J

```
"""
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
visit.add((row, col))
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()
```