<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:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """

        m = len(board)
        n = len(board[0])

        def dfs(i, j):
            dfs_stack = [(i, j)]
            marked = set(dfs_stack)

            while len(dfs_stack) > 0:
                curri, currj = dfs_stack.pop()
                if curri == 0 or curri == m-1 or currj == 0 or currj == n-1:
                    return False
                if board[curri-1][currj] == 'O' and (curri-1, currj) not in marked: 
                    dfs_stack.append((curri-1, currj))
                    marked.add((curri-1, currj))
                if board[curri+1][currj] == 'O' and (curri+1, currj) not in marked: 
                    dfs_stack.append((curri+1, currj))
                    marked.add((curri+1, currj))
                if board[curri][currj-1] == 'O' and (curri, currj-1) not in marked: 
                    dfs_stack.append((curri, currj-1))
                    marked.add((curri, currj-1))
                if board[curri][currj+1] == 'O' and (curri, currj+2) not in marked: 
                    dfs_stack.append((curri, currj+1))
                    marked.add((curri, currj+1))

            return True

        def dfsRewrite(i, j):
            dfs_stack = [(i, j)]
            board[i][j] = 'X'

            while len(dfs_stack) > 0:
                curri, currj = dfs_stack.pop()

                if board[curri-1][currj] == 'O': 
                    dfs_stack.append((curri-1, currj))
                    board[curri-1][currj] = 'X'
                if board[curri+1][currj] == 'O': 
                    dfs_stack.append((curri+1, currj))
                    board[curri+1][currj] = 'X'
                if board[curri][currj-1] == 'O': 
                    dfs_stack.append((curri, currj-1))
                    board[curri][currj-1] = 'X'
                if board[curri][currj+1] == 'O': 
                    dfs_stack.append((curri, currj+1))
                    board[curri][currj+1] = 'X'


        for i in range(m):
            for j in range(n):
                if board[i][j] == 'O': 
                    if dfs(i, j): dfsRewrite(i, j)
        
23150