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 :
- Traverse the matrix and find the starting index of the matrix.
- Create a recursive function that takes the index and visited matrix.
- Mark the current cell and check if the current cell is a destination or not. If the current cell is destination return true.
- Call the recursion function for all adjacent empty and unvisited cells.
- If any of the recursive function returns true then unmark the cell and return true else unmark the cell and return false.
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 |
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:
- Create an empty Graph having N*N node(Vertex), push all nodes into a graph, and note down the source and sink vertex.
- Now apply BFS on the graph, create a queue and insert the source node in the queue
- Run a loop till the size of the queue is greater than 0
- Remove the front node of the queue and check if the node is the destination if the destination returns true. mark the node
- Check all adjacent cells if unvisited and blank insert them in the queue.
- 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 |
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.
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