<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.
        """
        # 2d itr through board. For every "O", do a DFS for all the O's
        # If already visited, don't need DFS for that "O" again
        # If any of the included "O"'s touch the edge, all "O"'s part of the DFS are permenantly "O"s
        # Must complete the DFS
        n,m = len(board),len(board[0])

        q = collections.deque()
        curItr = []
        for i,row in enumerate(board):
            for j,cell in enumerate(row):
                if cell == "O":
                    # Perform DFS on cell
                    q.append((i,j))
                    touchedEdge = False
                    curItr = []
                    board[i][j] = "X"
                    while q:
                        #print(q)
                        a,b = q.pop()
                        curItr.append((a,b))
                        if a == 0 or b == 0 or a == n-1 or b == m-1:
                            touchedEdge = True
                        # add cur cell's neighbors to q if its "O" and check if neighbors 'exist' and not visited
                        if a > 0  and board[a-1][b] == "O":
                            board[a-1][b] = "X"
                            q.append((a-1,b))
                        if a < n - 1 and board[a+1][b] == "O":
                            board[a+1][b] = "X"
                            q.append((a+1,b))
                        if b > 0 and board[a][b-1] == "O":
                            board[a][b-1] = "X"
                            q.append((a,b-1))
                        if b < m - 1 and board[a][b+1] == "O":
                            board[a][b+1] = "X"
                            q.append((a,b+1))
                    # Need a way to mark all 'touched edge' as X! 
                    if touchedEdge: # If touch edge change all back to O!
                        for a,b in curItr:
                            board[a][b] = "O"

                   
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