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 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 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.
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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)) {
                        // 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 */

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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).
    The 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 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 notedown 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 queue is greater than 0
  4. Remove the front node of the queue and check if the node is destination, if destination return true. mark the node
  5. Check all adjacent cells if unvisited and blank insert them in the queue.
  6. If destination is not reached return true.

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])
{
    // 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 (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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Java

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// 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);
        adj.get(d).add(s);
    }
  
    // 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 (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
    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

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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

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).
    The space is required to store the visited array and to create the queue.

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