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 

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 <bits/stdc++.h>
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..

Javascript

<script>


// Javascript implementation of the approach
var R = 4
var 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
function 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 (var 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 (var 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 (var i = 1; i < R; i++) {
        for (var 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)
function numberOfPaths(m, n)
{
    // We have to calculate m+n-2 C n-1 here
    // which will be (m+n-2)! / (n-1)! (m-1)!
    var path = 1;
    for (var 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
function solve(maze)
{

    // Total count of paths - Total paths that do not
    // pass through any of the marked cell
    var ans = numberOfPaths(R, C) - countPaths(maze);

    // return answer
    return ans;
}

// Driver code
var maze = [ [ 0, 0, 0, 0 ],
                   [ 0, -1, 0, 0 ],
                   [ -1, 0, 0, 0 ],
                   [ 0, 0, 0, 0 ] ];
document.write( solve(maze));

// This code is contributed by rrrtnx.
</script>

Output:

16

Time Complexity: O(R*C)
Auxiliary Space: O(R*C)