<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.
        """
        n_rows, n_cols = len(board), len(board[0])
        dirs = [(-1,0), (1,0), (0,1), (0,-1)]

        def dfs(r, c):

            if (not(0<=r<n_rows)) or (not(0<=c<n_cols)) or (board[r][c] != 'O'):
                return
            
            board[r][c] = "Y"

            dfs(r+1,c)
            dfs(r-1,c)
            dfs(r,c+1)
            dfs(r,c-1)

            # for dr, dc in dirs:
            #     new_r, new_c = r+dr, c+dc
            #     if (0 <= new_r < n_rows - 1) and (0 <= new_c < n_cols-1) and (board[new_r][new_c] == "O"):
            #         # print(f'satisified bndry = {(new_r, new_c)}')
            #         dfs(new_r, new_c)
                    

        # convert all border-O's to say Y
        # use DFS from these border ones
        for r in range(n_rows):
            for c in range(n_cols):
                if (r==0 or c==0 or r==n_rows-1 or c==n_cols-1) and (board[r][c] == "O"):
                    # print(f'entering from {(r,c)}')
                    dfs(r,c)

        # print(f'now board = {board}')

        # Then convert all O's to X
        # and Y's to O
        for r in range(n_rows):
            for c in range(n_cols):
                if board[r][c] == "O": board[r][c] = "X"
                if board[r][c] == "Y": board[r][c] = "O"
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