Regions Cut By Slashes - Leetcode Solution

The problem "Regions Cut By Slashes" on LeetCode is about a 2D grid where each cell can be either empty or contain a slash (/) or a backslash (). The slashes divide the cell into four regions, and neighboring cells with slashes can either be connected or disconnected.

The task is to find the total number of regions in the grid. For example, in the following grid:

 /\
/  \
\  /
 \/

There are 5 regions, as indicated by the different colors:

 RR
GBBG
 GY
YY

To solve this problem, we can use a modified version of the Union-Find algorithm. We start by initializing a disjoint set for each cell, with the parent of each cell being itself. Then, we iterate over each cell in the grid and perform the following operations:

1. If the cell contains a backslash (), merge its bottom-left region with its top-right region, and its top-left region with its bottom-right region.
2. If the cell contains a slash (/), merge its top-left region with its top-right region, and its bottom-left region with its bottom-right region.
3. If the cell is empty, do nothing.

After all cells have been processed, we count the number of set representatives (i.e. the number of disjoint sets) in the grid to get the total number of regions.

The implementation of the algorithm can be done efficiently using the Union-Find data structure. Here is the example code in Python:

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))

    def find(self, i):
        if self.parent[i] != i:
            self.parent[i] = self.find(self.parent[i])
        return self.parent[i]

    def union(self, i, j):
        pi, pj = self.find(i), self.find(j)
        if pi != pj:
            self.parent[pi] = pj

class Solution:
    def regionsBySlashes(self, grid: List[str]) -> int:
        n = len(grid)
        uf = UnionFind(4*n*n)
        for i in range(n):
            for j in range(n):
                k = 4*(i*n+j)
                if grid[i][j] == ' ':
                    uf.union(k, k+1)
                    uf.union(k, k+2)
                    uf.union(k, k+3)
                elif grid[i][j] == '/':
                    uf.union(k, k+3)
                    uf.union(k+1, k+2)
                else:
                    uf.union(k, k+1)
                    uf.union(k+2, k+3)
                if i > 0:
                    uf.union(k, k-4*n+2)
                if i < n-1:
                    uf.union(k+2, k+4*n)
                if j > 0:
                    uf.union(k+3, k-3)
                if j < n-1:
                    uf.union(k+1, k+7)
        return sum(uf.parent[i] == i for i in range(4*n*n))

The time complexity of this algorithm is O(n^2 alpha(n)), where alpha(n) is the inverse Ackermann function, which is essentially a constant for any practical value of n. The space complexity is also O(n^2).

1