Find whether there is path between two cells in matrix
Last Updated :
23 Jul, 2025
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:
Find whether there is path between two cells in matrix using Recursion:
The idea is to find the source index of the cell in each matrix and then recursively find a path from the source index to the destination in the matrix. The algorithm involves recursively finding all the paths until a final path is found to the destination.
Follow the steps below to solve the problem:
- 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 the 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.
Below is the implementation of the above approach.
C++
// C++ program to find path between two
// cell in matrix
#include <bits/stdc++.h>
using namespace std;
#define N 4
// Method for checking boundaries
bool 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 isaPath(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 = isaPath(matrix, i - 1, j, visited);
// if path is found in up
// direction return true
if (up)
return true;
// traverse left
bool left = isaPath(matrix, i, j - 1, visited);
// if path is found in left
// direction return true
if (left)
return true;
// traverse down
bool down = isaPath(matrix, i + 1, j, visited);
// if path is found in down
// direction return true
if (down)
return true;
// traverse right
bool right = isaPath(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 not
void isPath(int matrix[][N])
{
// Defining visited array to keep
// track of already visited indexes
bool visited[N][N];
memset(visited, 0, sizeof(visited));
// 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 (isaPath(matrix, i, j, visited)) {
// if path exists
flag = true;
break;
}
}
}
if (flag)
cout << "YES";
else
cout << "NO";
}
// Driver program to
// check above function
int 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 program to find path between two
// cell in matrix
import java.io.*;
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 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 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 boundaries
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 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 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# 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 boundaries
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 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
<script>
// JavaScript program to find path between two
// cell in matrix
// Method for finding and printing
// whether the path exists or not
function 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 boundaries
function 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 function
let 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>
Time Complexity: O(N*M), In the worst case, we have to visit each cell only one time because we keep the visited array for not visiting the already visited cell.
Auxiliary Space: O(N*M), Space is required to store the visited array.
Find whether there is path between two cells in matrix using Breadth First Search:
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.
Follow the steps below to solve the problem:
- 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 reached return true.
Below is the implementation of the above approach.
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 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 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
// C# program to find path between two
// cell in matrix
using System;
using System.Collections.Generic;
class Graph {
public int V;
public List<List<int> > adj;
public Graph(int V)
{
this.V = V;
adj = new List<List<int>>();
for (int i = 0; i < V; i++) {
adj.Add(new List<int>());
}
}
// Add edge to graph
public void AddEdge(int s, int d) { adj[s].Add(d); }
// BFS function to find path
// from source to sink
public bool BFS(int s, int d)
{
// Base case
if (s == d)
return true;
// Mark all the vertices as not visited
bool[] visited = new bool[V];
// Create a queue for BFS
List<int> queue = new List<int>();
// Mark the current node as visited and
// enqueue it
visited[s] = true;
queue.Add(s);
// it will be used to get all adjacent
// vertices of a vertex
List<int> edges = new List<int>();
while (queue.Count != 0 ) {
// Dequeue a vertex from queue
s = queue[0];
queue.RemoveAt(0);
// 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];
foreach (int curr in 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.Add(curr);
}
}
}
// If BFS is complete without visiting d
return false;
}
static bool isSafe(int i, int j, int[, ] M)
{
if ((i < 0 || i >= M.GetLength(0)) || (j < 0 || j >= M.GetLength(1))
|| 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 bool findPath(int[, ] M)
{
// Source and destination
int s = -1, d = -1;
int N = M.GetLength(0);
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)
{
int[, ] M = { { 0, 3, 0, 1 },
{ 3, 0, 3, 3 },
{ 2, 3, 3, 3 },
{ 0, 3, 3, 3 } };
Console.WriteLine(((findPath(M)) ? "Yes" : "No"));
}
}
// This code is contributed by phasing17
<script>
// JavaScript program to find path between two
// cell in matrix
let V;
let adj=[];
function Graph(v)
{
V=v;
for (let i = 0; i < V; i++)
{
adj.push([]);
}
}
// add edge to graph
function addEdge(s,d)
{
adj[s].push(d);
}
// BFS function to find path
// from source to sink
function 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 function
let 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>
Time Complexity: O(N*M), Every cell of the matrix is visited only once so the time complexity is O(N*M).
Auxiliary Space: O(N*M), Space is required to store the visited array and to create the queue.
Find whether there is path between two cells in matrix using Breadth First Search (On matrix):
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.
Follow the steps below to solve the problem:
- Create BFS queue q
- scan the matrix, if there exists a cell in the matrix such that its value is 1 then push it to q
- Run BFS algorithm with q, skipping cells that are not valid. i.e: they are walls (value is 0) or outside the matrix bounds and marking them as walls upon successful visitation.
- If in the BFS algorithm process there was a vertex x=(i,j) such that M[i][j] is 2 stop and return true.
- BFS algorithm terminated without returning true then there was no element M[i][j] which is 2, then return false
Below is the implementation of the above approach:
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)
{
// 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 code
int 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;
}
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
class BFSElement:
def __init__(self, i, j):
self.i = i
self.j = j
R, C = 4, 4
def findPath(M):
# 1) Create BFS queue q
q = []
# 2)scan the matrix
for i in range(R):
for j in range(C):
# if there exists a cell in the matrix such
# that its value is 1 then append it to q
if (M[i][j] == 1):
q.append(BFSElement(i, j))
break
# 3) run BFS algorithm with q.
while (len(q) != 0):
x = q[0]
q = q[1:]
i = x.i
j = x.j
# skipping cells which are not valid.
# if outside the matrix bounds
if (i < 0 or i >= R or j < 0 or 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
# appending to queue u=(i,j+1),u=(i,j-1)
# u=(i+1,j),u=(i-1,j)
for k in range(-1, 2, 2):
q.append(BFSElement(i + k, j))
q.append(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
M = [[0, 3, 0, 1],
[3, 0, 3, 3],
[2, 3, 3, 3],
[0, 3, 3, 3]]
if(findPath(M) == True):
print("Yes")
else:
print("No")
# This code is contributed by shinjanpatra
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
<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 code
let 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>
Time Complexity: O(N*M), Every cell of the matrix is visited only once so the time complexity is O(N*M).
Auxiliary Space: O(N*M), Space is required to store the visited array and to create the queue.
Find whether path exist | DSA Problem
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem