An n x n grid is composed of 1 x 1 squares where each 1 x 1 square consists of a '/', '\', or blank space ' '. These characters divide the square into contiguous regions.

Given the grid grid represented as a string array, return the number of regions.

Note that backslash characters are escaped, so a '\' is represented as '\\'.

Example 1:

Input: grid = [" /","/ "]
Output: 2
Example 2:

Input: grid = [" /"," "]
Output: 1
Example 3:

Input: grid = ["/\\\\","\\\\/"]
Output: 5
Explanation: Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.

Constraints:

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

With the cutted sections, we will join the sections depending on the slashes. For example, '/' means that north joins west and south joins east. Note that we must also join the north region of a square to the south region of the square to the top of it. Similar for the left square's east region and current square's west region.

This is best approached by DSU, in the very end after joining all sections, we just have to find the number of components in the map regions.

Implementation

def regionsBySlashes(grid: List[str]) -> int:
    regions = {}
    def find(x):
        y = regions.get(x, x)
        if y != x:
            regions[x] = y = find(y)
        return y
    def union(x, y):
        regions[find(x)] = find(y)

    size = len(grid)
    for i in range(size):
        for j in range(size):
            if i > 0:           # connect square with top
                union((i - 1, j, 'S'), (i, j, 'N'))
            if j > 0:           # connect square with left
                union((i, j - 1, 'E'), (i, j, 'W'))
            if grid[i][j] != "/":             # ' ' or '\'
                union((i, j, 'N'), (i, j, 'E'))   # connect NE
                union((i, j, 'S'), (i, j, 'W'))   # connect SW
            if grid[i][j] != "\\":            # ' ' or '/'
                union((i, j, 'N'), (i, j, 'W'))   # connect NW
                union((i, j, 'S'), (i, j, 'E'))   # connect SE
    return len(set(map(find, regions)))

Alternatively, we can find the number of connected components as in Number of Coneected Components, but bare in mind that we should only decrease the component count when we know that two elements aren't from the same set.