# 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

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

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 ` `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 > 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 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 q = ``new` `Queue(); ` ` `  `    ``// 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  `

Output:

```2
```

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