Count number of ways to reach destination in a Maze

3.1

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 C++ implementation of above idea.

// 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;
}

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.

GATE CS Corner    Company Wise Coding Practice

Recommended Posts:



3.1 Average Difficulty : 3.1/5.0
Based on 31 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.