# 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 ## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem can be solved using recursion, for very 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:

 `# 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"``) `

Output:

```Yes
```

Time Complexity: O(2N)

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