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++

// 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;
}

Java

// 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

C#

// 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.

PHP


<?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.
?>


Output:

4

Time Complexity : O(R x C)

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Improved By : nitin mittal, vt_m




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