<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.
        """
        # Getting the number of rows and columns
        ROWS, COLS = len(board), len(board[0])

        # Setting up DFS function for matrix
        visit = set()
        def dfs(row: int, col: int) -> None:
            # Mark the current cell as visited and convert to "*"
            visit.add((row, col))
            board[row][col] = "*"

            # Getting ready to move down, up, right, and left
            for dx, dy in [[1, 0], [-1, 0], [0, 1], [0, -1]]:
                r, c = row + dx, col + dy
                # Move to the next cell if its in bounds, not visited, and it's "O"
                if (0 <= r < ROWS and 0 <= c < COLS and (r, c) not in visit and board[r][c] == "O"):
                    dfs(r, c)

            return
        
        # Call the DFS function on the top and bottom edges
        for r in range(ROWS):
            if board[r][0] == "O":
                dfs(r, 0)
            if board[r][COLS - 1] == "O":
                dfs(r, COLS - 1)
        
        # Call the DFS function of the left and right edges
        for c in range(COLS):
            if board[0][c] == "O":
                dfs(0, c)
            if board[ROWS - 1][c] == "O":
                dfs(ROWS - 1, c)
        
        # Convert all left over letters to what they are suppose to be
        for r in range(ROWS):
            for c in range(COLS):
                if board[r][c] == "O":
                    board[r][c] = "X"
                if board[r][c] == "*":
                    board[r][c] = "O"
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