Open In App

# Maze With N doors and 1 Key

Given an N * N binary maze where a 0 denotes that the position can be visited and a 1 denotes that the position cannot be visited without a key, the task is to find whether it is possible to visit the bottom-right cell from the top-left cell with only one key along the way. If possible then print “Yes” else print “No”.

Example:

Input: maze[][] = {
{0, 0, 1},
{1, 0, 1},
{1, 1, 0}}
Output: Yes

Approach: This problem can be solved using recursion, for every possible move, if the current cell is 0 then without altering the status of the key check whether it is the destination else move forward. If the current cell is 1 then the key must be used, now for the further moves the key will be set to false i.e. it’ll never be used again on the same path. If any path reaches the destination then print Yes else print No.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Recursive function to check whether there is``// a path from the top left cell to the``// bottom right cell of the maze``bool` `findPath(vector > maze, ``int` `xpos, ``int` `ypos,``              ``bool` `key)``{` `    ``// Check whether the current cell is``    ``// within the maze``    ``if` `(xpos < 0 || xpos >= maze.size() || ypos < 0``        ``|| ypos >= maze.size())``        ``return` `false``;` `    ``// If key is required to move further``    ``if` `(maze[xpos][ypos] == ``'1'``) {` `        ``// If the key hasn't been used before``        ``if` `(key == ``true``)` `            ``// If current cell is the destination``            ``if` `(xpos == maze.size() - 1``                ``&& ypos == maze.size() - 1)``                ``return` `true``;` `        ``// Either go down or right``        ``return` `findPath(maze, xpos + 1, ypos, ``false``)``               ``|| findPath(maze, xpos, ypos + 1, ``false``);` `        ``// Key has been used before``        ``return` `false``;``    ``}` `    ``// If current cell is the destination``    ``if` `(xpos == maze.size() - 1 && ypos == maze.size() - 1)``        ``return` `true``;` `    ``// Either go down or right``    ``return` `findPath(maze, xpos + 1, ypos, key)``           ``|| findPath(maze, xpos, ypos + 1, key);``}` `bool` `mazeProb(vector > maze, ``int` `xpos, ``int` `ypos)``{``    ``bool` `key = ``true``;``    ``if` `(findPath(maze, xpos, ypos, key))``        ``return` `true``;` `    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``vector > maze = { { ``'0'``, ``'0'``, ``'1'` `},``                                  ``{ ``'1'``, ``'0'``, ``'1'` `},``                                  ``{ ``'1'``, ``'1'``, ``'0'` `} };``    ``int` `n = maze.size();` `    ``// If there is a path from the cell (0, 0)``    ``if` `(mazeProb(maze, 0, 0))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;``}` `// This code is contributed by grand_master`

## Java

 `// Java implementation of the approach``import` `java.io.*;``import` `java.util.ArrayList;` `class` `GFG {` `    ``// Recursive function to check whether there``    ``// is a path from the top left cell to the``    ``// bottom right cell of the maze``    ``static` `boolean``    ``findPath(ArrayList > maze, ``int` `xpos,``             ``int` `ypos, ``boolean` `key)``    ``{` `        ``// Check whether the current cell is``        ``// within the maze``        ``if` `(xpos < ``0` `|| xpos >= maze.size() || ypos < ``0``            ``|| ypos >= maze.size())``            ``return` `false``;` `        ``// If key is required to move further``        ``if` `(maze.get(xpos).get(ypos) == ``'1'``) {` `            ``// If the key hasn't been used before``            ``if` `(key == ``true``)` `                ``// If current cell is the destination``                ``if` `(xpos == maze.size() - ``1``                    ``&& ypos == maze.size() - ``1``)``                    ``return` `true``;` `            ``// Either go down or right``            ``return` `findPath(maze, xpos + ``1``, ypos, ``false``)``                ``|| findPath(maze, xpos, ypos + ``1``, ``false``);``        ``}` `        ``// If current cell is the destination``        ``if` `(xpos == maze.size() - ``1``            ``&& ypos == maze.size() - ``1``)``            ``return` `true``;` `        ``// Either go down or right``        ``return` `findPath(maze, xpos + ``1``, ypos, key)``            ``|| findPath(maze, xpos, ypos + ``1``, key);``    ``}` `    ``static` `boolean``    ``mazeProb(ArrayList > maze, ``int` `xpos,``             ``int` `ypos)``    ``{``        ``boolean` `key = ``true``;` `        ``if` `(findPath(maze, xpos, ypos, key))``            ``return` `true``;` `        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `size = ``3``;``        ``ArrayList > maze``            ``= ``new` `ArrayList >(size);` `        ``for` `(``int` `i = ``0``; i < size; i++) {``            ``maze.add(``new` `ArrayList());``        ``}` `        ``// We are making these``        ``//{ { '0', '0', '1' },``        ``//  { '1', '0', '1' },``        ``//  { '1', '1', '0' } };``        ``maze.get(``0``).add(``0``);``        ``maze.get(``0``).add(``0``);``        ``maze.get(``0``).add(``1``);``        ``maze.get(``1``).add(``1``);``        ``maze.get(``1``).add(``0``);``        ``maze.get(``1``).add(``1``);``        ``maze.get(``2``).add(``1``);``        ``maze.get(``2``).add(``1``);``        ``maze.get(``2``).add(``0``);` `        ``// If there is a path from the cell (0, 0)``        ``if` `(mazeProb(maze, ``0``, ``0``))``            ``System.out.print(``"Yes"``);``        ``else``            ``System.out.print(``"No"``);``    ``}``}` `// This code is contributed by sujitmeshram`

## Python3

 `# Python3 implementation of the approach` `# Recursive function to check whether there is``# a path from the top left cell to the``# bottom right cell of the maze`  `def` `findPath(maze, xpos, ypos, key):` `    ``# Check whether the current cell is``    ``# within the maze``    ``if` `xpos < ``0` `or` `xpos >``=` `len``(maze) ``or` `ypos < ``0` `\``            ``or` `ypos >``=` `len``(maze):``        ``return` `False` `    ``# If key is required to move further``    ``if` `maze[xpos][ypos] ``=``=` `'1'``:` `        ``# If the key hasn't been used before``        ``if` `key ``=``=` `True``:` `            ``# If current cell is the destination``            ``if` `xpos ``=``=` `len``(maze)``-``1` `and` `ypos ``=``=` `len``(maze)``-``1``:``                ``return` `True` `            ``# Either go down or right``            ``return` `findPath(maze, xpos ``+` `1``, ypos, ``False``) ``or` `\``                ``findPath(maze, xpos, ypos ``+` `1``, ``False``)` `        ``# Key has been used before``        ``return` `False` `    ``# If current cell is the destination``    ``if` `xpos ``=``=` `len``(maze)``-``1` `and` `ypos ``=``=` `len``(maze)``-``1``:``        ``return` `True` `    ``# Either go down or right``    ``return` `findPath(maze, xpos ``+` `1``, ypos, key) ``or` `\``        ``findPath(maze, xpos, ypos ``+` `1``, key)`  `def` `mazeProb(maze, xpos, ypos):``    ``key ``=` `True``    ``if` `findPath(maze, xpos, ypos, key):``        ``return` `True``    ``return` `False`  `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``maze ``=` `[[``'0'``, ``'0'``, ``'1'``],``            ``[``'1'``, ``'0'``, ``'1'``],``            ``[``'1'``, ``'1'``, ``'0'``]]``    ``n ``=` `len``(maze)` `    ``# If there is a path from the cell (0, 0)``    ``if` `mazeProb(maze, ``0``, ``0``):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` `    ``// Recursive function to check whether there``    ``// is a path from the top left cell to the``    ``// bottom right cell of the maze``    ``static` `bool` `findPath(List > maze, ``int` `xpos,``                         ``int` `ypos, ``bool` `key)``    ``{` `        ``// Check whether the current cell is``        ``// within the maze``        ``if` `(xpos < 0 || xpos >= maze.Count || ypos < 0``            ``|| ypos >= maze.Count)``            ``return` `false``;` `        ``// If key is required to move further``        ``if` `(maze[xpos][ypos] == ``'1'``) {` `            ``// If the key hasn't been used before``            ``if` `(key == ``true``)` `                ``// If current cell is the destination``                ``if` `(xpos == maze.Count - 1``                    ``&& ypos == maze.Count - 1)``                    ``return` `true``;` `            ``// Either go down or right``            ``return` `findPath(maze, xpos + 1, ypos, ``false``)``                ``|| findPath(maze, xpos, ypos + 1, ``false``);``        ``}` `        ``// If current cell is the destination``        ``if` `(xpos == maze.Count - 1``            ``&& ypos == maze.Count - 1)``            ``return` `true``;` `        ``// Either go down or right``        ``return` `findPath(maze, xpos + 1, ypos, key)``            ``|| findPath(maze, xpos, ypos + 1, key);``    ``}` `    ``static` `bool` `mazeProb(List > maze, ``int` `xpos,``                         ``int` `ypos)``    ``{``        ``bool` `key = ``true``;` `        ``if` `(findPath(maze, xpos, ypos, key))``            ``return` `true``;` `        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `size = 3;``        ``List > maze = ``new` `List >(size);` `        ``for` `(``int` `i = 0; i < size; i++) {``            ``maze.Add(``new` `List<``int``>());``        ``}` `        ``// We are making these``        ``//{ { '0', '0', '1' },``        ``//  { '1', '0', '1' },``        ``//  { '1', '1', '0' } };``        ``maze[0].Add(0);``        ``maze[0].Add(0);``        ``maze[0].Add(1);``        ``maze[1].Add(1);``        ``maze[1].Add(0);``        ``maze[1].Add(1);``        ``maze[2].Add(1);``        ``maze[2].Add(1);``        ``maze[2].Add(0);` `        ``// If there is a path from the cell (0, 0)``        ``if` `(mazeProb(maze, 0, 0))``            ``Console.Write(``"Yes"``);``        ``else``            ``Console.Write(``"No"``);``    ``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(2N)

### Optimized Approach:

Dynamic Programming can be used to improve the time complexity

The main idea is for every cell the answer is dependent upon its previous row and col .

Here maze[1][1] is dependent on maze[1][0] or maze[0][1] if it is a possible path .

Hence using this approach we can compute result of maze[n-1][n-1] from its previous adjacent cells

And also there are some edge condition for 0th row and 0th col as the these cells are dependent on their previous col and row respectively.

Below is the implementation of above approach.

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `bool` `mazeProb(vector > maze, ``int` `n)``{` `    ``for` `(``int` `row = 0; row < n; ++row) {``        ``for` `(``int` `col = 0; col < n; ++col) {` `            ``if` `(row == 0 && col == 0) ``// Skip the first cell``                ``continue``;``            ``if` `(row == 0) {``              ``// for first row result depend on previous col``                ``maze[row][col] = min(``                    ``2, maze[row][col] + maze[row][col - 1]);``            ``}``            ``else` `if` `(col == 0) {``              ``// for first col result depends on previous row``                ``maze[row][col] = min(``                    ``2, maze[row][col] + maze[row - 1][col]);``            ``}``            ``else` `{``              ``// for other cells, result will be``              ``// minimum of previous row or col cell``                ``maze[row][col]``                    ``= min(2, maze[row][col]``                                 ``+ min(maze[row][col - 1],``                                       ``maze[row - 1][col]));``            ``}``        ``}``    ``}` `    ``return` `maze[n - 1][n - 1]``           ``!= 2; ``// if last cell value is 2 then there is no``                 ``// path available``}` `// Driver code``int` `main()``{``    ``vector > maze``        ``= { { 0, 0, 1 }, { 1, 0, 1 }, { 1, 1, 0 } };``    ``int` `n = maze.size();` `    ``// If there is a path from the cell (0, 0)``    ``if` `(mazeProb(maze, 3))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;``}` `// This code is contributed by pratham sonawane`

## Java

 `// Java implementation of the approach` `import` `java.util.ArrayList;` `public` `class` `GFG {``    ``static` `boolean` `mazeProb(ArrayList<``int``[]> maze, ``int` `n)``    ``{``        ``for` `(``int` `row = ``0``; row < n; ++row) {``            ``for` `(``int` `col = ``0``; col < n; ++col) {` `                ``if` `(row == ``0``                    ``&& col == ``0``) ``// Skip the first cell``                    ``continue``;``                ``if` `(row == ``0``) {``                    ``// for first row result depend on``                    ``// previous col``                    ``maze.get(row)[col] = Math.min(``                        ``2``, maze.get(row)[col]``                               ``+ maze.get(row)[col - ``1``]);``                ``}``                ``else` `if` `(col == ``0``) {``                    ``// for first col result depends on``                    ``// previous row``                    ``maze.get(row)[col] = Math.min(``                        ``2``, maze.get(row)[col]``                               ``+ maze.get(row - ``1``)[col]);``                ``}``                ``else` `{``                    ``// for other cells, result will be``                    ``// minimum of previous row or col cell``                    ``maze.get(row)[col] = Math.min(``                        ``2``, maze.get(row)[col]``                               ``+ Math.min(``                                   ``maze.get(row)[col - ``1``],``                                   ``maze.get(row - ``1``)[col]));``                ``}``            ``}``        ``}` `        ``return` `maze.get(n - ``1``)[n - ``1``]``            ``!= ``2``; ``// if last cell value is 2 then there is``                  ``// no path available``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``ArrayList<``int``[]> maze = ``new` `ArrayList<``int``[]>();``        ``maze.add(``new` `int``[] { ``0``, ``0``, ``1` `});``        ``maze.add(``new` `int``[] { ``1``, ``0``, ``1` `});``        ``maze.add(``new` `int``[] { ``1``, ``1``, ``0` `});` `        ``int` `n = maze.size();` `        ``// If there is a path from the cell (0, 0)``        ``if` `(mazeProb(maze, ``3``))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);``    ``}``}` `// This code is contributed by Lovely Jain`

## Python3

 `# Python implementation of the approach``def` `mazeProb(maze, n):` `    ``for` `row ``in` `range``(n):``        ``for` `col ``in` `range``(n):` `            ``if` `(row ``=``=` `0` `and` `col ``=``=` `0``): ``# Skip the first cell``                ``continue``            ``if` `(row ``=``=` `0``):``              ``# for first row result depend on previous col``                ``maze[row][col] ``=` `min``(``                    ``2``, maze[row][col] ``+` `maze[row][col ``-` `1``])``            ``elif` `(col ``=``=` `0``):``              ``# for first col result depends on previous row``                ``maze[row][col] ``=` `min``(``                    ``2``, maze[row][col] ``+` `maze[row ``-` `1``][col])``            ``else``:``              ``# for other cells, result will be``              ``# minimum of previous row or col cell``                ``maze[row][col]  ``=` `min``(``2``, maze[row][col]  ``+` `min``(maze[row][col ``-` `1``], maze[row ``-` `1``][col]))` `    ``return` `maze[n ``-` `1``][n ``-` `1``]!``=` `2` `# if last cell value is 2 then there is no``                 ``# path available` `# Driver code``maze ``=` `[ [ ``0``, ``0``, ``1` `], [ ``1``, ``0``, ``1` `], [ ``1``, ``1``, ``0` `] ]``n ``=` `len``(maze)` `# If there is a path from the cell (0, 0)``if` `(mazeProb(maze, ``3``)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed by shinjanpatra`

## C#

 `//C# code for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG {``  ``static` `bool` `mazeProb(List<``int``[]> maze, ``int` `n)``  ``{``    ``for` `(``int` `row = 0; row < n; ++row) {``      ``for` `(``int` `col = 0; col < n; ++col) {` `        ``if` `(row == 0``            ``&& col == 0) ``// Skip the first cell``          ``continue``;``        ``if` `(row == 0) {``          ``// for first row result depend on``          ``// previous col``          ``maze[row][col] = Math.Min(``            ``2, maze[row][col]``            ``+ maze[row][col - 1]);``        ``}``        ``else` `if` `(col == 0) {``          ``// for first col result depends on``          ``// previous row``          ``maze[row][col] = Math.Min(``            ``2, maze[row][col]``            ``+ maze[row - 1][col]);``        ``}``        ``else` `{``          ``// for other cells, result will be``          ``// minimum of previous row or col cell``          ``maze[row][col] = Math.Min(``            ``2,``            ``maze[row][col]``            ``+ Math.Min(maze[row][col - 1],``                       ``maze[row - 1][col]));``        ``}``      ``}``    ``}` `    ``return` `maze[n - 1][n - 1]``      ``!= 2; ``// if last cell value is 2 then there is``    ``// no path available``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main()``  ``{``    ``List<``int``[]> maze = ``new` `List<``int``[]>();``    ``maze.Add(``new` `int``[] { 0, 0, 1 });``    ``maze.Add(``new` `int``[] { 1, 0, 1 });``    ``maze.Add(``new` `int``[] { 1, 1, 0 });` `    ``int` `n = maze.Count;` `    ``// If there is a path from the cell (0, 0)``    ``if` `(mazeProb(maze, 3))``      ``Console.WriteLine(``"Yes"``);``    ``else``      ``Console.WriteLine(``"No"``);``  ``}``}` `// This code is contributed by pradeepkumarppk2003`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(N^2)

Space Complexity: O(1)