<p>You are given an <code>m x n</code> matrix <code>board</code> containing <strong>letters</strong> <code>&#39;X&#39;</code> and <code>&#39;O&#39;</code>, <strong>capture regions</strong> that are <strong>surrounded</strong>:</p>

<ul>
	<li><strong>Connect</strong>: A cell is connected to adjacent cells horizontally or vertically.</li>
	<li><strong>Region</strong>: To form a region <strong>connect every</strong> <code>&#39;O&#39;</code> cell.</li>
	<li><strong>Surround</strong>: The region is surrounded with <code>&#39;X&#39;</code> cells if you can <strong>connect the region </strong>with <code>&#39;X&#39;</code> cells and none of the region cells are on the edge of the <code>board</code>.</li>
</ul>

<p>A <strong>surrounded region is captured</strong> by replacing all <code>&#39;O&#39;</code>s with <code>&#39;X&#39;</code>s in the input matrix <code>board</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]</span></p>

<p><strong>Explanation:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/xogrid.jpg" style="width: 367px; height: 158px;" />
<p>In the above diagram, the bottom region is not captured because it is on the edge of the board and cannot be surrounded.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[&quot;X&quot;]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[[&quot;X&quot;]]</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == board.length</code></li>
	<li><code>n == board[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>board[i][j]</code> is <code>&#39;X&#39;</code> or <code>&#39;O&#39;</code>.</li>
</ul>

class Solution {
public:
    int dx[4] = {1, 0, -1, 0};
    int dy[4] = {0, -1, 0, 1};

    bool check(int i, int j, int n, int m) {
        return i < n && j < m && i >= 0 && j >= 0;
    }

    void dfs(int i, int j, vector<vector<char>> &grid, vector<vector<int>> &vis) {
        int n = grid.size();
        int m = grid[0].size();
        vis[i][j] = 1;
        for (int k = 0; k < 4; k++) {
            int ni = i + dx[k];
            int nj = j + dy[k];
            if (check(ni, nj, n, m)) {
                if (grid[ni][nj] == 'O' && !vis[ni][nj]) {
                    dfs(ni, nj, grid, vis);
                }
            }
        }
    }
    void dfs2(int i, int j, vector<vector<char>> &grid, vector<vector<int>> &vis) {
    int n = grid.size();
    int m = grid[0].size();
    vis[i][j] = 1;
    grid[i][j] = 'X';
    for (int k = 0; k < 4; k++) {
        int ni = i + dx[k];
        int nj = j + dy[k];
        if (check(ni, nj, n, m)) {
            if (grid[ni][nj] == 'O' && !vis[ni][nj]) {
                dfs2(ni, nj, grid, vis);
            }
        }
    }
}
    void solve(vector<vector<char>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> vis(n, vector<int>(m, 0));

        // Run DFS from the boundary cells
        for (int i = 0; i < n; i++) {
            if (grid[i][0] == 'O' && !vis[i][0]) {
                dfs(i, 0, grid, vis);
            }
            if (grid[i][m - 1] == 'O' && !vis[i][m - 1]) {
                dfs(i, m - 1, grid, vis);
            }
        }
        for (int i = 0; i < m; i++) {
            if (grid[0][i] == 'O' && !vis[0][i]) {
                dfs(0, i, grid, vis);
            }
            if (grid[n - 1][i] == 'O' && !vis[n - 1][i]) {
                dfs(n - 1, i, grid, vis);
            }
        }

        // Process remaining cells
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 'O' && !vis[i][j]) {
                    dfs2(i, j, grid, vis);
                }
            }
        }
    }
};
18143