cell in bottommost
        // row, there is no way of visiting any cell
        // directly below it.
        else
            break;
    }
  
    // The only difference is that if a cell is -1,
    // simply ignore it else recursively compute
    // count value maze[i][j]
    for (int i=1; i<R; i++)
    {
        for (int j=1; j<C; j++)
        {
            // If blockage is found, ignore this cell 
            if (maze[i][j] == -1)
                continue;
  
            // If we can reach maze[i][j] from maze[i-1][j]
            // then increment count.
            if (maze[i-1][j] > 0)
                maze[i][j] = (maze[i][j] + maze[i-1][j]);
  
            // If we can reach maze[i][j] from maze[i][j-1]
            // then increment count.
            if (maze[i][j-1] > 0)
                maze[i][j] = (maze[i][j] + maze[i][j-1]);
        }
    }
  
    // If the final cell is blocked, output 0, otherwise
    // the answer
    return (maze[R-1][C-1] > 0)? maze[R-1][C-1] : 0;
}
  
// Driver code
int main()
{
    int maze[R][C] =  {{0,  0, 0, 0},
                       {0, -1, 0, 0},
                       {-1, 0, 0, 0},
                       {0,  0, 0, 0}};
    cout << countPaths(maze);
    return 0;
}

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Java














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// Java program to count number of paths in a maze 


// with obstacles. 


import java.io.*; 


  


class GFG  


{ 


    static int R = 4; 


    static int C = 4; 


      


    // Returns count of possible paths in  


    // a maze[R][C] from (0,0) to (R-1,C-1) 


    static int countPaths(int maze[][]) 


    { 


        // If the initial cell is blocked,  


        // there is no way of moving anywhere 


        if (maze[0][0]==-1) 


            return 0; 


      


        // Initializing the leftmost column 


        for (int i = 0; i < R; i++) 


        { 


            if (maze[i][0] == 0) 


                maze[i][0] = 1; 


      


            // If we encounter a blocked cell  


            // in leftmost row, there is no way  


            // of visiting any cell directly below it. 


            else


                break; 


        } 


      


        // Similarly initialize the topmost row 


        for (int i =1 ; i< C ; i++) 


        { 


            if (maze[0][i] == 0) 


                maze[0][i] = 1; 


      


            // If we encounter a blocked cell in  


            // bottommost row, there is no way of  


            // visiting any cell directly below it. 


            else


                break; 


        } 


      


        // The only difference is that if a cell  


        // is -1, simply ignore it else recursively  


        // compute count value maze[i][j] 


        for (int i = 1; i < R; i++) 


        { 


            for (int j = 1; j <C ; j++) 


            { 


                // If blockage is found,  


                // ignore this cell  


                if (maze[i][j] == -1) 


                    continue; 


      


                // If we can reach maze[i][j] from  


                // maze[i-1][j] then increment count. 


                if (maze[i - 1][j] > 0) 


                    maze[i][j] = (maze[i][j] +  


                                 maze[i - 1][j]); 


      


                // If we can reach maze[i][j] from 


                //  maze[i][j-1] then increment count. 


                if (maze[i][j - 1] > 0) 


                    maze[i][j] = (maze[i][j] +  


                                  maze[i][j - 1]); 


            } 


        } 


      


        // If the final cell is blocked,  


        // output 0, otherwise the answer 


        return (maze[R - 1][C - 1] > 0) ?  


                maze[R - 1][C - 1] : 0; 


    } 


      


    // Driver code 


  


    public static void main (String[] args)  


    { 


        int maze[][] = {{0, 0, 0, 0}, 


                       {0, -1, 0, 0}, 


                       {-1, 0, 0, 0}, 


                       {0, 0, 0, 0}}; 


        System.out.println (countPaths(maze)); 


      


    } 


  


} 


  


// This code is contributed by vt_m 



















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Python3














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# Python 3 program to count number of paths  


# in a maze with obstacles. 


  


R = 4


C = 4


  


# Returns count of possible paths in a 


# maze[R][C] from (0,0) to (R-1,C-1) 


def countPaths(maze): 


      


    # If the initial cell is blocked,  


    # there is no way of moving anywhere 


    if (maze[0][0] == -1): 


        return 0


  


    # Initializing the leftmost column 


    for i in range(R): 


        if (maze[i][0] == 0): 


            maze[i][0] = 1


  


        # If we encounter a blocked cell in  


        # leftmost row, there is no way of  


        # visiting any cell directly below it. 


        else: 


            break


  


    # Similarly initialize the topmost row 


    for i in range(1, C, 1): 


        if (maze[0][i] == 0): 


            maze[0][i] = 1


  


        # If we encounter a blocked cell in  


        # bottommost row, there is no way of 


        # visiting any cell directly below it. 


        else: 


            break


  


    # The only difference is that if a cell is -1, 


    # simply ignore it else recursively compute 


    # count value maze[i][j] 


    for i in range(1, R, 1): 


        for j in range(1, C, 1): 


              


            # If blockage is found, ignore this cell  


            if (maze[i][j] == -1): 


                continue


  


            # If we can reach maze[i][j] from  


            # maze[i-1][j] then increment count. 


            if (maze[i - 1][j] > 0): 


                maze[i][j] = (maze[i][j] +


                              maze[i - 1][j]) 


  


            # If we can reach maze[i][j] from  


            # maze[i][j-1] then increment count. 


            if (maze[i][j - 1] > 0): 


                maze[i][j] = (maze[i][j] +


                              maze[i][j - 1]) 


  


    # If the final cell is blocked,  


    # output 0, otherwise the answer 


    if (maze[R - 1][C - 1] > 0): 


        return maze[R - 1][C - 1]  


    else: 


        return 0


  


# Driver code 


if __name__ == '__main__': 


    maze = [[0, 0, 0, 0], 


            [0, -1, 0, 0], 


            [-1, 0, 0, 0], 


            [0, 0, 0, 0 ]] 


    print(countPaths(maze)) 


  


# This code is contributed by 


# Surendra_Gangwar 



















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C#














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// C# program to count number of paths in a maze 


// with obstacles. 


using System; 


  


class GFG { 


      


    static int R = 4; 


    static int C = 4; 


      


    // Returns count of possible paths in  


    // a maze[R][C] from (0,0) to (R-1,C-1) 


    static int countPaths(int [,]maze) 


    { 


          


        // If the initial cell is blocked,  


        // there is no way of moving anywhere 


        if (maze[0,0]==-1) 


            return 0; 


      


        // Initializing the leftmost column 


        for (int i = 0; i < R; i++) 


        { 


            if (maze[i,0] == 0) 


                maze[i,0] = 1; 


      


            // If we encounter a blocked cell  


            // in leftmost row, there is no way  


            // of visiting any cell directly below it. 


            else


                break; 


        } 


      


        // Similarly initialize the topmost row 


        for (int i =1 ; i< C ; i++) 


        { 


            if (maze[0,i] == 0) 


                maze[0,i] = 1; 


      


            // If we encounter a blocked cell in  


            // bottommost row, there is no way of  


            // visiting any cell directly below it. 


            else


                break; 


        } 


      


        // The only difference is that if a cell  


        // is -1, simply ignore it else recursively  


        // compute count value maze[i][j] 


        for (int i = 1; i < R; i++) 


        { 


            for (int j = 1; j <C ; j++) 


            { 


                // If blockage is found,  


                // ignore this cell  


                if (maze[i,j] == -1) 


                    continue; 


      


                // If we can reach maze[i][j] from  


                // maze[i-1][j] then increment count. 


                if (maze[i - 1,j] > 0) 


                    maze[i,j] = (maze[i,j] +  


                                maze[i - 1,j]); 


      


                // If we can reach maze[i][j] from 


                // maze[i][j-1] then increment count. 


                if (maze[i,j - 1] > 0) 


                    maze[i,j] = (maze[i,j] +  


                                maze[i,j - 1]); 


            } 


        } 


      


        // If the final cell is blocked,  


        // output 0, otherwise the answer 


        return (maze[R - 1,C - 1] > 0) ?  


                maze[R - 1,C - 1] : 0; 


    } 


      


    // Driver code 


    public static void Main ()  


    { 


        int [,]maze = { {0, 0, 0, 0}, 


                        {0, -1, 0, 0}, 


                        {-1, 0, 0, 0}, 


                        {0, 0, 0, 0}}; 


                          


        Console.Write (countPaths(maze)); 


    } 


  


} 


  


// This code is contributed by nitin mittal. 



















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PHP














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<?php 


// PHP program to count number 


// of paths in a maze with obstacles. 


  


$R = 4; 


$C = 4; 


  


// Returns count of possible  


// paths in a maze[R][C] 


// from (0,0) to (R-1,C-1) 


function countPaths( $maze) 


{ 


    global $R, $C; 


      


    // If the initial cell is 


    // blocked, there is no 


    // way of moving anywhere 


    if ($maze[0][0] == - 1) 


        return 0; 


  


    // Initializing the  


    // leftmost column 


    for ( $i = 0; $i < $R; $i++) 


    { 


        if ($maze[$i][0] == 0) 


            $maze[$i][0] = 1; 


  


        // If we encounter a blocked 


        // cell in leftmost row,  


        // there is no way of  


        // visiting any cell 


        // directly below it. 


        else


            break; 


    } 


  


    // Similarly initialize  


    // the topmost row 


    for($i = 1; $i < $C; $i++) 


    { 


        if ($maze[0][$i] == 0) 


            $maze[0][$i] = 1; 


  


        // If we encounter a blocked 


        // cell in bottommost row,  


        // there is no way of  


        // visiting any cell 


        // directly below it. 


        else


            break; 


    } 


  


    // The only difference is  


    // that if a cell is -1, 


    // simply ignore it else 


    // recursively compute 


    // count value maze[i][j] 


    for($i = 1; $i < $R; $i++) 


    { 


        for($j = 1; $j < $C; $j++) 


        { 


              


            // If blockage is found,  


            // ignore this cell  


            if ($maze[$i][$j] == -1) 


                continue; 


  


            // If we can reach maze[i][j]  


            // from maze[i-1][j] 


            // then increment count. 


            if ($maze[$i - 1][$j] > 0) 


                $maze[$i][$j] = ($maze[$i][$j] +  


                           $maze[$i - 1][$j]); 


  


            // If we can reach maze[i][j] 


            // from maze[i][j-1] 


            // then increment count. 


            if ($maze[$i][$j - 1] > 0) 


                $maze[$i][$j] = ($maze[$i][$j] +  


                             $maze[$i][$j - 1]); 


        } 


    } 


  


    // If the final cell is  


    // blocked, output 0,  


    // otherwise the answer 


    return ($maze[$R - 1][$C - 1] > 0) ? 


            $maze[$R - 1][$C - 1] : 0; 


} 


  


    // Driver Code 


    $maze = array(array(0, 0, 0, 0), 


                  array(0, -1, 0, 0), 


                  array(-1, 0, 0, 0), 


                  array(0, 0, 0, 0)); 


    echo countPaths($maze); 


  


// This code is contributed by anuj_67. 


?> 



















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Output:


4



Time Complexity : O(R x C)


This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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