# 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:

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 a 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.

## 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`

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:

1. Create an empty Graph having N*N node(Vertex), push all nodes into a graph, and note down the 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 the queue is greater than 0
4. Remove the front node of the queue and check if the node is the destination if the destination returns true. mark the node
5. Check all adjacent cells if unvisited and blank insert them in the queue.
6. If the destination is not reached return true.

## C++

 `// C++ program to find path` `// between two cell in matrix` `#include ` `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 > 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 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 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`

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.

This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.