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.

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.

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