130. Surrounded Regions

/*
 * @lc app=leetcode id=130 lang=cpp
 *
 * [130] Surrounded Regions
 *
 * https://leetcode.com/problems/surrounded-regions/description/
 *
 * algorithms
 * Medium (41.16%)
 * Likes:    9078
 * Dislikes: 2037
 * Total Accepted:    893.7K
 * Total Submissions: 2.1M
 * Testcase Example:  '[["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]'
 *
 * You are given an m x n matrix board containing letters 'X' and 'O', capture
 * regions that are surrounded:
 *
 *
 * Connect: A cell is connected to adjacent cells horizontally or
 * vertically.
 * Region: To form a region connect every 'O' cell.
 * Surround: The region is surrounded with 'X' cells if you can connect the
 * region with 'X' cells and none of the region cells are on the edge of the
 * board.
 *
 *
 * To capture a surrounded region, replace all 'O's with 'X's in-place within
 * the original board. You do not need to return anything.
 *
 *
 * Example 1:
 *
 *
 * Input: board =
 * [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
 *
 * Output:
 * [["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
 *
 * Explanation:
 *
 * In the above diagram, the bottom region is not captured because it is on the
 * edge of the board and cannot be surrounded.
 *
 *
 * Example 2:
 *
 *
 * Input: board = [["X"]]
 *
 * Output: [["X"]]
 *
 *
 *
 * Constraints:
 *
 *
 * m == board.length
 * n == board[i].length
 * 1 <= m, n <= 200
 * board[i][j] is 'X' or 'O'.
 *
 *
 */

// @lc code=start
class Solution {
public:
    void solve(vector<vector<char>>& board)
    {
        int rows = board.size(), cols = board[0].size();
        for (int i = 0; i < rows; i++)
        {
            for (int j = 0; j < cols; j++)
            {
                if (board[i][j] == 'O' && (i == 0 || j == 0 || i == rows - 1 || j == cols - 1))
                {
                    dfs(board, i, j);
                }
            }
        }
        for (int i = 0; i < rows; i++)
        {
            for (int j = 0; j < cols; j++)
            {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == '#') board[i][j] = 'O';
            }
        }
    }
    void dfs(vector<vector<char>>& board, int i, int j)
    {
        int rows = board.size(), cols = board[0].size();
        if (i < 0 || j < 0 || i >= rows || j >= cols || board[i][j] != 'O')
        {
            return;
        }
        board[i][j] = '#';
        vector<vector<int>> directions = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
        for(auto dir : directions)
        {
            dfs(board, i + dir[0], j + dir[1]);
        }
    }
};
// @lc code=end

• T: $O()$
• S: $O()$