#### Count Paths

p
```"""
Given a grid, where you start from the top left position [0, 0],
you want to find how many paths you can take to get to the bottom right position.

start here  ->   0  0  0  0
1  1  0  0
0  0  0  1
0  1  0  0  <- finish here
how many 'distinct' paths can you take to get to the finish?
Using a recursive depth-first search algorithm below, you are able to
find the number of distinct unique paths (count).

'*' will demonstrate a path
In the example above, there are two distinct paths:
1.                2.
*  *  *  0      *  *  *  *
1  1  *  0      1  1  *  *
0  0  *  1      0  0  *  1
0  1  *  *      0  1  *  *
"""

def depth_first_search(grid: list[list[int]], row: int, col: int, visit: set) -> int:
"""
Recursive Backtracking Depth First Search Algorithm

Starting from top left of a matrix, count the number of
paths that can reach the bottom right of a matrix.
1 represents a block (inaccessible)
0 represents a valid space (accessible)

0  0  0  0
1  1  0  0
0  0  0  1
0  1  0  0
>>> grid = [[0, 0, 0, 0], [1, 1, 0, 0], [0, 0, 0, 1], [0, 1, 0, 0]]
>>> depth_first_search(grid, 0, 0, set())
2

0  0  0  0  0
0  1  1  1  0
0  1  1  1  0
0  0  0  0  0
>>> grid = [[0, 0, 0, 0, 0], [0, 1, 1, 1, 0], [0, 1, 1, 1, 0], [0, 0, 0, 0, 0]]
>>> depth_first_search(grid, 0, 0, set())
2
"""
row_length, col_length = len(grid), len(grid[0])
if (
min(row, col) < 0
or row == row_length
or col == col_length
or (row, col) in visit
or grid[row][col] == 1
):
return 0
if row == row_length - 1 and col == col_length - 1:
return 1

count = 0
count += depth_first_search(grid, row + 1, col, visit)
count += depth_first_search(grid, row - 1, col, visit)
count += depth_first_search(grid, row, col + 1, visit)
count += depth_first_search(grid, row, col - 1, visit)

visit.remove((row, col))
return count

if __name__ == "__main__":
import doctest

doctest.testmod()
```