Find whether there is path between two cells in matrix

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 is only single source and single destination(sink).

Examples:



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

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

Asked in: Adobe Interview

Simple solution is that find the source index of cell in matrix and then recursively find a path from source index to destination in matrix .
algorithm :

Find source index in matrix , let we consider ( i , j )
After that Find path from source(1) to sink(2)
FindPathUtil ( M[][N] , i  , j )

   IF we seen M[i][j] == 0 (wall) ||
      ((i, j) is out of matrix index)
     return false ;
   IF we seen destination M[i][j] == 2
     return true ;
   
   Next move in path by traverse all 4 adjacent  
   cell of current cell
   IF (FindPathUtil(M[][N], i+1, j) ||
       FindPathUtil(M[][N], i-1, j) ||
       FindPathUtil(M[][N], i, j+1) ||
       FindPathUtil(M[][N], i, j-1));
      return true ;

 return false ;

Below is the implementation of above approach.

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// 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))
                {
                        flag=true; // if path exists
                        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;
        } 
        return false; // no path has been found 
    } 
      
    // 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}};
                          
    isPath(matrix, 4);                 // calling isPath method 
    }
}
  
/* This code is contributed by Madhu Priya */

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Efficient solution (Graph)

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 total number of Node is N*N.


 1. Create an empty Graph having N*N node ( Vertex ).
 2. push all  node into graph.
 3. notedown source and sink vertex
 4. Now Appling BFS to find is there is path between
    to vertex or not in  graph
        IF Path found return true
        Else return False



below are the implementations of above idea.




C++















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// 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); 


    adj[d].push_back(s); 


} 


  


// 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]) 


{ 


    int s , d ; // source and destination 


    int V = N*N+2; 


    Graph g(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 (j< 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; 


} 



















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Python3














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# 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 



















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Output:


 Yes



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