<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.
        """
        # options: recurse and have return value to X if valid region or O if not
        #   issue is return can only apply to root nodes in the same path, not tree
        # could also post process at end using visited ds, biasing toward O or X
        # could also process border areas first, marking dfs of found O's as visited but don't modify
        #   subsequently iterate through interior cells
        def bfs(i,j,isBorder):
            if board[i][j] == 'O':
                stack = [[i,j]]
                while stack:
                    ii,jj = stack.pop()
                    if not isBorder:
                        board[ii][jj] = 'X'
                    visited.add((ii,jj))

                    directions = [[-1,0],[1,0],[0,-1],[0,1]]
                    for di,dj in directions:
                        ni,nj = ii+di,jj+dj
                        if 0 <= ni < m and 0 <= nj < n and board[ni][nj] == 'O' and (ni,nj) not in visited:
                            stack.append([ni,nj])

        # initial edge case here applies to any matrix without interior cells
        if len(board) < 3 or len(board[0]) < 3:
            return board
        m, n = len(board), len(board[0])

        # identify regions connected to the edge of the board
        border_coords = ([[i,0] for i in range(1,m)]+
                 [[0,j] for j in range(n-1)]+
                 [[i,n-1] for i in range(m-1)]+
                 [[m-1,j] for j in range(1,n)])

        visited = set()

        for i,j in border_coords:
            bfs(i,j,True)

        for i in range(1,m-1):
            for j in range(1,n-1):
                if (i,j) not in visited:
                    bfs(i,j,False)

        return board


        

        
25130