# Count number of ways to reach destination in a maze

Given a maze of 0 and -1 cells, the task is to find all the paths from (0, 0) to (n-1, m-1), and every path should pass through at least one cell which contains -1. From a given cell, we are allowed to move to cells (i+1, j) and (i, j+1) only.
This problem is a variation of the problem published here.

Examples:

Input: maze[][] = {
{0, 0, 0, 0},
{0, -1, 0, 0},
{-1, 0, 0, 0},
{0, 0, 0, 0}}
Output: 16

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: To find all the paths which go through at least one marked cell (cell containing -1). If we find the paths that do not go through any of the marked cells and all the possible paths from (0, 0) to (n-1, m-1) then we can find all the paths that go through at least one of the marks cells.
Number of paths that pass through at least one marked cell = (Total number of paths – Number of paths that do not pass through any marked cell)
We will use the approach mentioned in this article to find the total number of paths that do not pass through any marked cell and the total number of paths from source to destination will be (m + n – 2)! / (n – 1)! * (m – 1)! where m and n are the number of rows and columns.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std; #define R 4 #define C 4    // Function to return the count of possible paths // in a maze[R][C] from (0, 0) to (R-1, C-1) that // do not pass through any of the marked cells 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; } // Function to return the count of all possible // paths from (0, 0) to (n - 1, m - 1) int numberOfPaths(int m, int n) {     // We have to calculate m+n-2 C n-1 here     // which will be (m+n-2)! / (n-1)! (m-1)!     int path = 1;     for (int i = n; i < (m + n - 1); i++) {         path *= i;         path /= (i - n + 1);     }     return path; }    // Function to return the total count of paths // from (0, 0) to (n - 1, m - 1) that pass // through at least one of the marked cells int solve(int maze[][C]) {        // Total count of paths - Total paths that do not     // pass through any of the marked cell     int ans = numberOfPaths(R, C) - countPaths(maze);        // return answer     return ans; }    // 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 << solve(maze);        return 0; }

## Java

 // Java implementation of the approach import java.io.*;    class GFG  { static int R = 4; static int C = 4;    // Function to return the count of possible paths // in a maze[R][C] from (0, 0) to (R-1, C-1) that // do not pass through any of the marked cells 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; }    // Function to return the count of all possible // paths from (0, 0) to (n - 1, m - 1) static int numberOfPaths(int m, int n) {     // We have to calculate m+n-2 C n-1 here     // which will be (m+n-2)! / (n-1)! (m-1)!     int path = 1;     for (int i = n; i < (m + n - 1); i++)     {         path *= i;         path /= (i - n + 1);     }     return path; }    // Function to return the total count of paths // from (0, 0) to (n - 1, m - 1) that pass // through at least one of the marked cells static int solve(int maze[][]) {        // Total count of paths - Total paths that do not     // pass through any of the marked cell     int ans = numberOfPaths(R, C) - countPaths(maze);        // return answer     return ans; }    // 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(solve(maze)); } }    // This code is contributed by anuj_67..

## Python3

 # Python3 implementation of the approach R = 4 C = 4    # Function to return the count of  # possible paths in a maze[R][C]  # from (0, 0) to (R-1, C-1) that # do not pass through any of  # the marked cells 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):         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):         for j in range(1, C):                            # 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    # Function to return the count of  # all possible paths from  # (0, 0) to (n - 1, m - 1) def numberOfPaths(m, n):        # We have to calculate m+n-2 C n-1 here     # which will be (m+n-2)! / (n-1)! (m-1)!     path = 1     for i in range(n, m + n - 1):            path *= i         path //= (i - n + 1)        return path    # Function to return the total count  # of paths from (0, 0) to (n - 1, m - 1)  # that pass through at least one  # of the marked cells def solve(maze):            # Total count of paths - Total paths      # that do not pass through any of      # the marked cell     ans = (numberOfPaths(R, C) -            countPaths(maze))        # return answer     return ans    # Driver code maze = [[ 0, 0, 0, 0 ],         [ 0, -1, 0, 0 ],         [ -1, 0, 0, 0 ],         [ 0, 0, 0, 0 ]]    print(solve(maze))    # This code is contributed # by Mohit Kumar

## C#

 // C# implementation of the approach using System; class GFG  { static int R = 4; static int C = 4;    // Function to return the count of possible paths // in a maze[R][C] from (0, 0) to (R-1, C-1) that // do not pass through any of the marked cells 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; }    // Function to return the count of all possible // paths from (0, 0) to (n - 1, m - 1) static int numberOfPaths(int m, int n) {     // We have to calculate m+n-2 C n-1 here     // which will be (m+n-2)! / (n-1)! (m-1)!     int path = 1;     for (int i = n; i < (m + n - 1); i++)     {         path *= i;         path /= (i - n + 1);     }     return path; }    // Function to return the total count of paths // from (0, 0) to (n - 1, m - 1) that pass // through at least one of the marked cells static int solve(int [,]maze) {        // Total count of paths - Total paths that do not     // pass through any of the marked cell     int ans = numberOfPaths(R, C) -                countPaths(maze);        // return answer     return ans; }    // 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(solve(maze)); } }    // This code is contributed by anuj_67..

Output:

16

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