<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.
        """
        m = len(board)
        n = len(board[0])
        regions = {}
        visited = [[0 for _ in range(n)] for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if board[i][j] == "O" and visited[i][j] == 0:
                    [isBorder, regions, visited] = self.bfs(board, (i,j), regions, visited)
                    print(isBorder)
                    print(regions)
                    print(visited)
                    if isBorder == False:
                        for cell in regions[(i,j)]:
                            board[cell[0]][cell[1]] = "X"


    def bfs(self, board, t, regions, visited):
        sp = list(t)
        q = deque()
        q.append(sp)
        regions[t] = [sp]
        isBorder = self.isBorderCell(board, sp[0], sp[1])
        visited[sp[0]][sp[1]] = 1
        while len(q) > 0:
            cell = q.popleft()
            row = cell[0]
            col = cell[1]
            for diff in [-1, 1]:
                nrow = row + diff
                ncol = col + diff
                if 0 <= nrow <= len(board)-1 and board[nrow][col] == "O" and visited[nrow][col] == 0:
                    q.append([nrow, col])
                    regions[t].append([nrow, col])
                    visited[nrow][col] = 1
                    isBorder = isBorder or self.isBorderCell(board, nrow, col)
                if 0 <= ncol <= len(board[0])-1 and board[row][ncol] == "O" and visited[row][ncol] == 0:
                    q.append([row, ncol])
                    regions[t].append([row, ncol])
                    visited[row][ncol] = 1
                    isBorder = isBorder or self.isBorderCell(board, row, ncol)
        return [isBorder, regions, visited]

    def isBorderCell(self, board, i, j):
        if i == 0 or j == 0 or i == len(board)-1 or j == len(board[0])-1:
            return True
        else:
            return False










        
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