C# Solution for Regions Cut By Slashes Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires counting distinct regions formed by slashes (/), backslashes (\), 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires counting distinct regions formed by slashes (/), backslashes (\\), and blank spaces ( ) in a grid. The solution involves transforming the grid into a larger grid where each cell is expanded into a 3x3 block. Each block is filled based on the content of the corresponding cell in the original grid, representing how slashes and backslashes divide the regions.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tGrid Expansion:\n\t\u2022\tEach cell in the original n x n grid is transformed into a 3x3 block in a new grid. This block represents the divisions caused by slashes and backslashes:\n\t    \u2022\t/ creates two distinct regions within the 3x3 block.\n\t    \u2022\t\\ also creates two distinct regions within the block.\n\t    \u2022\t  creates a single connected region within the block.\n\t\u2022\tFill the 3x3 blocks accordingly based on the character in the original grid.\n2.\tDFS Traversal:\n\t\u2022\tAfter expanding the grid, perform Depth-First Search (DFS) to count the number of connected components in the expanded grid.\n\t\u2022\tEach unvisited cell in the expanded grid starts a new DFS, representing a new region.\n3.\tCount Regions:\n\t\u2022\tCount the number of DFS initiations. Each initiation corresponds to a new distinct region.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\u2022\tGrid Expansion:\n\t\u2022\tExpanding the grid takes O(n^2) time, where n is the number of rows/columns in the original grid. Each cell in the original grid is transformed into a 3x3 block.\n\u2022\tDFS Traversal:\n\t\u2022\tEach cell in the expanded 3n x 3n grid is visited once. Therefore, the time complexity of DFS is O((3n)^2), which simplifies to O(n^2).\n\u2022\tOverall Time Complexity:\n\t\u2022\tThe time complexity is O(n^2) due to both the grid expansion and DFS traversal.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\u2022\tExpanded Grid:\n\t\u2022\tThe space required for the expanded grid is O((3n)^2), which simplifies to O(n^2).\n\u2022\tDFS Stack:\n\t\u2022\tThe space required for the DFS stack is proportional to the number of cells in the expanded grid, which is also O(n^2).\n\u2022\tOverall Space Complexity:\n\t\u2022\tThe space complexity is O(n^2) due to the expanded grid and the DFS stack.\n\n# Code\n```\npublic class Solution {\n    public int RegionsBySlashes(string[] grid) {\n        int n = grid.Length;\n        int size = n * 3;\n        int[,] expandedGrid = new int[size, size];\n\n        // Expand the grid by filling the expandedGrid based on the slashes in the original grid\n        for (int i = 0; i < n; i++) {\n            for (int j = 0; j < n; j++) {\n                char c = grid[i][j];\n                int r = i * 3;\n                int c1 = j * 3;\n                if (c == \'/\') {\n                    expandedGrid[r, c1 + 2] = 1;\n                    expandedGrid[r + 1, c1 + 1] = 1;\n                    expandedGrid[r + 2, c1] = 1;\n                } else if (c == \'\\\\\') {\n                    expandedGrid[r, c1] = 1;\n                    expandedGrid[r + 1, c1 + 1] = 1;\n                    expandedGrid[r + 2, c1 + 2] = 1;\n                }\n            }\n        }\n\n        // Now count the regions using DFS\n        int regions = 0;\n        for (int i = 0; i < size; i++) {\n            for (int j = 0; j < size; j++) {\n                if (expandedGrid[i, j] == 0) {\n                    regions++;\n                    DFS(expandedGrid, i, j, size);\n                }\n            }\n        }\n\n        return regions;\n    }\n\n    private void DFS(int[,] grid, int r, int c, int size) {\n        if (r < 0 || r >= size || c < 0 || c >= size || grid[r, c] != 0) {\n            return;\n        }\n\n        grid[r, c] = 1; // Mark the cell as visited\n\n        // Visit all 4 adjacent cells\n        DFS(grid, r - 1, c, size);\n        DFS(grid, r + 1, c, size);\n        DFS(grid, r, c - 1, size);\n        DFS(grid, r, c + 1, size);\n    }\n}\n```
['C#']