<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:
    void dfs(int r, int c, bool mark, vector<vector<char>>& board, vector<vector<bool>>& visited) {
        // base cases: not 'O' or visited, leave
        if (board.at(r).at(c)!='O' or visited.at(r).at(c)) {
            return;
        }
        // process current node
        if (mark) {
            board[r][c] = 'X';
        }
        visited[r][c] = true;

        // neighbor nodes
        if (r>0) {dfs(r-1,c,mark,board,visited);}
        if (r<board.size()-1) {dfs(r+1,c,mark,board,visited);}
        if (c>0) {dfs(r,c-1,mark,board,visited);}
        if (c<board.at(0).size()-1) {dfs(r,c+1,mark,board,visited);}
    }
    void solve(vector<vector<char>>& board) {
        // sizes
        int row_num = board.size();
        int col_num = board.at(0).size();
        // create visited matrix
        vector<vector<bool>> visited(row_num,vector<bool>(col_num,false));

        // visit all edge nodes with dfs
        // 1) first the up and bottom rows
        for (auto r: {0,row_num-1}) {
            for (int c=0; c<col_num; ++c) {
                dfs(r,c,false,board,visited);
            }
        }
        // 2) next the left and right columns
        for (int r=1; r<row_num-1; ++r) {
            for (auto c: {0,col_num-1}) {
                dfs(r,c,false,board,visited);
            }
        }
        // then visit inner nodes with dfs
        for (int r=1; r<row_num-1; ++r) {
            for (int c=1; c<col_num-1; ++c) {
                dfs(r,c,true,board,visited);
            }
        }
        
    }
};
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