human
I found an interesting problem on https://leetcode.com/problems/regions-cut-by-slashes/:
In a N x N grid composed of 1 x 1 squares, each 1 x 1 square consists of a /, \, or blank space.  These characters divide the square into contiguous regions.
(Note that backslash characters are escaped, so a \ is represented as "\\".)
Return the number of regions.
 









Example 1:
Input:
[
  " /",
  "/ "
]
Output: 2
Explanation: The 2x2 grid is as follows:


Example 2:
Input:
[
  " /",
  "  "
]
Output: 1
Explanation: The 2x2 grid is as follows:


Example 3:
Input:
[
  "\\/",
  "/\\"
]
Output: 4
Explanation: (Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.)
The 2x2 grid is as follows:


Example 4:
Input:
[
  "/\\",
  "\\/"
]
Output: 5
Explanation: (Recall that because \ characters are escaped, "/\\" refers to /\, and "\\/" refers to \/.)
The 2x2 grid is as follows:


Example 5:
Input:
[
  "//",
  "/ "
]
Output: 3
Explanation: The 2x2 grid is as follows:

 
Note:

1 <= grid.length == grid[0].length <= 30
grid[i][j] is either '/', '\', or ' '.

I tried it in Python, but could not do it. Can you solve it?

=====
What is a the Python program you requested?
assistant
#Here is the Python program you requested.

class Solution:
    def regionsBySlashes(self, grid: List[str]) -> int:
        
        N = len(grid)
        base = N * (N + 1)
        parent = [i for i in range(2 * base)]
        
        def find(x):
            if parent[x] != parent[parent[x]]:
                parent[x] = find(parent[x])
            return parent[x]
        
        def union(x, y):
            parent[find(x)] = find(y)
            return find(y)
        
        for i in range(N):
            for j in range(N):
                up, down, left, right = i * N + j, (i + 1) * N + j, base + i * (N + 1) + j, base + i * (N + 1) + j + 1
                
                if grid[i][j] == '/':
                    union(up, left)
                    union(down, right)
                elif grid[i][j] == '\\\\':
                    union(up, right)
                    union(down, left)
                else:
                    union(up, left)
                    union(down, right)
                    union(left, right)
        
        comp = set()
        for i in range(2 * base):
            comp.add(find(i))
        
        return len(comp)