<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>

# True means that it is surrounded
def dfs(board, i, j, n, m, visited):
    if i < 0 or i >= n or j < 0 or j >= m:
        # if i go out of bounds it's not surrounded
        return False
    elif board[j][i] in ('#', "X"):
        # if i have already visited stop 
        return True
    elif board[j][i] == 'O':
        # if its part of the same its possibly surrounded
        board[j][i] = '#'
        visited.append((i,j))
    # if surrounded on all sides it is surrounded
    return all (
        (dfs(board, i-1, j, n, m, visited),
         dfs(board, i+1, j, n, m, visited),
         dfs(board, i, j-1, n, m, visited),
         dfs(board, i, j+1, n, m, visited)))

def print_board(board):
    print("\n".join(" ".join(row) for row in board), end="\n\n")

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])
        for j in range(m):
            for i in range(n):
                # print(f"{i=} {j=}")
                # print_board(board)
                if board[j][i] == "O":
                    visited = []
                    is_surrounded = dfs(board, i, j, n, m, visited)
                    char = "X" if is_surrounded else "O"
                    for i_v,j_v in visited:
                        board[j_v][i_v] = char
        # print_board(board)

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