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 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:

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# 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")

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Output:

Yes

Time Complexity: O(2N)



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Improved By : Akanksha_Rai