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).
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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 functions 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 boundariesbool 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 boundaries, 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 notvoid 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 functionint 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 matrixclass 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 boundaries 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 boundaries, 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 notdef 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 boundariesdef 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 boundaries, 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 codeif __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 matrixusing System;class GFG{// Method for finding and printing// whether the path exists or notstatic 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 boundariespublic 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 boundaries, 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 |
Javascript
<script>// JavaScript program to find path between two// cell in matrix // Method for finding and printing // whether the path exists or notfunction isPath(matrix,n){ // Defining visited array to keep // track of already visited indexes let visited = new Array(n); for(let i=0;i<n;i++) { visited[i]=new Array(n); for(let j=0;j<n;j++) { visited[i][j]=false; } } // Flag to indicate whether the // path exists or not let flag = false; for (let i = 0; i < n; i++) { for (let 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 (checkPath( matrix, i, j, visited)) { // if path exists flag = true; break; } } } if (flag) document.write("YES<br>"); else document.write("NO<br>");}// Method for checking boundariesfunction isSafe(i,j,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)function checkPath(matrix,i,j,visited){ // Checking the boundaries, 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 let up = checkPath( matrix, i - 1, j, visited); // if path is found in up // direction return true if (up) return true; // traverse left let left = checkPath( matrix, i, j - 1, visited); // if path is found in left // direction return true if (left) return true; // traverse down let down = checkPath( matrix, i + 1, j, visited); // if path is found in down // direction return true if (down) return true; // traverse right let 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 program to// check above functionlet 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 ab2127</script> |
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:
- 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 4class 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 graphvoid Graph::addEdge(int s, int d){ adj[s].push_back(d);}// BFS function to find path// from source to sinkbool 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 functionint 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 matriximport 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 matrixfrom collections import defaultdictclass 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 Falsedef 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 codeif __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 |
Javascript
<script>// JavaScript program to find path between two// cell in matrixlet V;let adj=[];function Graph(v){ V=v; for (let i = 0; i < V; i++) { adj.push([]); }}// add edge to graphfunction addEdge(s,d){ adj[s].push(d);}// BFS function to find path // from source to sinkfunction BFS(s,d){ // Base case if (s == d) return true; // Mark all the vertices as not visited let visited = new Array(V); for(let i=0;i<V;i++) { visited[i]=false; } // Create a queue for BFS let queue=[]; // Mark the current node as visited and // enqueue it visited[s] = true; queue.push(s); // it will be used to get all adjacent // vertices of a vertex let edges; while (queue.length!=0) { // Dequeue a vertex from queue s = queue.shift(); // 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[s]; for (let curr=0;curr< edges.length;curr++) { // If this adjacent node is the // destination node, then return true if (edges[curr] == d) return true; // Else, continue to do BFS if (!visited[edges[curr]]) { visited[edges[curr]] = true; queue.push(edges[curr]); } } } // If BFS is complete without visiting d return false;}function isSafe(i,j,M){ let 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)function findPath(M){ // Source and destination let s = -1, d = -1; let N = M.length; let V = N * N + 2; Graph(V); // Create graph with n*n node // each cell consider as node let k = 1; // Number of current vertex for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) { if (M[i][j] != 0) { // connect all 4 adjacent // cell to current cell if (isSafe(i, j + 1, M)) addEdge(k, k + 1); if (isSafe(i, j - 1, M)) addEdge(k, k - 1); if (i < N - 1 && isSafe(i + 1, j, M)) addEdge(k, k + N); if (i > 0 && isSafe(i - 1, j, M)) 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 BFS(s, d);} // Driver program to check above functionlet M = [[ 0, 3, 0, 1 ], [ 3, 0, 3, 3 ], [ 2, 3, 3, 3 ], [ 0, 3, 3, 3 ]];document.write(((findPath(M)) ? "Yes" : "No"));// This code is contributed by patel2127</script> |
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 codeint 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 |
Javascript
<script>class BFSElement{ constructor(i,j) { this.i=i; this.j=j; }}let R = 4, C = 4;let b;function findPath(M){ // 1) Create BFS queue q let q = []; // 2)scan the matrix for (let i = 0; i < R; ++i) { for (let 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(new BFSElement(i, j)); break; } } } // 3) run BFS algorithm with q. while (q.length != 0) { let x = q.shift(); let i = x.i; let 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 (let k = -1; k <= 1; k += 2) { q.push(new BFSElement(i + k, j)); q.push(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 codelet M=[[ 0, 3, 0, 1 ], [ 3, 0, 3, 3 ], [ 2, 3, 3, 3 ], [ 0, 3, 3, 3 ]];if(findPath(M) == true) document.write("Yes");else document.write("No");// This code is contributed by unknown2108</script> |
Output
Yes
Time Complexity: O(n*m).
Space Complexity: O(n*m).
The improvement is contributed by Ephi F.
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