<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.
        """
        global m,n
        def bfs(a,b):
            global m,n
            visit = dict()
            visit[(a,b)]= True
            queue = []
            queue.append((a,b))
            while len(queue)>0:
                x, y = queue.pop(0)
                if x>=m-1 or y>=n-1 or x == 0 or y == 0:
                    return False
                if board[x][y+1] =='O' and (x,y+1) not in visit:
                    visit[(x,y+1)] = True
                    queue.append((x,y+1))
                if board[x][y-1] =='O' and (x,y-1) not in visit:
                    visit[(x,y-1)] = True
                    queue.append((x,y-1))
                if board[x+1][y] =='O' and (x+1,y) not in visit:
                    visit[(x+1,y)] = True
                    queue.append((x+1,y))
                if board[x-1][y] =='O' and (x-1,y) not in visit:
                    visit[(x-1,y)] = True
                    queue.append((x-1,y))
            
            return True
        def surround(i,j):
            if i>=m-1 or j >=n-1:
                return
            board[i][j] = 'X'
            if board[i][j+1] == 'O':
                surround(i,j+1)
            if board[i][j-1] == 'O':
                surround(i,j-1)
            if board[i+1][j] == 'O':
                surround(i+1,j)
            if board[i-1][j] == 'O':
                surround(i-1,j)

        m = len(board)
        n = len(board[0])
        for k in range(m):
            for l in range(n):
                if bfs(k,l):
                    surround(k,l)
        
25850