<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.
        """

        '''
        x x x x
        x o o x
        x x o x
        x o x x

        for each of the 4 orders, have a dfs loop that goes from all X. If ypu find a O, add it to a []. Keep a visited for each border.

        Common across the 4 sets can be changed to X

        x o x
        o x o
        x o x
        '''
        # consists of sets of all O's that can be reached from a O in the border
        left = set()
        right = set()
        top = set()
        bottom = set()

        r = len(board)
        c = len(board[0])

        visited_l = [[False for _ in range(c)] for _ in range(r)]
        visited_r = [[False for _ in range(c)] for _ in range(r)]
        visited_t = [[False for _ in range(c)] for _ in range(r)]
        visited_b = [[False for _ in range(c)] for _ in range(r)]

        def valid(x,y):
            if x>=0 and x<r and y>=0 and y<c:
                return True

            return False

        direc = [(0,1),(1,0),(0,-1),(-1,0)]

        def dfs(x,y,visited,list_add):

            for i in direc:
                tx = x+i[0]
                ty = y+i[1]

                if valid(tx,ty) and visited[tx][ty]==False and board[tx][ty]=='O':
                    visited[tx][ty]=True
                    list_add.add((tx,ty))
                    dfs(tx,ty,visited,list_add)


        for i in range(r):
            if board[i][0]=='O':
                dfs(i,0,visited_l,left)

            if board[i][c-1]=='O':
                dfs(i,c-1,visited_r,right)


        for i in range(c):
            if board[0][i]=='O':
                dfs(0,i,visited_t,top)

            if board[r-1][i]=='O':
                dfs(r-1,i,visited_b,bottom)

        for i in range(r):
            for j in range(c):
                if board[i][j]=='O':
                    if (i,j) not in left and (i,j) not in right and (i,j) not in top and (i,j) not in bottom:
                        if i!=0 and j!=0 and i!=r-1 and j!=c-1:
                            board[i][j]='X'



        
        

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