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.
// 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++
// 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
# 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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