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Find whether there is path between two cells in matrix
  • Difficulty Level : Medium
  • Last Updated : 10 Mar, 2021

Given N X N matrix filled with 1, 0, 2, 3. Find whether there is a path possible from source to destination, traversing through blank cells only. You can traverse up, down, right, and left. 

  • A value of cell 1 means Source.
  • A value of cell 2 means Destination.
  • A value of cell 3 means Blank cell.
  • A value of cell 0 means Blank Wall.

Note: there are an only a single source and single destination(sink). 
Examples: 

Input: 
M[3][3] = {{ 0, 3, 2 }, 
{ 3, 3, 0 }, 
{ 1, 3, 0 }}; 
Output : Yes 
Explanation: 
 

Input: 
M[4][4] = {{ 0, 3, 1, 0 }, 
{ 3, 0, 3, 3 }, 
{ 2, 3, 0, 3 }, 
{ 0, 3, 3, 3 }}; 
Output: Yes 
Explanation: 
 



 

Asked in: Adobe Interview 
 

Simple Solution: Recursion.
Approach: Find the source index of the cell in each matrix and then recursively find a path from source index to destination in the matrix. The algorithm involves recursively finding all the paths until a final path is found to the destination.
Algorithm :  

  1. Traverse the matrix and find the starting index of the matrix.
  2. Create a recursive function that takes the index and visited matrix.
  3. Mark the current cell and check if the current cell is a destination or not. If the current cell is destination return true.
  4. Call the recursion function for all adjacent empty and unvisited cells.
  5. If any of the recursive function returns true then unmark the cell and return true else unmark the cell and return false.

C++




// C++ program to find path between two
// cell in matrix
#include <iostream>
using namespace std;
#define N 4
 
// Method for checking boundries
bool isSafe(int i, int j, int matrix[][N])
{
  if (i >= 0 && i < N && j >= 0 && j < N)
    return true;
  return false;
}
 
// Returns true if there is a
// path from a source (a
// cell with value 1) to a
// destination (a cell with
// value 2)
bool isPath(int matrix[][N], int i, int j,
            bool visited[][N])
{
  // Checking the boundries, walls and
  // whether the cell is unvisited
  if (isSafe(i, j, matrix) && matrix[i][j] != 0
      && !visited[i][j])
  {
    // Make the cell visited
    visited[i][j] = true;
 
    // if the cell is the required
    // destination then return true
    if (matrix[i][j] == 2)
      return true;
 
    // traverse up
    bool up = isPath(matrix, i - 1, j, visited);
 
    // if path is found in up
    // direction return true
    if (up)
      return true;
 
    // traverse left
    bool left = isPath(matrix, i, j - 1, visited);
 
    // if path is found in left
    // direction return true
    if (left)
      return true;
 
    // traverse down
    bool down = isPath(matrix, i + 1, j, visited);
 
    // if path is found in down
    // direction return true
    if (down)
      return true;
 
    // traverse right
    bool right = isPath(matrix, i, j + 1, visited);
 
    // if path is found in right
    // direction return true
    if (right)
      return true;
  }
 
  // no path has been found
  return false;
}
 
// Method for finding and printing
// whether the path exists or not
void isPath(int matrix[][N])
{
 
  // Defining visited array to keep
  // track of already visited indexes
  bool visited[N][N];
 
  // Flag to indicate whether the
  // path exists or not
  bool flag = false;
  for (int i = 0; i < N; i++)
  {
    for (int j = 0; j < N; j++)
    {
      // if matrix[i][j] is source
      // and it is not visited
      if (matrix[i][j] == 1 && !visited[i][j])
 
        // Starting from i, j and
        // then finding the path
        if (isPath(matrix, i, j, visited))
        {
 
          // if path exists
          flag = true;
          break;
        }
    }
  }
  if (flag)
    cout << "YES";
  else
    cout << "NO";
}
 
// Driver program to
// check above function
int main()
{
  int matrix[N][N] = { { 0, 3, 0, 1 },
                      { 3, 0, 3, 3 },
                      { 2, 3, 3, 3 },
                      { 0, 3, 3, 3 } };
 
  // calling isPath method
  isPath(matrix);
  return 0;
}
 
// This code is contributed by sudhanshugupta2019a.

Java




// Java program to find path between two
// cell in matrix
class Path {
 
    // Method for finding and printing
    // whether the path exists or not
    public static void isPath(
        int matrix[][], int n)
    {
        // Defining visited array to keep
        // track of already visited indexes
        boolean visited[][]
            = new boolean[n][n];
 
        // Flag to indicate whether the
        // path exists or not
        boolean flag = false;
 
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                // if matrix[i][j] is source
                // and it is not visited
                if (
                    matrix[i][j] == 1
                    && !visited[i][j])
 
                    // Starting from i, j and
                    // then finding the path
                    if (isPath(
                            matrix, i, j, visited)) {
                        // if path exists
                        flag = true;
                        break;
                    }
            }
        }
        if (flag)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
 
    // Method for checking boundries
    public static boolean isSafe(
        int i, int j,
        int matrix[][])
    {
 
        if (
            i >= 0 && i < matrix.length
            && j >= 0
            && j < matrix[0].length)
            return true;
        return false;
    }
 
    // Returns true if there is a
    // path from a source (a
    // cell with value 1) to a
    // destination (a cell with
    // value 2)
    public static boolean isPath(
        int matrix[][],
        int i, int j,
        boolean visited[][])
    {
 
        // Checking the boundries, walls and
        // whether the cell is unvisited
        if (
            isSafe(i, j, matrix)
            && matrix[i][j] != 0
            && !visited[i][j]) {
            // Make the cell visited
            visited[i][j] = true;
 
            // if the cell is the required
            // destination then return true
            if (matrix[i][j] == 2)
                return true;
 
            // traverse up
            boolean up = isPath(
                matrix, i - 1,
                j, visited);
 
            // if path is found in up
            // direction return true
            if (up)
                return true;
 
            // traverse left
            boolean left
                = isPath(
                    matrix, i, j - 1, visited);
 
            // if path is found in left
            // direction return true
            if (left)
                return true;
 
            // traverse down
            boolean down = isPath(
                matrix, i + 1, j, visited);
 
            // if path is found in down
            // direction return true
            if (down)
                return true;
 
            // traverse right
            boolean right
                = isPath(
                    matrix, i, j + 1,
                    visited);
 
            // if path is found in right
            // direction return true
            if (right)
                return true;
        }
        // no path has been found
        return false;
    }
 
    // driver program to
    // check above function
    public static void main(String[] args)
    {
 
        int matrix[][] = { { 0, 3, 0, 1 },
                           { 3, 0, 3, 3 },
                           { 2, 3, 3, 3 },
                           { 0, 3, 3, 3 } };
 
        // calling isPath method
        isPath(matrix, 4);
    }
}
 
/* This code is contributed by Madhu Priya */

Python3




# Python3 program to find
# path between two cell in matrix
 
# Method for finding and printing
# whether the path exists or not
def isPath(matrix, n):
 
    # Defining visited array to keep
    # track of already visited indexes
    visited = [[False for x in range (n)]
                      for y in range (n)]
    
    # Flag to indicate whether the
    # path exists or not
    flag = False
 
    for i in range (n):
        for j in range (n):
           
            # If matrix[i][j] is source
            # and it is not visited
            if (matrix[i][j] == 1 and not
                visited[i][j]):
 
                # Starting from i, j and
                # then finding the path
                if (checkPath(matrix, i,
                              j, visited)):
                   
                    # If path exists
                    flag = True
                    break
    if (flag):
        print("YES")
    else:
        print("NO")
 
# Method for checking boundries
def isSafe(i, j, matrix):
   
    if (i >= 0 and i < len(matrix) and
        j >= 0 and j < len(matrix[0])):
        return True
    return False
 
# Returns true if there is a
# path from a source(a
# cell with value 1) to a
# destination(a cell with
# value 2)
def checkPath(matrix, i, j,
              visited):
 
    # Checking the boundries, walls and
    # whether the cell is unvisited
    if (isSafe(i, j, matrix) and
        matrix[i][j] != 0 and not
        visited[i][j]):
       
        # Make the cell visited
        visited[i][j] = True
 
        # If the cell is the required
        # destination then return true
        if (matrix[i][j] == 2):
           return True
 
        # traverse up
        up = checkPath(matrix, i - 1,
                       j, visited)
 
        # If path is found in up
        # direction return true
        if (up):
           return True
 
        # Traverse left
        left = checkPath(matrix, i,
                         j - 1, visited)
 
        # If path is found in left
        # direction return true
        if (left):
           return True
 
        # Traverse down
        down = checkPath(matrix, i + 1,
                         j, visited)
 
        # If path is found in down
        # direction return true
        if (down):
           return True
 
        # Traverse right
        right = checkPath(matrix, i,
                          j + 1, visited)
 
        # If path is found in right
        # direction return true
        if (right):
           return True
     
    # No path has been found
    return False
 
# Driver code
if __name__ == "__main__":
   
    matrix = [[0, 3, 0, 1],
              [3, 0, 3, 3],
              [2, 3, 3, 3],
              [0, 3, 3, 3]]
 
    # calling isPath method
    isPath(matrix, 4)
 
# This code is contributed by Chitranayal

C#




// C# program to find path between two
// cell in matrix
using System;
 
class GFG{
 
// Method for finding and printing
// whether the path exists or not
static void isPath(int[,] matrix, int n)
{
     
    // Defining visited array to keep
    // track of already visited indexes
    bool[,] visited = new bool[n, n];
     
    // Flag to indicate whether the
    // path exists or not
    bool flag = false;
 
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
             
            // If matrix[i][j] is source
            // and it is not visited
            if (matrix[i, j] == 1 &&
              !visited[i, j])
               
                // Starting from i, j and
                // then finding the path
                if (isPath(matrix, i, j,
                           visited))
                {
                     
                    // If path exists
                    flag = true;
                    break;
                }
        }
    }
    if (flag)
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
 
// Method for checking boundries
public static bool isSafe(int i, int j,
                          int[,] matrix)
{
    if (i >= 0 && i < matrix.GetLength(0) &&
        j >= 0 && j < matrix.GetLength(1))
        return true;
         
    return false;
}
 
// Returns true if there is a path from
// a source (a cell with value 1) to a
// destination (a cell with value 2)
public static bool isPath(int[,] matrix, int i,
                          int j, bool[,] visited)
{
     
    // Checking the boundries, walls and
    // whether the cell is unvisited
    if (isSafe(i, j, matrix) &&
           matrix[i, j] != 0 &&
         !visited[i, j])
    {
         
        // Make the cell visited
        visited[i, j] = true;
 
        // If the cell is the required
        // destination then return true
        if (matrix[i, j] == 2)
            return true;
 
        // Traverse up
        bool up = isPath(matrix, i - 1,
                         j, visited);
 
        // If path is found in up
        // direction return true
        if (up)
            return true;
 
        // Traverse left
        bool left = isPath(matrix, i,
                           j - 1, visited);
 
        // If path is found in left
        // direction return true
        if (left)
            return true;
 
        // Traverse down
        bool down = isPath(matrix, i + 1,
                           j, visited);
 
        // If path is found in down
        // direction return true
        if (down)
            return true;
 
        // Traverse right
        bool right = isPath(matrix, i, j + 1,
                            visited);
 
        // If path is found in right
        // direction return true
        if (right)
            return true;
    }
     
    // No path has been found
    return false;
}
 
// Driver code  
static void Main()
{
    int[,] matrix = { { 0, 3, 0, 1 },
                      { 3, 0, 3, 3 },
                      { 2, 3, 3, 3 },
                      { 0, 3, 3, 3 } };
 
    // Calling isPath method
    isPath(matrix, 4);
}
}
 
// This code is contributed by divyeshrabadiya07
Output
YES

Complexity Analysis:  

  • Time Complexity: O(4n*m). 
    For every cell, there can be 4 adjacent unvisited cells so the time complexity is O(4n*m).
  • Space Complexity: O(n*m). 
    Space is required to store the visited array.

Efficient solution: Graph.
Approach: The idea is to use Breadth-First Search. Consider each cell as a node and each boundary between any two adjacent cells be an edge. so the total number of Node is N * N. 
So the idea is to do a breadth-first search from the starting cell till the ending cell is found.
Algorithm:  

  1. Create an empty Graph having N*N node(Vertex), push all nodes into a graph, and note down the source and sink vertex.
  2. Now apply BFS on the graph, create a queue and insert the source node in the queue
  3. Run a loop till the size of the queue is greater than 0
  4. Remove the front node of the queue and check if the node is the destination if the destination returns true. mark the node
  5. Check all adjacent cells if unvisited and blank insert them in the queue.
  6. If the destination is not reached return true.

C++




// C++ program to find path
// between two cell in matrix
#include <bits/stdc++.h>
using namespace std;
#define N 4
 
class Graph {
    int V;
    list<int>* adj;
 
public:
    Graph(int V)
    {
        this->V = V;
        adj = new list<int>[V];
    }
    void addEdge(int s, int d);
    bool BFS(int s, int d);
};
 
// add edge to graph
void Graph::addEdge(int s, int d)
{
    adj[s].push_back(d);
}
 
// BFS function to find path
// from source to sink
bool Graph::BFS(int s, int d)
{
    // Base case
    if (s == d)
        return true;
 
    // Mark all the vertices as not visited
    bool* visited = new bool[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;
 
    // Create a queue for BFS
    list<int> queue;
 
    // Mark the current node as visited and
    // enqueue it
    visited[s] = true;
    queue.push_back(s);
 
    // it will be used to get all adjacent
    // vertices of a vertex
    list<int>::iterator i;
 
    while (!queue.empty()) {
        // Dequeue a vertex from queue
        s = queue.front();
        queue.pop_front();
 
        // Get all adjacent vertices of the
        // dequeued vertex s. If a adjacent has
        // not been visited, then mark it visited
        // and enqueue it
        for (
            i = adj[s].begin(); i != adj[s].end(); ++i) {
            // If this adjacent node is the
            // destination node, then return true
            if (*i == d)
                return true;
 
            // Else, continue to do BFS
            if (!visited[*i]) {
                visited[*i] = true;
                queue.push_back(*i);
            }
        }
    }
 
    // If BFS is complete without visiting d
    return false;
}
 
bool isSafe(int i, int j, int M[][N])
{
    if (
        (i < 0 || i >= N)
        || (j < 0 || j >= N)
        || M[i][j] == 0)
        return false;
    return true;
}
 
// Returns true if there is
// a path from a source (a
// cell with value 1) to a
// destination (a cell with
// value 2)
bool findPath(int M[][N])
{
    // source and destination
    int s, d;
    int V = N * N + 2;
    Graph g(V);
 
    // create graph with n*n node
    // each cell consider as node
    // Number of current vertex
    int k = 1;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (M[i][j] != 0) {
                // connect all 4 adjacent
                // cell to current cell
                if (isSafe(i, j + 1, M))
                    g.addEdge(k, k + 1);
                if (isSafe(i, j - 1, M))
                    g.addEdge(k, k - 1);
                if (i < N - 1 && isSafe(i + 1, j, M))
                    g.addEdge(k, k + N);
                if (i > 0 && isSafe(i - 1, j, M))
                    g.addEdge(k, k - N);
            }
 
            // Source index
            if (M[i][j] == 1)
                s = k;
 
            // Destination index
            if (M[i][j] == 2)
                d = k;
            k++;
        }
    }
 
    // find path Using BFS
    return g.BFS(s, d);
}
 
// driver program to check
// above function
int main()
{
    int M[N][N] = { { 0, 3, 0, 1 },
                    { 3, 0, 3, 3 },
                    { 2, 3, 3, 3 },
                    { 0, 3, 3, 3 } };
 
    (findPath(M) == true) ? cout << "Yes" : cout << "No" << endl;
 
    return 0;
}

Java




// Java program to find path between two
// cell in matrix
import java.util.*;
 
class Graph {
    int V;
    List<List<Integer> > adj;
 
    Graph(int V)
    {
        this.V = V;
        adj = new ArrayList<>(V);
        for (int i = 0; i < V; i++) {
            adj.add(i, new ArrayList<>());
        }
    }
 
    // add edge to graph
    void addEdge(int s, int d)
    {
        adj.get(s).add(d);
    }
 
    // BFS function to find path
    // from source to sink
    boolean BFS(int s, int d)
    {
        // Base case
        if (s == d)
            return true;
 
        // Mark all the vertices as not visited
        boolean[] visited = new boolean[V];
 
        // Create a queue for BFS
        Queue<Integer> queue
            = new LinkedList<>();
 
        // Mark the current node as visited and
        // enqueue it
        visited[s] = true;
        queue.offer(s);
 
        // it will be used to get all adjacent
        // vertices of a vertex
        List<Integer> edges;
 
        while (!queue.isEmpty()) {
            // Dequeue a vertex from queue
            s = queue.poll();
 
            // Get all adjacent vertices of the
            // dequeued vertex s. If a adjacent has
            // not been visited, then mark it visited
            // and enqueue it
            edges = adj.get(s);
            for (int curr : edges) {
                // If this adjacent node is the
                // destination node, then return true
                if (curr == d)
                    return true;
 
                // Else, continue to do BFS
                if (!visited[curr]) {
                    visited[curr] = true;
                    queue.offer(curr);
                }
            }
        }
 
        // If BFS is complete without visiting d
        return false;
    }
 
    static boolean isSafe(
        int i, int j, int[][] M)
    {
        int N = M.length;
        if (
            (i < 0 || i >= N)
            || (j < 0 || j >= N)
            || M[i][j] == 0)
            return false;
        return true;
    }
 
    // Returns true if there is a
    // path from a source (a
    // cell with value 1) to a
    // destination (a cell with
    // value 2)
    static boolean findPath(int[][] M)
    {
        // Source and destination
        int s = -1, d = -1;
        int N = M.length;
        int V = N * N + 2;
        Graph g = new Graph(V);
 
        // Create graph with n*n node
        // each cell consider as node
        int k = 1; // Number of current vertex
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                if (M[i][j] != 0) {
 
                    // connect all 4 adjacent
                    // cell to current cell
                    if (isSafe(i, j + 1, M))
                        g.addEdge(k, k + 1);
                    if (isSafe(i, j - 1, M))
                        g.addEdge(k, k - 1);
                    if (i < N - 1
                        && isSafe(i + 1, j, M))
                        g.addEdge(k, k + N);
                    if (i > 0 && isSafe(i - 1, j, M))
                        g.addEdge(k, k - N);
                }
 
                // source index
                if (M[i][j] == 1)
                    s = k;
 
                // destination index
                if (M[i][j] == 2)
                    d = k;
                k++;
            }
        }
 
        // find path Using BFS
        return g.BFS(s, d);
    }
 
    // Driver program to check above function
    public static void main(
        String[] args) throws Exception
    {
        int[][] M = { { 0, 3, 0, 1 },
                      { 3, 0, 3, 3 },
                      { 2, 3, 3, 3 },
                      { 0, 3, 3, 3 } };
 
        System.out.println(
            ((findPath(M)) ? "Yes" : "No"));
    }
}
 
// This code is contributed by abhay379201

Python3




# Python3 program to find path between two
# cell in matrix
from collections import defaultdict
class Graph:
    def __init__(self):
        self.graph = defaultdict(list)
     
    # add edge to graph
    def addEdge(self, u, v):
        self.graph[u].append(v)
 
    # BFS function to find path from source to sink    
    def BFS(self, s, d):
         
        # Base case
        if s == d:
            return True
             
        # Mark all the vertices as not visited
        visited = [False]*(len(self.graph) + 1)
 
        # Create a queue for BFS
        queue = []
        queue.append(s)
 
        # Mark the current node as visited and
        # enqueue it
        visited[s] = True
        while(queue):
 
            # Dequeue a vertex from queue
            s = queue.pop(0)
 
            # Get all adjacent vertices of the
            # dequeued vertex s. If a adjacent has
            # not been visited, then mark it visited
            # and enqueue it
            for i in self.graph[s]:
                 
                # If this adjacent node is the destination
                # node, then return true
                if i == d:
                    return True
 
                # Else, continue to do BFS
                if visited[i] == False:
                    queue.append(i)
                    visited[i] = True
 
        # If BFS is complete without visiting d
        return False
 
def isSafe(i, j, matrix):
    if i >= 0 and i <= len(matrix) and j >= 0 and j <= len(matrix[0]):
        return True
    else:
        return False
 
# Returns true if there is a path from a source (a
# cell with value 1) to a destination (a cell with
# value 2)
def findPath(M):
    s, d = None, None # source and destination
    N = len(M)
    g = Graph()
 
    # create graph with n * n node
    # each cell consider as node
    k = 1 # Number of current vertex
    for i in range(N):
        for j in range(N):
            if (M[i][j] != 0):
 
                # connect all 4 adjacent cell to
                # current cell
                if (isSafe(i, j + 1, M)):
                    g.addEdge(k, k + 1)
                if (isSafe(i, j - 1, M)):
                    g.addEdge(k, k - 1)
                if (isSafe(i + 1, j, M)):
                    g.addEdge(k, k + N)
                if (isSafe(i - 1, j, M)):
                    g.addEdge(k, k - N)
             
            if (M[i][j] == 1):
                s = k
 
            # destination index    
            if (M[i][j] == 2):
                d = k
            k += 1
 
    # find path Using BFS
    return g.BFS(s, d)
 
# Driver code
if __name__=='__main__':
    M =[[0, 3, 0, 1], [3, 0, 3, 3], [2, 3, 3, 3], [0, 3, 3, 3]]
    if findPath(M):
        print("Yes")
    else:
        print("No")
 
# This Code is Contributed by Vikash Kumar 37
Output



Yes

Complexity Analysis:  

  • Time Complexity: O(n*m). 
    Every cell of the matrix is visited only once so the time complexity is O(n*m).
  • Space Complexity: O(n*m). 
    Space is required to store the visited array and to create the queue.

Simple and Efficient solution: The graph is the matrix in itself.

Approach: The idea is to use Breadth-First Search on the matrix itself.

Consider a cell=(i,j) as a vertex v in the BFS queue. A new vertex u is placed in the BFS queue if u=(i+1,j) or u=(i-1,j) or u=(i,j+1) or u=(i,j-1). Starting the BFS algorithm from cell=(i,j) such that M[i][j] is 1 and stopping either if there was a reachable vertex u=(i,j) such that M[i][j] is 2 and returning true or every cell was covered and there was no such a cell and returning false.

Algorithm:  

1) Create BFS queue q

2) scan the matrix, if there exists a cell in the matrix such that its value is 1 then push it to q

3) run BFS algorithm with q, skipping cells which are not valid. i.e: they are walls (value is 0) or outside the matrix bounds and marking them as walls upon successful visitation.

   3.1) if in the BFS algorithm process there was a vertex x=(i,j) such that M[i][j] is 2 stop and return true

4) BFS algorithm terminated without returning true then there was no element M[i][j] which is 2, then return false

C++




#include <iostream>
#include <queue>
using namespace std;
#define R 4
#define C 4
 
// Structure to define a vertex u=(i,j)
typedef struct BFSElement {
    BFSElement(int i, int j)
    {
        this->i = i;
        this->j = j;
    }
    int i;
    int j;
} BFSElement;
 
bool findPath(int M[R][C])
{
    // 1) Create BFS queue q
    queue<BFSElement> q;
 
    // 2)scan the matrix
    for (int i = 0; i < R; ++i) {
        for (int j = 0; j < C; ++j) {
           
            // if there exists a cell in the matrix such
            // that its value is 1 then push it to q
            if (M[i][j] == 1) {
                q.push(BFSElement(i, j));
                break;
            }
        }
    }
   
    // 3) run BFS algorithm with q.
    while (!q.empty()) {
        BFSElement x = q.front();
        q.pop();
        int i = x.i;
        int j = x.j;
       
        // skipping cells which are not valid.
        // if outside the matrix bounds
        if (i < 0 || i > R || j < 0 || j > C)
            continue;
       
        // if they are walls (value is 0).
        if (M[i][j] == 0)
            continue;
 
        // 3.1) if in the BFS algorithm process there was a
        // vertex x=(i,j) such that M[i][j] is 2 stop and
        // return true
        if (M[i][j] == 2)
            return true;
       
        // marking as wall upon successful visitation
        M[i][j] = 0;
 
        // pushing to queue u=(i,j+1),u=(i,j-1)
        //                 u=(i+1,j),u=(i-1,j)
        for (int k = -1; k <= 1; k += 2) {
            q.push(BFSElement(i + k, j));
            q.push(BFSElement(i, j + k));
        }
    }
   
    // BFS algorithm terminated without returning true
    // then there was no element M[i][j] which is 2, then
    // return false
    return false;
}
 
// Main Driver code
int main()
{
 
    int M[R][C] = { { 0, 3, 0, 1 },
                    { 3, 0, 3, 3 },
                    { 2, 3, 3, 3 },
                    { 0, 3, 3, 3 } };
 
    (findPath(M) == true) ? cout << "Yes"
                          : cout << "No" << endl;
 
    return 0;
}

Java




import java.io.*;
import java.util.*;
 
class BFSElement
{
    int i, j;
    BFSElement(int i, int j)
    {
        this.i = i;
        this.j = j;
    }
}
 
class GFG {
    static int R = 4, C = 4;
    BFSElement b;
     
    static boolean findPath(int M[][])
    {
       
        // 1) Create BFS queue q
        Queue<BFSElement> q = new LinkedList<>();
       
        // 2)scan the matrix
        for (int i = 0; i < R; ++i)
        {
            for (int j = 0; j < C; ++j)
            {
                
                // if there exists a cell in the matrix such
                // that its value is 1 then push it to q
                if (M[i][j] == 1) {
                    q.add(new BFSElement(i, j));
                    break;
                }
            }
         
        }
     
        // 3) run BFS algorithm with q.
        while (q.size() != 0)
        {
            BFSElement x = q.peek();
            q.remove();
            int i = x.i;
            int j = x.j;
           
            // skipping cells which are not valid.
            // if outside the matrix bounds
            if (i < 0 || i >= R || j < 0 || j >= C)
                continue;
            
            // if they are walls (value is 0).
            if (M[i][j] == 0)
                continue;
      
            // 3.1) if in the BFS algorithm process there was a
            // vertex x=(i,j) such that M[i][j] is 2 stop and
            // return true
            if (M[i][j] == 2)
                return true;
            
            // marking as wall upon successful visitation
            M[i][j] = 0;
      
            // pushing to queue u=(i,j+1),u=(i,j-1)
            // u=(i+1,j),u=(i-1,j)
            for (int k = -1; k <= 1; k += 2)
            {
                q.add(new BFSElement(i + k, j));
                q.add(new BFSElement(i, j + k));
            }
        }
             
        // BFS algorithm terminated without returning true
        // then there was no element M[i][j] which is 2, then
        // return false
        return false;
     
    }
   
    // Main Driver code
    public static void main (String[] args)
    {
        int M[][] = { { 0, 3, 0, 1 },
                    { 3, 0, 3, 3 },
                    { 2, 3, 3, 3 },
                    { 0, 3, 3, 3 } };
         
        if(findPath(M) == true)
            System.out.println("Yes");
        else
            System.out.println("No");     
    }
}
 
// This code is contributed by avanitrachhadiya2155

C#




using System;
using System.Collections.Generic;
 
public class BFSElement
{
  public int i, j;
  public BFSElement(int i, int j)
  {
    this.i = i;
    this.j = j;
  }
}
 
public class GFG
{   
  static int R = 4, C = 4;  
  static bool findPath(int[,] M)
  {
 
    // 1) Create BFS queue q
    Queue<BFSElement> q = new Queue<BFSElement>();
 
    // 2)scan the matrix
    for (int i = 0; i < R; ++i)
    {
      for (int j = 0; j < C; ++j)
      {
 
        // if there exists a cell in the matrix such
        // that its value is 1 then push it to q
        if (M[i, j] == 1) {
          q.Enqueue(new BFSElement(i, j));
          break;
        }
      }
 
    }
 
    // 3) run BFS algorithm with q.
    while (q.Count != 0)
    {
      BFSElement x = q.Peek();
      q.Dequeue();
      int i = x.i;
      int j = x.j;
 
      // skipping cells which are not valid.
      // if outside the matrix bounds
      if (i < 0 || i >= R || j < 0 || j >= C)
        continue;
 
      // if they are walls (value is 0).
      if (M[i, j] == 0)
        continue;
 
      // 3.1) if in the BFS algorithm process there was a
      // vertex x=(i,j) such that M[i][j] is 2 stop and
      // return true
      if (M[i, j] == 2)
        return true;
 
      // marking as wall upon successful visitation
      M[i, j] = 0;
 
      // pushing to queue u=(i,j+1),u=(i,j-1)
      // u=(i+1,j),u=(i-1,j)
      for (int k = -1; k <= 1; k += 2)
      {
        q.Enqueue(new BFSElement(i + k, j));
        q.Enqueue(new BFSElement(i, j + k));
      }
    }
 
    // BFS algorithm terminated without returning true
    // then there was no element M[i][j] which is 2, then
    // return false
    return false;
 
  }
 
  // Main Driver code
  static public void Main (){
    int[,] M = { { 0, 3, 0, 1 },
                { 3, 0, 3, 3 },
                { 2, 3, 3, 3 },
                { 0, 3, 3, 3 } };
 
    if(findPath(M) == true)
      Console.WriteLine("Yes");
    else
      Console.WriteLine("No");  
  }
}
 
// This code is contributed by rag2127
Output
Yes

Time Complexity:  O(n*m).

Space Complexity: O(n*m).

The improvement is contributed by Ephi F.

This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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