<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(vector<vector<char>>& board, int i, int j) {
        vector<pair<int, int>> dir{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        stack<pair<int, int>> s;
        vector<pair<int, int>> convertedRegion;
        s.push(make_pair(i, j));
        int borderRight = board[0].size() - 1, borderBottom = board.size() - 1;
        unordered_set<int> mySet;
        while (!s.empty()) {
            int tempi = s.top().first;
            int tempj = s.top().second;
            s.pop();
            if (tempi == 0 || tempi == borderBottom || tempj == 0 ||
                tempj == borderRight) {
                return;
            }
            int mathIndex = tempi * (borderRight + 1) + tempj;
            if (mySet.find(mathIndex) == mySet.end()) {
                mySet.insert(mathIndex);
                convertedRegion.push_back({tempi, tempj});
                for (auto p : dir) {
                    int neighborI = tempi + p.first;
                    int neighborJ = tempj + p.second;

                    if (neighborI >= 0 && neighborI <= borderBottom &&
                        neighborJ >= 0 && neighborJ <= borderRight &&
                        board[neighborI][neighborJ] == 'O') {
                        int neighborIndex =
                            neighborI * (borderRight + 1) + neighborJ;
                        if (mySet.find(neighborIndex) == mySet.end()) {
                            s.push({neighborI, neighborJ});
                        }
                    }
                }
            }
        }
        for (auto p : convertedRegion) {
            board[p.first][p.second] = 'X';
        }
    }
    void solve(vector<vector<char>>& board) {
        int row = board.size();
        int col = board[0].size();
        for (int i = 1; i < row - 1; i++) {
            for (int j = 1; j < col - 1; j++) {
                if (board[i][j] == 'O') {
                    dfs(board, i, j);
                }
            }
        }
    }
};
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