Count number of ways to reach destination in a Maze using BFS

Given a maze with obstacles, count number of paths to reach rightmost-bottom most cell from the topmost-leftmost cell. A cell in the 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: mat[][] = { 
{1, 0, 0, 1}, 
{1, 1, 1, 1}, 
{1, 0, 1, 1}} 
Output: 2

Input: mat[][] = { 
{1, 1, 1, 1}, 
{1, 0, 1, 1}, 
{0, 1, 1, 1}, 
{1, 1, 1, 1}} 
Output: 4 

Approach: The idea is to use a queue and apply bfs and use a variable count to store the number of possible paths. Make a pair of row and column and insert (0, 0) into the queue. Now keep popping pairs from queue, if the popped value is the end of matrix then increment count, otherwise check if the next column can give a valid move or the next row can give a valid move and according to that, insert the corresponding row, column pair into the queue.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

#define m 4
#define n 3

// Function to return the number of valid
// paths in the given maze
int Maze(int matrix[n][m])
{
    queue<pair<int, int> > q;

    // Insert the starting point i.e.
    // (0, 0) in the queue
    q.push(make_pair(0, 0));

    // To store the count of possible paths
    int count = 0;

    while (!q.empty()) {
        pair<int, int> p = q.front();
        q.pop();

        // Increment the count of paths since
        // it is the destination
        if (p.first == n - 1 && p.second == m - 1)
            count++;

        // If moving to the next row is a valid move
        if (p.first + 1 < n
            && matrix[p.first + 1][p.second] == 1) {
            q.push(make_pair(p.first + 1, p.second));
        }

        // If moving to the next column is a valid move
        if (p.second + 1 < m
            && matrix[p.first][p.second + 1] == 1) {
            q.push(make_pair(p.first, p.second + 1));
        }
    }

    return count;
}

// Driver code
int main()
{
    // Matrix to represent maze
    int matrix[n][m] = { { 1, 0, 0, 1 },
                         { 1, 1, 1, 1 },
                         { 1, 0, 1, 1 } };

    cout << Maze(matrix);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
class GFG
{
static int m = 4;
static int n = 3;
static class pair
{
    int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to return the number of valid
// paths in the given maze
static int Maze(int matrix[][])
{
    Queue<pair> q = new LinkedList<>();

    // Insert the starting point i.e.
    // (0, 0) in the queue
    q.add(new pair(0, 0));

    // To store the count of possible paths
    int count = 0;

    while (!q.isEmpty())
    {
        pair p = q.peek();
        q.remove();

        // Increment the count of paths since
        // it is the destination
        if (p.first == n - 1 && p.second == m - 1)
            count++;

        // If moving to the next row is a valid move
        if (p.first + 1 < n &&
            matrix[p.first + 1][p.second] == 1)
        {
            q.add(new pair(p.first + 1, p.second));
        }

        // If moving to the next column is a valid move
        if (p.second + 1 < m &&
            matrix[p.first][p.second + 1] == 1)
        {
            q.add(new pair(p.first, p.second + 1));
        }
    }
    return count;
}

// Driver code
public static void main(String[] args)
{
    // Matrix to represent maze
    int matrix[][] = {{ 1, 0, 0, 1 },
                      { 1, 1, 1, 1 },
                      { 1, 0, 1, 1 }};

    System.out.println(Maze(matrix));
}
}

// This code is contributed by Princi Singh

Python3

# Python3 implementation of the approach
from collections import deque
m = 4
n = 3

# Function to return the number of valid
# paths in the given maze
def Maze(matrix):
    q = deque()

    # Insert the starting poi.e.
    # (0, 0) in the queue
    q.append((0, 0))

    # To store the count of possible paths
    count = 0

    while (len(q) > 0):
        p = q.popleft()

        # Increment the count of paths since
        # it is the destination
        if (p[0] == n - 1 and p[1] == m - 1):
            count += 1

        # If moving to the next row is a valid move
        if (p[0] + 1 < n and
            matrix[p[0] + 1][p[1]] == 1):
            q.append((p[0] + 1, p[1]))

        # If moving to the next column is a valid move
        if (p[1] + 1 < m and
            matrix[p[0]][p[1] + 1] == 1):
            q.append((p[0], p[1] + 1))

    return count

# Driver code

# Matrix to represent maze
matrix = [ [ 1, 0, 0, 1 ],
           [ 1, 1, 1, 1 ],
           [ 1, 0, 1, 1 ] ]

print(Maze(matrix))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{
static int m = 4;
static int n = 3;
class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function to return the number of valid
// paths in the given maze
static int Maze(int [,]matrix)
{
    Queue<pair> q = new Queue<pair>();

    // Insert the starting point i.e.
    // (0, 0) in the queue
    q.Enqueue(new pair(0, 0));

    // To store the count of possible paths
    int count = 0;

    while (q.Count != 0)
    {
        pair p = q.Peek();
        q.Dequeue();

        // Increment the count of paths since
        // it is the destination
        if (p.first == n - 1 && p.second == m - 1)
            count++;

        // If moving to the next row is a valid move
        if (p.first + 1 < n &&
            matrix[p.first + 1, p.second] == 1)
        {
            q.Enqueue(new pair(p.first + 1, p.second));
        }

        // If moving to the next column is a valid move
        if (p.second + 1 < m &&
            matrix[p.first, p.second + 1] == 1)
        {
            q.Enqueue(new pair(p.first, p.second + 1));
        }
    }
    return count;
}

// Driver code
public static void Main(String[] args)
{
    // Matrix to represent maze
    int [,]matrix = {{ 1, 0, 0, 1 },
                     { 1, 1, 1, 1 },
                     { 1, 0, 1, 1 }};

    Console.WriteLine(Maze(matrix));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript implementation of the approach
var m = 4;
var n = 3;

// Function to return the number of valid
// paths in the given maze
function Maze(matrix)
{
    var q = [];

    // Insert the starting point i.e.
    // (0, 0) in the queue
    q.push([0, 0]);

    // To store the count of possible paths
    var count = 0;

    while (q.length != 0)
    {
        var p = q[0];
        q.shift();

        // Increment the count of paths since
        // it is the destination
        if (p[0] == n - 1 && p[1] == m - 1)
            count++;

        // If moving to the next row is a valid move
        if (p[0] + 1 < n &&
            matrix[p[0] + 1][p[1]] == 1)
        {
            q.push([p[0] + 1, p[1]]);
        }

        // If moving to the next column is a valid move
        if (p[1] + 1 < m &&
            matrix[p[0]][p[1] + 1] == 1)
        {
            q.push([p[0], p[1] + 1]);
        }
    }
    return count;
}

// Driver code

// Matrix to represent maze
var matrix = [ [ 1, 0, 0, 1 ],
               [ 1, 1, 1, 1 ],
               [ 1, 0, 1, 1 ] ];

document.write( Maze(matrix));

// This code is contributed by rutvik_56

</script>

Output:

2

Time Complexity: O(N * M). 
Auxiliary Space: O(N * M).