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

14 Aug, 2021
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++
Java
Python3
C#
Javascript
// 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 mazeint 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 codeint 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 implementation of the approachimport 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 mazestatic 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 codepublic 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 implementation of the approachfrom collections import dequem = 4n = 3 # Function to return the number of valid# paths in the given mazedef 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 mazematrix = [ [ 1, 0, 0, 1 ],           [ 1, 1, 1, 1 ],           [ 1, 0, 1, 1 ] ] print(Maze(matrix))     # This code is contributed by Mohit Kumar
// C# implementation of the approachusing 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 mazestatic 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 codepublic 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
<script> // Javascript implementation of the approachvar m = 4;var n = 3; // Function to return the number of valid// paths in the given mazefunction 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 mazevar 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>
2
 
Time Complexity: O(N * M). Auxiliary Space: O(N * M).  
mohit kumar 29
princi singh
princiraj1992
rutvik_56
pankajsharmagfg
BFS
Data Structures
Matrix
Queue
Data Structures
Matrix
Queue
BFS
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