<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>

from collections import deque

class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        N = len(board)
        if N <= 2:
            return
        M = len(board[0])
        if M<=2:
            return
        
        visited = [ [False] * M  for _ in range(N) ]
        for i in range(1, N-1):
            for j in range(1, M-1):
                if board[i][j] == "X" or visited[i][j]:
                    continue
                
                coords = []
                q = deque()
                q.append((i,j))
                coords.append((i,j))
                visited[i][j] = True
                toColor = True
                # BFS((i,j))
                while q:
                    x, y = q.popleft()
                    print(x, y)
                    # обойти все клетки, имеющие общее ребро
                    for (nx, ny) in ((x-1, y), (x+1, y), (x, y-1), (x, y+1)):
                        #if  nx >=0 and ny >= 0 and nx < N and ny < M and visited[nx][ny] and board[nx][ny] != "X":
                        if  nx >=0 and ny >= 0 and nx < N and ny < M and not visited[nx][ny] and board[nx][ny] != "X":
                            # если это граничная клетка без воды
                            if nx==0 or ny==0 or nx == N-1 or ny == M-1:  
                                toColor = False
                                coords = [] # есть соединение с сушей, не красим

                            visited[nx][ny] = True
                            q.append((nx, ny))
                            if toColor:
                                coords.append((nx, ny))
                if toColor:
                    for x, y in coords:
                        board[x][y] = 'X'


        
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