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LeetCode 959: Regions Cut By Slashes Solution

Master LeetCode problem 959 (Regions Cut By Slashes), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we can treat each grid cell as divided into 4 sub-cells by the slashes ('/', '', ' '). Then, we can perform a depth-first search (DFS) to identify the regions formed by these sub-cells. We will mark visited sub-cells and count the number of regions as we traverse the grid.

  1. For each cell in the grid:

    • Divide the cell into 4 sub-cells (top-left, top-right, bottom-left, bottom-right).
    • Mark visited sub-cells as we traverse the grid.
    • Perform DFS to identify regions within each cell.

  2. In the DFS:

    • Start from an unvisited sub-cell and explore all neighboring unvisited sub-cells.
    • Mark visited sub-cells and increment the region count.
    • Continue until all sub-cells are visited.

  3. Return the total number of regions found in the grid.

Time Complexity: O(n^2) where n is the size of the grid. Space Complexity: O(n^2) for storing visited information and recursion stack.

:

Solution Code

class Solution {
    public int regionsBySlashes(String[] grid) {
        int n = grid.length;
        int[][] visited = new int[3 * n][3 * n];
        int regions = 0;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i].charAt(j) == '/') {
                    mark(visited, i * 3, j * 3 + 2);
                    mark(visited, i * 3 + 1, j * 3 + 1);
                    mark(visited, i * 3 + 2, j * 3);
                } else if (grid[i].charAt(j) == '\\') {
                    mark(visited, i * 3, j * 3);
                    mark(visited, i * 3 + 1, j * 3 + 1);
                    mark(visited, i * 3 + 2, j * 3 + 2);
                }
            }
        }
        
        for (int i = 0; i < 3 * n; i++) {
            for (int j = 0; j < 3 * n; j++) {
                if (visited[i][j] == 0) {
                    regions++;
                    dfs(visited, i, j);
                }
            }
        }
        
        return regions;
    }
    
    private void mark(int[][] visited, int i, int j) {
        if (i >= 0 && i < visited.length && j >= 0 && j < visited[0].length) {
            visited[i][j] = 1;
        }
    }
    
    private void dfs(int[][] visited, int i, int j) {
        if (i >= 0 && i < visited.length && j >= 0 && j < visited[0].length && visited[i][j] == 0) {
            visited[i][j] = 1;
            int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
            for (int[] dir : dirs) {
                int ni = i + dir[0];
                int nj = j + dir[1];
                dfs(visited, ni, nj);
            }
        }
    }
}

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