Given a square maze containing positive numbers, find all paths from a corner cell (any of the extreme four corners) to the middle cell. We can move exactly n steps from a cell in 4 directions i.e. North, East, West and South where n is value of the cell,
We can move to mat[i+n][j], mat[i-n][j], mat[i][j+n], and mat[i][j-n] from a cell mat[i][j] where n is value of mat[i][j].
Example
Input: 9 x 9 maze [ 3, 5, 4, 4, 7, 3, 4, 6, 3 ] [ 6, 7, 5, 6, 6, 2, 6, 6, 2 ] [ 3, 3, 4, 3, 2, 5, 4, 7, 2 ] [ 6, 5, 5, 1, 2, 3, 6, 5, 6 ] [ 3, 3, 4, 3, 0, 1, 4, 3, 4 ] [ 3, 5, 4, 3, 2, 2, 3, 3, 5 ] [ 3, 5, 4, 3, 2, 6, 4, 4, 3 ] [ 3, 5, 1, 3, 7, 5, 3, 6, 4 ] [ 6, 2, 4, 3, 4, 5, 4, 5, 1 ] Output: (0, 0) -> (0, 3) -> (0, 7) -> (6, 7) -> (6, 3) -> (3, 3) -> (3, 4) -> (5, 4) -> (5, 2) -> (1, 2) -> (1, 7) -> (7, 7) -> (7, 1) -> (2, 1) -> (2, 4) -> (4, 4) -> MID
The idea is to use backtracking. We start with each corner cell of the maze and recursively checks if it leads to the solution or not. Following is the Backtracking algorithm –
If destination is reached
- print the path
Else
- Mark current cell as visited and add it to path array.
- Move forward in all 4 allowed directions and recursively check if any of them leads to a solution.
- If none of the above solutions work then mark this cell as not visited and remove it from path array.
Below is the implementation of the above approach:
// C++ program to find a path from corner cell to // middle cell in maze containing positive numbers #include <bits/stdc++.h> using namespace std;
// Rows and columns in given maze #define N 9 // check whether given cell is a valid cell or not. bool isValid(set<pair< int , int > > visited,
pair< int , int > pt)
{ // check if cell is not visited yet to
// avoid cycles (infinite loop) and its
// row and column number is in range
return (pt.first >= 0) && (pt.first < N) &&
(pt.second >= 0) && (pt.second < N) &&
(visited.find(pt) == visited.end());
} // Function to print path from source to middle coordinate void printPath(list<pair< int , int > > path)
{ for ( auto it = path.begin(); it != path.end(); it++)
cout << "(" << it->first << ", "
<< it->second << ") -> " ;
cout << "MID" << endl << endl;
} // For searching in all 4 direction int row[] = {-1, 1, 0, 0};
int col[] = { 0, 0, -1, 1};
// Coordinates of 4 corners of matrix int _row[] = { 0, 0, N-1, N-1};
int _col[] = { 0, N-1, 0, N-1};
void findPathInMazeUtil( int maze[N][N],
list<pair< int , int > > &path,
set<pair< int , int > > &visited,
pair< int , int > &curr)
{ // If we have reached the destination cell.
// print the complete path
if (curr.first == N / 2 && curr.second == N / 2)
{
printPath(path);
return ;
}
// consider each direction
for ( int i = 0; i < 4; ++i)
{
// get value of current cell
int n = maze[curr.first][curr.second];
// We can move N cells in either of 4 directions
int x = curr.first + row[i]*n;
int y = curr.second + col[i]*n;
// Constructs a pair object with its first element
// set to x and its second element set to y
pair< int , int > next = make_pair(x, y);
// if valid pair
if (isValid(visited, next))
{
// mark cell as visited
visited.insert(next);
// add cell to current path
path.push_back(next);
// recurse for next cell
findPathInMazeUtil(maze, path, visited, next);
// backtrack
path.pop_back();
// remove cell from current path
visited.erase(next);
}
}
} // Function to find a path from corner cell to // middle cell in maze containing positive numbers void findPathInMaze( int maze[N][N])
{ // list to store complete path
// from source to destination
list<pair< int , int > > path;
// to store cells already visited in current path
set<pair< int , int > > visited;
// Consider each corners as the starting
// point and search in maze
for ( int i = 0; i < 4; ++i)
{
int x = _row[i];
int y = _col[i];
// Constructs a pair object
pair< int , int > pt = make_pair(x, y);
// mark cell as visited
visited.insert(pt);
// add cell to current path
path.push_back(pt);
findPathInMazeUtil(maze, path, visited, pt);
// backtrack
path.pop_back();
// remove cell from current path
visited.erase(pt);
}
} int main()
{ int maze[N][N] =
{
{ 3, 5, 4, 4, 7, 3, 4, 6, 3 },
{ 6, 7, 5, 6, 6, 2, 6, 6, 2 },
{ 3, 3, 4, 3, 2, 5, 4, 7, 2 },
{ 6, 5, 5, 1, 2, 3, 6, 5, 6 },
{ 3, 3, 4, 3, 0, 1, 4, 3, 4 },
{ 3, 5, 4, 3, 2, 2, 3, 3, 5 },
{ 3, 5, 4, 3, 2, 6, 4, 4, 3 },
{ 3, 5, 1, 3, 7, 5, 3, 6, 4 },
{ 6, 2, 4, 3, 4, 5, 4, 5, 1 }
};
findPathInMaze(maze);
return 0;
} |
# Python program to find a path from corner cell to # middle cell in maze containing positive numbers # Rows and columns in given maze N = 9
# check whether given cell is a valid cell or not. def isValid(visited, pt):
# check if cell is not visited yet to
# avoid cycles (infinite loop) and its
# row and column number is in range
return (pt[ 0 ] > = 0 ) and (pt[ 0 ] < N) and (pt[ 1 ] > = 0 ) and (pt[ 1 ] < N) and (pt not in visited)
# Function to print path from source to middle coordinate def printPath(path):
for i in path:
print ( "({}, {}) -> " . format (i[ 0 ], i[ 1 ]), end = "")
print ( "MID" )
print ()
# For searching in all 4 direction row = [ - 1 , 1 , 0 , 0 ]
col = [ 0 , 0 , - 1 , 1 ]
# Coordinates of 4 corners of matrix _row = [ 0 , 0 , N - 1 , N - 1 ]
_col = [ 0 , N - 1 , 0 , N - 1 ]
def findPathInMazeUtil(maze, path, visited, curr):
# If we have reached the destination cell.
# print the complete path
if curr[ 0 ] = = N / / 2 and curr[ 1 ] = = N / / 2 :
printPath(path)
return
# consider each direction
for i in range ( 4 ):
# get value of current cell
n = maze[curr[ 0 ]][curr[ 1 ]]
# We can move N cells in either of 4 directions
x = curr[ 0 ] + row[i] * n
y = curr[ 1 ] + col[i] * n
next = (x, y)
# if valid pair
if isValid(visited, next ):
# mark cell as visited
visited.append( next )
# add cell to current path
path.append( next )
# recurse for next cell
findPathInMazeUtil(maze, path, visited, next )
# backtrack
# remove cell from current path
path.pop()
visited.remove( next )
# Function to find a path from corner cell to # middle cell in maze containing positive numbers def findPathInMaze(maze):
# list to store complete path
# from source to destination
path = []
# to store cells already visited in current path
visited = []
# Consider each corners as the starting
# point and search in maze
for i in range ( 4 ):
x = _row[i]
y = _col[i]
pt = (x, y)
# mark cell as visited
visited.append(pt)
# add cell to current path
path.append(pt)
findPathInMazeUtil(maze, path, visited, pt)
# backtrack
# remove cell from current path
path.pop()
visited.remove(pt)
if __name__ = = "__main__" :
maze = [
[ 3 , 5 , 4 , 4 , 7 , 3 , 4 , 6 , 3 ],
[ 6 , 7 , 5 , 6 , 6 , 2 , 6 , 6 , 2 ],
[ 3 , 3 , 4 , 3 , 2 , 5 , 4 , 7 , 2 ],
[ 6 , 5 , 5 , 1 , 2 , 3 , 6 , 5 , 6 ],
[ 3 , 3 , 4 , 3 , 0 , 1 , 4 , 3 , 4 ],
[ 3 , 5 , 4 , 3 , 2 , 2 , 3 , 3 , 5 ],
[ 3 , 5 , 4 , 3 , 2 , 6 , 4 , 4 , 3 ],
[ 3 , 5 , 1 , 3 , 7 , 5 , 3 , 6 , 4 ],
[ 6 , 2 , 4 , 3 , 4 , 5 , 4 , 5 , 1 ]
]
findPathInMaze(maze)
# This code is contributed by Vikram_Shirsat |
using System;
using System.Collections.Generic;
namespace MazePathFinder
{ class Program
{
// Rows and columns in given maze
const int N = 9;
// check whether given cell is a valid cell or not.
static bool IsValid(List<( int , int )> visited, ( int , int ) pt)
{
// check if cell is not visited yet to
// avoid cycles (infinite loop) and its
// row and column number is in range
return (pt.Item1 >= 0) && (pt.Item1 < N) && (pt.Item2 >= 0) && (pt.Item2 < N) && (!visited.Contains(pt));
}
// Function to print path from source to middle coordinate
static void PrintPath(List<( int , int )> path)
{
foreach ( var i in path)
{
Console.Write($ "({i.Item1}, {i.Item2}) -> " );
}
Console.WriteLine( "MID" );
Console.WriteLine();
}
// For searching in all 4 direction
static int [] row = { -1, 1, 0, 0 };
static int [] col = { 0, 0, -1, 1 };
// Coordinates of 4 corners of matrix
static int [] _row = { 0, 0, N - 1, N - 1 };
static int [] _col = { 0, N - 1, 0, N - 1 };
static void FindPathInMazeUtil( int [][] maze, List<( int , int )> path, List<( int , int )> visited, ( int , int ) curr)
{
// If we have reached the destination cell.
// print the complete path
if (curr.Item1 == N / 2 && curr.Item2 == N / 2)
{
PrintPath(path);
return ;
}
// consider each direction
for ( int i = 0; i < 4; i++)
{
// get value of current cell
int n = maze[curr.Item1][curr.Item2];
// We can move N cells in either of 4 directions
int x = curr.Item1 + row[i] * n;
int y = curr.Item2 + col[i] * n;
var next = (x, y);
// if valid pair
if (IsValid(visited, next))
{
// mark cell as visited
visited.Add(next);
// add cell to current path
path.Add(next);
// recurse for next cell
FindPathInMazeUtil(maze, path, visited, next);
// backtrack
// remove cell from current path
path.RemoveAt(path.Count - 1);
visited.Remove(next);
}
}
}
// Function to find a path from corner cell to
// middle cell in maze containing positive numbers
static void FindPathInMaze( int [][] maze)
{
// list to store complete path
// from source to destination
var path = new List<( int , int )>();
// to store cells already visited in current path
var visited = new List<( int , int )>();
// Consider each corners as the starting
// point and search in maze
for ( int i = 0; i < 4; i++)
{
int x = _row[i];
int y = _col[i];
var pt = (x, y);
// mark cell as visited
visited.Add(pt);
// add cell to current path
path.Add(pt);
FindPathInMazeUtil(maze, path, visited, pt);
// backtrack
// remove cell from current path
path.RemoveAt(path.Count - 1);
visited.Remove(pt);
}
}
static void Main( string [] args)
{
int [][] maze = new int [][] {
new int [] {3, 5, 4, 4, 7, 3, 4, 6, 3},
new int [] {6, 7, 5, 6, 6, 2, 6, 6, 2},
new int [] {3, 3, 4, 3, 2, 5, 4, 7, 2},
new int [] {6, 5, 5, 1, 2, 3, 6, 5, 6},
new int [] {3, 3, 4, 3, 0, 1, 4, 3, 4},
new int [] {3, 5, 4, 3, 2, 2, 3, 3, 5},
new int [] {3, 5, 4, 3, 2, 6, 4, 4, 3},
new int [] {3, 5, 1, 3, 7, 5, 3, 6, 4},
new int [] {6, 2, 4, 3, 4, 5, 4, 5, 1}
};
FindPathInMaze(maze);
}
}
} |
// JavaScript program to find a path from corner cell to // middle cell in maze containing positive numbers // Rows and columns in given maze let N = 9; // check whether given cell is a valid cell or not. function isValid(visited, pt) {
// check if cell is not visited yet to
// avoid cycles (infinite loop) and its
// row and column number is in range
return (pt[0] >= 0) && (pt[0] < N) && (pt[1] >= 0) && (pt[1] < N) &&
(!visited.some(item => item[0] === pt[0] && item[1] === pt[1]));
} // Function to print path from source to middle coordinate function printPath(path) {
let pathStr = "" ;
for (let i of path) {
pathStr += "(" + i[0] + ", " + i[1] + ") -> " ;
}
pathStr += "MID" ;
console.log(pathStr);
console.log();
} // For searching in all 4 direction const row = [-1, 1, 0, 0]; const col = [0, 0, -1, 1]; // Coordinates of 4 corners of matrix const _row = [0, 0, N-1, N-1]; const _col = [0, N-1, 0, N-1]; function findPathInMazeUtil(maze, path, visited, curr) {
// If we have reached the destination cell.
// print the complete path
if (curr[0] === Math.floor(N / 2) && curr[1] === Math.floor(N / 2)) {
printPath(path);
return ;
}
// consider each direction
for (let i = 0; i < 4; i++) {
// get value of current cell
let n = maze[curr[0]][curr[1]];
// We can move N cells in either of 4 directions
let x = curr[0] + row[i]*n;
let y = curr[1] + col[i]*n;
// Constructs a pair object with its first element
// set to x and its second element set to y
let next = [x, y];
// if valid pair
if (isValid(visited, next)) {
// mark cell as visited
visited.push(next);
// add cell to current path
path.push(next);
// recurse for next cell
findPathInMazeUtil(maze, path, visited, next);
// backtrack
path.pop();
// remove cell from current path
visited = visited.filter(item => item[0] !== next[0] || item[1] !== next[1]);
}
}
} // Function to find a path from corner cell to // middle cell in maze containing positive numbers function findPathInMaze(maze) {
// list to store complete path
// from source to destination
let path = [];
// to store cells already visited in current path
let visited = [];
// Consider each corners as the starting
// point and search in maze
for (let i = 0; i < 4; i++) {
let x = _row[i];
let y = _col[i];
// Constructs a pair object
let pt = [x, y];
// mark cell as visited
visited.push(pt);
// add cell to current path
path.push(pt);
findPathInMazeUtil(maze, path, visited, pt);
// backtrack
path.pop();
// remove cell from current path
visited = visited.filter(item => item[0] !== pt[0] || item[1] !== pt[1]);
}
} let maze = [ [3, 5, 4, 4, 7, 3, 4, 6, 3],
[6, 7, 5, 6, 6, 2, 6, 6, 2],
[3, 3, 4, 3, 2, 5, 4, 7, 2],
[6, 5, 5, 1, 2, 3, 6, 5, 6],
[3, 3, 4, 3, 0, 1, 4, 3, 4],
[3, 5, 4, 3, 2, 2, 3, 3, 5],
[3, 5, 4, 3, 2, 6, 4, 4, 3],
[3, 5, 1, 3, 7, 5, 3, 6, 4],
[6, 2, 4, 3, 4, 5, 4, 5, 1]
]
findPathInMaze(maze) // This code is contributed by prasad264 |
import java.util.*;
public class MazePathFinder {
// Rows and columns in given maze
private static int N = 9 ;
// For searching in all 4 direction
private static int [] row = { - 1 , 1 , 0 , 0 };
private static int [] col = { 0 , 0 , - 1 , 1 };
// Coordinates of 4 corners of matrix
private static int [] _row = { 0 , 0 , N - 1 , N - 1 };
private static int [] _col = { 0 , N - 1 , 0 , N - 1 };
// check whether given cell is a valid cell or not.
private static boolean isValid(ArrayList< int []> visited,
int [] pt)
{
// check if cell is not visited yet to
// avoid cycles (infinite loop) and its
// row and column number is in range
return (pt[ 0 ] >= 0 ) && (pt[ 0 ] < N) && (pt[ 1 ] >= 0 )
&& (pt[ 1 ] < N)
&& (!visited.stream().anyMatch(
item
-> item[ 0 ] == pt[ 0 ] && item[ 1 ] == pt[ 1 ]));
}
// Function to print path from source to middle
// coordinate
private static void printPath(ArrayList< int []> path)
{
String pathStr = "" ;
for ( int [] i : path) {
pathStr += "(" + i[ 0 ] + ", " + i[ 1 ] + ") -> " ;
}
pathStr += "MID" ;
System.out.println(pathStr);
System.out.println();
}
private static void
findPathInMazeUtil( int [][] maze, ArrayList< int []> path,
ArrayList< int []> visited, int [] curr)
{
// If we have reached the destination cell.
// print the complete path
if (curr[ 0 ] == (N / 2 ) && curr[ 1 ] == (N / 2 )) {
printPath(path);
return ;
}
// consider each direction
for ( int i = 0 ; i < 4 ; i++) {
// get value of current cell
int n = maze[curr[ 0 ]][curr[ 1 ]];
// We can move N cells in either of 4 directions
int x = curr[ 0 ] + row[i] * n;
int y = curr[ 1 ] + col[i] * n;
int [] next = { x, y };
// if valid pair
if (isValid(visited, next)) {
// mark cell as visited
visited.add(next);
// add cell to current path
path.add(next);
// recurse for next cell
findPathInMazeUtil(maze, path, visited,
next);
// backtrack
// remove cell from current path
path.remove(path.size() - 1 );
visited.removeIf(item
-> item[ 0 ] == next[ 0 ]
&& item[ 1 ]
== next[ 1 ]);
}
}
}
// Function to find a path from corner cell to
// middle cell in maze containing positive numbers
public static void findPathInMaze( int [][] maze)
{
// list to store complete path
// from source to destination
ArrayList< int []> path = new ArrayList<>();
// to store cells already visited in current path
ArrayList< int []> visited = new ArrayList<>();
// Consider each corners as the starting
// point and search in maze
for ( int i = 0 ; i < 4 ; i++) {
int x = _row[i];
int y = _col[i];
int [] pt = { x, y };
// mark cell as visited
visited.add(pt);
// add cell to current path
path.add(pt);
findPathInMazeUtil(maze, path, visited, pt);
// backtrack
// remove cell from current path
path.remove(path.size() - 1 );
visited.removeIf(item
-> item[ 0 ] == pt[ 0 ]
&& item[ 1 ] == pt[ 1 ]);
}
}
public static void main(String[] args)
{
// input maze
int [][] maze = { { 3 , 5 , 4 , 4 , 7 , 3 , 4 , 6 , 3 },
{ 6 , 7 , 5 , 6 , 6 , 2 , 6 , 6 , 2 },
{ 3 , 3 , 4 , 3 , 2 , 5 , 4 , 7 , 2 },
{ 6 , 5 , 5 , 1 , 2 , 3 , 6 , 5 , 6 },
{ 3 , 3 , 4 , 3 , 0 , 1 , 4 , 3 , 4 },
{ 3 , 5 , 4 , 3 , 2 , 2 , 3 , 3 , 5 },
{ 3 , 5 , 4 , 3 , 2 , 6 , 4 , 4 , 3 },
{ 3 , 5 , 1 , 3 , 7 , 5 , 3 , 6 , 4 },
{ 6 , 2 , 4 , 3 , 4 , 5 , 4 , 5 , 1 } };
// find path from corner cell to middle cell in the
// maze
findPathInMaze(maze);
}
} |
(0, 0) -> (0, 3) -> (0, 7) -> (6, 7) -> (6, 3) -> (3, 3) -> (3, 4) -> (5, 4) -> (5, 2) -> (1, 2) -> (1, 7) -> (7, 7) -> (7, 1) -> (2, 1) -> (2, 4) -> (4, 4) -> MID
Output :
(0, 0) -> (0, 3) -> (0, 7) -> (6, 7) -> (6, 3) -> (3, 3) -> (3, 4) -> (5, 4) -> (5, 2) -> (1, 2) -> (1, 7) -> (7, 7) -> (7, 1) -> (2, 1) -> (2, 4) -> (4, 4) -> MID
Time Complexity: O(4^N2), where N is the size of the maze and the program may need to explore all possible paths in the maze before finding the path from the corner to the middle cell.
Space Complexity: O(N2), where N*N is the size of the maze.
A better approach:
// C++ program for the above approach #include<bits/stdc++.h> using namespace std;
void printPath(vector<vector< int >>&maze, int i, int j, string ans){
// If we reach the center cell
if (i == maze.size()/2 && j==maze.size()/2){
// Make the final answer, Print the
// final answer and Return
ans += "(" ;
ans += to_string(i);
ans += ", " ;
ans += to_string(j);
ans += ") -> MID" ;
cout<<ans<<endl;
return ;
}
// If the element at the current position
// in maze is 0, simply Return as it has
// been visited before.
if (maze[i][j]==0){
return ;
}
// If element is non-zero, then note
// the element in variable 'k'
int k = maze[i][j];
// Mark the cell visited by making the
// element 0. Don't worry, the element
// is safe in 'k'
maze[i][j]=0;
// Make recursive calls in all 4
// directions pro-actively i.e. if the next
// cell lies in maze or not. Right call
ans += "(" ;
ans += to_string(i);
ans += ", " ;
ans += to_string(j);
ans += ") -> " ;
if (j+k<maze.size()){
printPath(maze, i, j+k, ans);
}
// down call
if (i+k<maze.size()){
printPath(maze, i+k, j,ans);
}
// left call
if (j-k>0){
printPath(maze, i, j-k, ans);
}
// up call
if (i-k>0){
printPath(maze, i-k, j, ans);
}
// Unmark the visited cell by substituting
// its original value from 'k'
maze[i][j] = k;
} int main () {
// Creating the maze
vector<vector< int >>maze = {
{ 3, 5, 4, 4, 7, 3, 4, 6, 3 },
{ 6, 7, 5, 6, 6, 2, 6, 6, 2 },
{ 3, 3, 4, 3, 2, 5, 4, 7, 2 },
{ 6, 5, 5, 1, 2, 3, 6, 5, 6 },
{ 3, 3, 4, 3, 0, 1, 4, 3, 4 },
{ 3, 5, 4, 3, 2, 2, 3, 3, 5 },
{ 3, 5, 4, 3, 2, 6, 4, 4, 3 },
{ 3, 5, 1, 3, 7, 5, 3, 6, 4 },
{ 6, 2, 4, 3, 4, 5, 4, 5, 1 }
};
// Calling the printPath function
printPath(maze,0,0, "" );
} // This code is contributed by shinjanpatra |
// Java program to find a path from corner cell to // middle cell in maze containing positive numbers import java.io.*;
class GFG {
public static void main (String[] args) {
// Creating the maze
int [][] maze = {
{ 3 , 5 , 4 , 4 , 7 , 3 , 4 , 6 , 3 },
{ 6 , 7 , 5 , 6 , 6 , 2 , 6 , 6 , 2 },
{ 3 , 3 , 4 , 3 , 2 , 5 , 4 , 7 , 2 },
{ 6 , 5 , 5 , 1 , 2 , 3 , 6 , 5 , 6 },
{ 3 , 3 , 4 , 3 , 0 , 1 , 4 , 3 , 4 },
{ 3 , 5 , 4 , 3 , 2 , 2 , 3 , 3 , 5 },
{ 3 , 5 , 4 , 3 , 2 , 6 , 4 , 4 , 3 },
{ 3 , 5 , 1 , 3 , 7 , 5 , 3 , 6 , 4 },
{ 6 , 2 , 4 , 3 , 4 , 5 , 4 , 5 , 1 }
};
// Calling the printPath function
printPath(maze, 0 , 0 , "" );
}
public static void printPath( int [][] maze, int i, int j, String ans){
// If we reach the center cell
if (i == maze.length/ 2 && j==maze.length/ 2 ){
// Make the final answer, Print the
// final answer and Return
ans += "(" +i+ ", " +j+ ") -> MID" ;
System.out.println(ans);
return ;
}
// If the element at the current position
// in maze is 0, simply Return as it has
// been visited before.
if (maze[i][j]== 0 ){
return ;
}
// If element is non-zero, then note
// the element in variable 'k'
int k = maze[i][j];
// Mark the cell visited by making the
// element 0. Don't worry, the element
// is safe in 'k'
maze[i][j]= 0 ;
// Make recursive calls in all 4
// directions pro-actively i.e. if the next
// cell lies in maze or not. Right call
if (j+k<maze.length){
printPath(maze, i, j+k, ans+ "(" +i+ ", " +j+ ") -> " );
}
// down call
if (i+k<maze.length){
printPath(maze, i+k, j, ans+ "(" +i+ ", " +j+ ") -> " );
}
// left call
if (j-k> 0 ){
printPath(maze, i, j-k, ans+ "(" +i+ ", " +j+ ") -> " );
}
// up call
if (i-k> 0 ){
printPath(maze, i-k, j, ans+ "(" +i+ ", " +j+ ") -> " );
}
// Unmark the visited cell by substituting
// its original value from 'k'
maze[i][j] = k;
}
} |
# Python program to find a path from corner cell to # middle cell in maze containing positive numbers def printPath(maze, i, j, ans):
# If we reach the center cell
if (i = = len (maze) / / 2 and j = = len (maze) / / 2 ):
# Make the final answer, Print
# final answer and Return
ans + = "(" + str (i) + ", " + str (j) + ") -> MID" ;
print (ans);
return ;
# If the element at the current position
# in maze is 0, simply Return as it has
# been visited before.
if (maze[i][j] = = 0 ):
return ;
# If element is non-zero, then note
# the element in variable 'k'
k = maze[i][j];
# Mark the cell visited by making the
# element 0. Don't worry, the element
# is safe in 'k'
maze[i][j] = 0 ;
# Make recursive calls in all 4
# directions pro-actively i.e. if the next
# cell lies in maze or not. Right call
if (j + k < len (maze)):
printPath(maze, i, j + k, ans + "(" + str (i) + ", " + str (j) + ") -> " );
# down call
if (i + k < len (maze)):
printPath(maze, i + k, j, ans + "(" + str (i) + ", " + str (j) + ") -> " );
# left call
if (j - k > 0 ):
printPath(maze, i, j - k, ans + "(" + str (i) + ", " + str (j) + ") -> " );
# up call
if (i - k > 0 ):
printPath(maze, i - k, j, ans + "(" + str (i) + ", " + str (j) + ") -> " );
# Unmark the visited cell by substituting
# its original value from 'k'
maze[i][j] = k;
# Driver code
if __name__ = = '__main__' :
# Creating the maze
maze = [[ 3 , 5 , 4 , 4 , 7 , 3 , 4 , 6 , 3 ],[ 6 , 7 , 5 , 6 , 6 , 2 , 6 , 6 , 2 ],[ 3 , 3 , 4 , 3 , 2 , 5 , 4 , 7 , 2 ],
[ 6 , 5 , 5 , 1 , 2 , 3 , 6 , 5 , 6 ],[ 3 , 3 , 4 , 3 , 0 , 1 , 4 , 3 , 4 ],[ 3 , 5 , 4 , 3 , 2 , 2 , 3 , 3 , 5 ],
[ 3 , 5 , 4 , 3 , 2 , 6 , 4 , 4 , 3 ],[ 3 , 5 , 1 , 3 , 7 , 5 , 3 , 6 , 4 ],[ 6 , 2 , 4 , 3 , 4 , 5 , 4 , 5 , 1 ]] ;
# Calling the printPath function
printPath(maze, 0 , 0 , "");
# This code contributed by gauravrajput1 |
// C# program to find a path from corner // cell to middle cell in maze containing // positive numbers using System;
class GFG{
// Driver Code public static void Main(String[] args)
{ // Creating the maze
int [,] maze = {
{ 3, 5, 4, 4, 7, 3, 4, 6, 3 },
{ 6, 7, 5, 6, 6, 2, 6, 6, 2 },
{ 3, 3, 4, 3, 2, 5, 4, 7, 2 },
{ 6, 5, 5, 1, 2, 3, 6, 5, 6 },
{ 3, 3, 4, 3, 0, 1, 4, 3, 4 },
{ 3, 5, 4, 3, 2, 2, 3, 3, 5 },
{ 3, 5, 4, 3, 2, 6, 4, 4, 3 },
{ 3, 5, 1, 3, 7, 5, 3, 6, 4 },
{ 6, 2, 4, 3, 4, 5, 4, 5, 1 }
};
// Calling the printPath function
printPath(maze, 0, 0, "" );
} public static void printPath( int [,] maze, int i,
int j, String ans)
{ // If we reach the center cell
if (i == maze.GetLength(0) / 2 &&
j == maze.GetLength(1) / 2)
{
// Make the readonly answer, Print the
// readonly answer and Return
ans += "(" + i + ", " + j + ") -> MID" ;
Console.WriteLine(ans);
return ;
}
// If the element at the current position
// in maze is 0, simply Return as it has
// been visited before.
if (maze[i, j] == 0)
{
return ;
}
// If element is non-zero, then note
// the element in variable 'k'
int k = maze[i, j];
// Mark the cell visited by making the
// element 0. Don't worry, the element
// is safe in 'k'
maze[i, j] = 0;
// Make recursive calls in all 4
// directions pro-actively i.e. if the next
// cell lies in maze or not. Right call
if (j + k < maze.GetLength(1))
{
printPath(maze, i, j + k,
ans + "(" + i +
", " + j + ") -> " );
}
// Down call
if (i + k < maze.GetLength(0))
{
printPath(maze, i + k, j,
ans + "(" + i +
", " + j + ") -> " );
}
// Left call
if (j - k > 0)
{
printPath(maze, i, j - k,
ans + "(" + i +
", " + j + ") -> " );
}
// Up call
if (i - k > 0)
{
printPath(maze, i - k, j,
ans + "(" + i +
", " + j + ") -> " );
}
// Unmark the visited cell by substituting
// its original value from 'k'
maze[i, j] = k;
} } // This code is contributed by gauravrajput1 |
<script> // Javascript program to find a path from corner cell to // middle cell in maze containing positive numbers function printPath( maze,i,j,ans)
{ // If we reach the center cell
if (i == Math.floor(maze.length/2) && j == Math.floor(maze.length/2))
{
// Make the final answer, Print the
// final answer and Return
ans += "(" +i+ ", " +j+ ") -> MID" ;
document.write(ans+ "<br>" );
return ;
}
// If the element at the current position
// in maze is 0, simply Return as it has
// been visited before.
if (maze[i][j] == 0){
return ;
}
// If element is non-zero, then note
// the element in variable 'k'
let k = maze[i][j];
// Mark the cell visited by making the
// element 0. Don't worry, the element
// is safe in 'k'
maze[i][j] = 0;
// Make recursive calls in all 4
// directions pro-actively i.e. if the next
// cell lies in maze or not. Right call
if (j + k < maze.length){
printPath(maze, i, j+k, ans+ "(" +i+ ", " +j+ ") -> " );
}
// down call
if (i + k < maze.length){
printPath(maze, i+k, j, ans+ "(" +i+ ", " +j+ ") -> " );
}
// left call
if (j-k>0){
printPath(maze, i, j-k, ans+ "(" +i+ ", " +j+ ") -> " );
}
// up call
if (i-k>0){
printPath(maze, i-k, j, ans+ "(" +i+ ", " +j+ ") -> " );
}
// Unmark the visited cell by substituting
// its original value from 'k'
maze[i][j] = k;
} let maze = [[ 3, 5, 4, 4, 7, 3, 4, 6, 3 ],[ 6, 7, 5, 6, 6, 2, 6, 6, 2 ],[ 3, 3, 4, 3, 2, 5, 4, 7, 2 ], [ 6, 5, 5, 1, 2, 3, 6, 5, 6 ],[ 3, 3, 4, 3, 0, 1, 4, 3, 4 ],[ 3, 5, 4, 3, 2, 2, 3, 3, 5 ],
[ 3, 5, 4, 3, 2, 6, 4, 4, 3 ],[ 3, 5, 1, 3, 7, 5, 3, 6, 4 ],[ 6, 2, 4, 3, 4, 5, 4, 5, 1 ]] ;
printPath(maze, 0, 0, "" );
// This code is contributed by unknown2108 </script> |
Output :
(0, 0) -> (0, 3) -> (0, 7) -> (6, 7) -> (6, 3) -> (3, 3) -> (3, 4) -> (5, 4) -> (5, 2) -> (1, 2) -> (1, 7) -> (7, 7) -> (7, 1) -> (2, 1) -> (2, 4) -> (4, 4) -> MID
Time Complexity: O(4^N2), where N is the size of the maze and the function makes recursive calls in all four directions, resulting in a branching factor of 4.
Space Complexity: O(N2), where N is the size of the maze. This is because the function uses a 2D vector to represent the maze, which has N^2 elements.