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: 2Input: 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 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
// 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
# 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#
// 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 |
Javascript
<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> |
Output:
2
Time Complexity: O(N * M).
Auxiliary Space: O(N * M).
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
