<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:
    # We need to capture O's that are surrounded.
    # Observation: if any O in a region is on an edge, it's safe
    # BFS or DFS? Either can probably work
    # We have a safe set. We go around edges and do a DFS if we find an O and add all to safe set
    # We then traverse rest of matrix and capture any O's that we find that aren't in the safe set
    # Time complexity: O(mn)
    # Space complexity: O(2mn) -> O(mn)
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        def safeDFS(row, col):
            if row < 0 or col < 0 or row >= len(board) or col >= len(board[row]) or board[row][col] == 'X' or (row, col) in safe:
                return
            safe.add((row, col))
            safeDFS(row + 1, col)
            safeDFS(row - 1, col)
            safeDFS(row, col + 1)
            safeDFS(row, col - 1)
        
        def dfs(row, col):
            if row < 0 or col < 0 or row >= len(board) or col >= len(board[row]) or board[row][col] == 'X' or (row, col) in safe:
                return
            board[row][col] = 'X'
            dfs(row + 1, col)
            dfs(row - 1, col)
            dfs(row, col + 1)
            dfs(row, col - 1)

        safe = set()
        for i in range(len(board)):
            if board[i][0] == 'O': 
                safeDFS(i, 0)
            if board[i][len(board[i]) - 1] == 'O':
                safeDFS(i, len(board[i]) - 1)
        for j in range(len(board[0])):
            if board[0][j] == 'O':
                safeDFS(0, j)
            if board[len(board) - 1][j] == 'O':
                safeDFS(len(board) - 1, j)
                        
        for x in range(1, len(board) - 1):
            for y in range(1, len(board[x]) - 1): 
                if board[x][y] == 'O':
                    dfs(x, y)
        
        
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