# Teaching Kids Programming – Nearest Exit from Entrance in Maze via Depth First Search Algorithm

Teaching Kids Programming – Nearest Exit from Entrance in Maze via Depth First Search Algorithm | ninjasquad

Teaching Kids Programming: Videos on Data Structures and Algorithms

You are given an m x n matrix maze (0-indexed) with empty cells (represented as ‘.’) and walls (represented as ‘+’). You are also given the entrance of the maze, where entrance = [entrancerow, entrancecol] denotes the row and column of the cell you are initially standing at.

In one step, you can move one cell up, down, left, or right. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the nearest exit from the entrance. An exit is defined as an empty cell that is at the border of the maze. The entrance does not count as an exit.

Return the number of steps in the shortest path from the entrance to the nearest exit, or -1 if no such path exists.

Example 1:
Input: maze = [[“+”,”+”,”.”,”+”],[“.”,”.”,”.”,”+”],[“+”,”+”,”+”,”.”]], entrance = [1,2]
Output: 1
Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3].
Initially, you are at the entrance cell [1,2].
– You can reach [1,0] by moving 2 steps left.
– You can reach [0,2] by moving 1 step up.
It is impossible to reach [2,3] from the entrance.
Thus, the nearest exit is [0,2], which is 1 step away.

Example 2:
Input: maze = [[“+”,”+”,”+”],[“.”,”.”,”.”],[“+”,”+”,”+”]], entrance = [1,0]
Output: 2
Explanation: There is 1 exit in this maze at [1,2].
[1,0] does not count as an exit since it is the entrance cell.
Initially, you are at the entrance cell [1,0].
– You can reach [1,2] by moving 2 steps right.
Thus, the nearest exit is [1,2], which is 2 steps away.

Example 3:
Input: maze = [[“.”,”+”]], entrance = [0,0]
Output: -1
Explanation: There are no exits in this maze.

Constraints:
maze.length == m
maze[i].length == n
1 <= m, n <= 100
maze[i][j] is either ‘.’ or ‘+’.
entrance.length == 2
0 <= entrancerow < m
0 <= entrancecol < n
entrance will always be an empty cell.

Hints:
Which type of traversal lets you find the distance from a point?
Try using a Breadth First Search.

### Nearest Exit from Entrance in Maze via Depth First Search Algorithm

Depth First Search Algorithm (DFS) is not suitable in finding the shortest path in a unweighted/weighted Graph, however, we can exhaust all the paths in order to know the optimal path (shortest).

Since this is undirected Graph, we need to remember the nodes that we have visited, thus we can use a hash set. And when we find a path, we don’t immediately terminate here, as we need to find all the exits and decide which one is the nearest. We also need to unmark the visited paths in DFS otherwise other possible routes may be blocked.

 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 ``` ```class Solution:     def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:         rows = len(maze)         assert rows > 0         cols = len(maze)         assert cols > 0                 # @cache         def isExit(r, c):             if r == 0 or c == 0 or r == rows - 1 or c == cols - 1:                 return maze[r][c] == '.'             return False                         self.ans = inf           def dfs(r, c, d, seen=set()):             if (r, c) != tuple(entrance) and isExit(r, c):                 self.ans = min(self.ans, d)                                 return True             if (r, c) in seen:                 return False             seen.add((r, c))             for dr, dc in ((0, 1), (1, 0), (0, -1), (-1, 0)):                 nr, nc = r + dr, c + dc                 if 0 <= nr < rows and 0 <= nc < cols and maze[nr][nc] == '.':                                         dfs(nr, nc, d + 1)                                 seen.remove((r, c))             return False           dfs(entrance, entrance, 0)         return self.ans if self.ans < inf else -1```
```class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows = len(maze)
assert rows > 0
cols = len(maze)
assert cols > 0

# @cache
def isExit(r, c):
if r == 0 or c == 0 or r == rows - 1 or c == cols - 1:
return maze[r][c] == '.'
return False

self.ans = inf

def dfs(r, c, d, seen=set()):
if (r, c) != tuple(entrance) and isExit(r, c):
self.ans = min(self.ans, d)
return True
if (r, c) in seen:
return False
for dr, dc in ((0, 1), (1, 0), (0, -1), (-1, 0)):
nr, nc = r + dr, c + dc
if 0 <= nr < rows and 0 <= nc < cols and maze[nr][nc] == '.':
dfs(nr, nc, d + 1)
seen.remove((r, c))
return False

dfs(entrance, entrance, 0)
return self.ans if self.ans < inf else -1```

The following is also a Python’s implementation using DFS algorithm. However, instead of using hash set, we can mark/unmark the visited empty cells walls.

 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 ``` ```class Solution:     def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:         rows = len(maze)         assert rows > 0         cols = len(maze)         assert cols > 0                 # @cache         def isExit(r, c):             if r == 0 or c == 0 or r == rows - 1 or c == cols - 1:                 return maze[r][c] == '.'             return False                         self.ans = inf           def dfs(r, c, d):             if (r, c) != tuple(entrance) and isExit(r, c):                 self.ans = min(self.ans, d)                                 return True             maze[r][c] = "+"             for dr, dc in ((0, 1), (1, 0), (0, -1), (-1, 0)):                 nr, nc = r + dr, c + dc                 if 0 <= nr < rows and 0 <= nc < cols and maze[nr][nc] == '.':                                         dfs(nr, nc, d + 1)                                 maze[r][c] = "."             return False           dfs(entrance, entrance, 0)         return self.ans if self.ans < inf else -1```
```class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
rows = len(maze)
assert rows > 0
cols = len(maze)
assert cols > 0

# @cache
def isExit(r, c):
if r == 0 or c == 0 or r == rows - 1 or c == cols - 1:
return maze[r][c] == '.'
return False

self.ans = inf

def dfs(r, c, d):
if (r, c) != tuple(entrance) and isExit(r, c):
self.ans = min(self.ans, d)
return True
maze[r][c] = "+"
for dr, dc in ((0, 1), (1, 0), (0, -1), (-1, 0)):
nr, nc = r + dr, c + dc
if 0 <= nr < rows and 0 <= nc < cols and maze[nr][nc] == '.':
dfs(nr, nc, d + 1)
maze[r][c] = "."
return False

dfs(entrance, entrance, 0)
return self.ans if self.ans < inf else -1```

Usually, we implement the DFS in Recursion which is trival compared to iterative implementation using Stack.

#### Nearest Exit from Entrance in Maze

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