Beginner Friendly Solution with Depth First Search + Step Explanation 😊😊
Intuition\nThe code is designed to count the number of regions formed by slashes in a grid. It does this by creating a larger matrix representation of the grid,
# Intuition\nThe code is designed to count the number of regions formed by slashes in a grid. It does this by creating a larger matrix representation of the grid, where each cell in the grid is expanded into a 3x3 block in the matrix. Slashes are then represented by \'1\'s in the matrix, effectively creating walls. The code then uses a Depth-First Search (DFS) algorithm to traverse the matrix, marking visited cells as \'1\' and incrementing a counter for each new region (or \'island\') found.\n\n# Approach\n1. Matrix Initialization: Create a larger matrix with dimensions 3 times the size of the original grid.\n2. Populate Matrix with Slashes: Iterate through the original grid, and for each cell:\n    - If the cell contains a backslash, set the corresponding 3x3 block in the matrix to \'1\'s in a diagonal pattern.\n    - If the cell contains a forward slash, set the corresponding 3x3 block in the matrix to \'1\'s in a reverse diagonal pattern.\n3. Region Counting:\n    - Iterate through the matrix.\n    - If a cell is \'0\', it represents the start of a new region.\n    - Perform a DFS starting at this cell, marking all reachable \'0\' cells as \'1\'.\n    - Increment the region counter for each new region found.\n4. Return Count: Return the total count of regions.\n\n# Complexity\n- Time complexity: O(N^2)\n\n- Space complexity: O(N^2)\n\n# Steps\n- Create matrix that with slashes as shown from the leetcode editorial (https://leetcode.com/problems/regions-cut-by-slashes/editorial/):\n![Screenshot (308).png](https://leetcode.com/problems/regions-cut-by-slashes/Figures/959_re/image2.png)\n- Search/count the regions by doing Depth First Search as "Number of Islands" problem (https://leetcode.com/problems/number-of-islands/): \n    - The code to do the counting of regions is really similar with "Number of Islands" problem. So, I recommend to do that problem first if you are struggling with this question.\n    - My solution for "Number of Islands" problem (https://leetcode.com/problems/number-of-islands/solutions/5616162/beginner-friendly-solution-with-matrix-and-depth-first-search-dfs/)\n\nThat\'s all, thank you and good luck. \uD83D\uDE0A\uD83D\uDE0A\n\n# Code\n```\n/**\n * @param {string[]} grid\n * @return {number}\n */\nconst directions = [\n  [-1, 0], // up\n  [0, 1], // right\n  [1, 0], // down\n  [0, -1], //left\n];\n\n// this solution is taken from number of islands solution\nfunction dfs(matrix, row, col) {\n  matrix[row][col] = 1;\n\n  for (let direction of directions) {\n    const nextRow = row + direction[0];\n    const nextCol = col + direction[1];\n\n    if (\n      nextRow >= 0 &&\n      nextRow <= matrix.length - 1 &&\n      nextCol >= 0 &&\n      nextCol <= matrix[0].length - 1 &&\n      matrix[nextRow][nextCol] === 0\n    ) {\n      dfs(matrix, nextRow, nextCol);\n    }\n  }\n}\n\nvar regionsBySlashes = function (grid) {\n  // initialize length of grid and matrix\n  const gridSize = grid.length;\n  const matrixSize = gridSize * 3;\n  const matrix = new Array(gridSize * 3)\n    .fill(null)\n    .map(() => new Array(gridSize * 3).fill(0));\n\n  // populate matrix with slashes\n  for (let i = 0; i < gridSize; i++) {\n    for (let j = 0; j < gridSize; j++) {\n      const row = i * 3;\n      const col = j * 3;\n\n      // backslash  slash\n      if (grid[i][j] === "\\\\") {\n        matrix[row][col] = 1;\n        matrix[row + 1][col + 1] = 1;\n        matrix[row + 2][col + 2] = 1;\n      }\n\n      // forward slash\n      if (grid[i][j] === "/") {\n        matrix[row][col + 2] = 1;\n        matrix[row + 1][col + 1] = 1;\n        matrix[row + 2][col] = 1;\n      }\n    }\n  }\n\n  // count regions\n  let count = 0;\n  // do sequential search\n  for (let row = 0; row < matrixSize; row++) {\n    for (let col = 0; col < matrixSize; col++) {\n      // if we find 0 meaning we find new region and switch all the zero to 1 (like "number of islands")\n      if (matrix[row][col] === 0) {\n        dfs(matrix, row, col); // this solution is taken from number of islands solution\n        count++;\n      }\n    }\n  }\n\n  return count;\n};\n```
['Array', 'Depth-First Search', 'Matrix', 'JavaScript']