<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.
        """
        
        # Find all positions with O on inner portion of matrix?
        # Find all positions on border of matrix with O
        # Traverse from these outside positions and find paths to inner O and if path exists
        # We remove the O's from the inner list signaling that these O's must remain unchanged
        # At the end, whatever positions are left in the inner portion must be changed to an X? 


        # What constitutes an O being on the border??? If row = 0 or col = 0 or row = 3 or col = 3

        rows = len(board)
        cols = len(board[0])

        inner_pos = set()
        outer_pos = set()
        outer_queue = deque()
        visited = set()

        for i in range(0, rows):
            for j in range(0, cols):
                if (i == 0 or i == rows - 1 or j == 0 or j == cols - 1):
                    if board[i][j] == "O":
                        outer_pos.add((i, j))
                        outer_queue.append((i, j))
                        visited.add((i, j))
                else:
                    if board[i][j] == "O":
                        inner_pos.add((i, j))
        
        def bfs():
            directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]

            while outer_queue:
                for _ in range(len(outer_queue)):
                    cr, cc = outer_queue.popleft()

                    for dr, dc in directions:
                        r = cr + dr
                        c = cc + dc
                        if r in range(rows) and c in range(cols) and (r, c) not in visited and board[r][c] == "O":
                            outer_queue.append((r, c))
                            visited.add((r, c))
                            if (r, c) in inner_pos:
                                inner_pos.remove((r, c))

        

        
        print(outer_pos)
        print(inner_pos)
        bfs()
        print(inner_pos)
        for pos in inner_pos:
            row, col = pos
            board[row][col] = "X"


    

22970