<p>You are given an <code>m x n</code> binary grid <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water. An <strong>island</strong> is a maximal <strong>4-directionally</strong> (horizontal or vertical) connected group of <code>1</code>&#39;s.</p>

<p>The grid is said to be <strong>connected</strong> if we have <strong>exactly one island</strong>, otherwise is said <strong>disconnected</strong>.</p>

<p>In one day, we are allowed to change <strong>any </strong>single land cell <code>(1)</code> into a water cell <code>(0)</code>.</p>

<p>Return <em>the minimum number of days to disconnect the grid</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land1.jpg" style="width: 500px; height: 169px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]

<strong>Output:</strong> 2
<strong>Explanation:</strong> We need at least 2 days to get a disconnected grid.
Change land grid[1][1] and grid[0][2] to water and get 2 disconnected island.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land2.jpg" style="width: 404px; height: 85px;" />
<pre>
<strong>Input:</strong> grid = [[1,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Grid of full water is also disconnected ([[1,1]] -&gt; [[0,0]]), 0 islands.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution:
    def minDays(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid[0])


        
     
        ans = False
        time = 0
        time_in = dict()
        ret = dict()
        used_v = set()

        def find_edges(i, j):
            rev_edg = []
            edges_white = []
            if i - 1 >= 0:
             #   print(i - 1, j)
              #  print(grid[i - 1][j])
               # print(used_v)
                if grid[i - 1][j] == 2 and   ((i - 1) * 30 + j) * 900 + (i * 30 + j)  not in used_v:
                    rev_edg.append([i - 1, j])
                elif  grid[i - 1][j] == 1:
                    edges_white.append([i - 1, j])
                    
                    
            if i + 1 < n:
                if grid[i + 1][j] == 2 and   ((i * 30 + j) * 900 + ((i + 1) * 30 + j) )  not in used_v:
                    rev_edg.append([i + 1, j])
                elif  grid[i + 1][j] == 1:
                    edges_white.append([i + 1, j])


            if j - 1 >= 0:
                if grid[i][j - 1] == 2 and   ((i * 30 + (j - 1)) * 900 + (i * 30 + j))   not in used_v:
                    rev_edg.append([i , j - 1])
                elif  grid[i][j - 1] == 1:
                    edges_white.append([i, j - 1])

            if j + 1 < m:
                if grid[i][j + 1] == 2 and  ((i * 30 + j) * 900 + (i * 30 + j + 1))   not in used_v:
                    rev_edg.append([i , j + 1])
                elif  grid[i][j + 1] == 1:
                    edges_white.append([i, j + 1])
            
            return [rev_edg, edges_white]


        def dfs (i, j):
            nonlocal ans
            nonlocal time
            time += 1
            time_in[i * 30 + j] = time
            ret[i * 30 + j] = time
           # nonlocal cnt_snd
            #cnt_snd += 1
            grid[i][j] = 2
            
            r_ed, w_ed = find_edges(i, j)
            #print('r_ed ', r_ed, ' w_ed ',  w_ed)
            #print(i, j, r_ed, w_ed, ret)
            for k in r_ed:
                #print('ret ', ret, ret[i * 30 + j], ret[k[0] * 30 + k[1]])

                ret[i * 30 + j] = min(ret[i * 30 + j], time_in[k[0] * 30 + k[1]])
                #print(ret[i * 30 + j])
            
            for k in w_ed:
                if grid[k[0]][k[1]] == 2:
                    continue
                used_v.add(min((i * 30 + j) * 900 + (k[0] * 30 + k[1]), (k[0] * 30 + k[1]) * 900 + (i * 30 + j)))
                
                dfs(k[0], k[1])

                    
                #print('sravn_r ', ret[i * 30 + j], ret[k[0] * 30 + k[1]])
                #print(ret[k[0] * 30 + k[1]] , time_in[i * 30 + j])
                ret[i * 30 + j] = min(ret[i * 30 + j], ret[k[0] * 30 + k[1]])
                ans = ans or (ret[k[0] * 30 + k[1]] > time_in[i * 30 + j])
                #print(ret)
                #print('dfs ', i, j, k[0], k[1])
                    

                    
                    
            

           
        
        for i in range(n):
            for j in range(m):
                if grid[i][j] == 1:
                    dfs(i, j)
                    
                    break
            if time > 0:
                break
        #print('time_in ', time_in,  'ret ',  ret)
        
        #print('ij ', i,j)
        for ii in range(i, n):
            for jj in range( m):
                if grid[ii][jj] == 1:
                    #print('2 syshi')
                    return 0
        
        if time <= 2:
            #print('tyt time')
            return time

        #print(n,m)


        if n == 3 and m == 3:
            if grid[0][0] == 0 and grid[2][2] == 0: 
                if grid[0][1] == 2 and grid[0][2] == 2: 
                    if grid[2][0] == 2 and grid[2][1] == 2:
                        if grid[1][0] == 2 and grid[1][1] == 2 and grid[1][2] == 2:
                            return 1
        
        if ans:
            ##print('most')
            return 1
        else:
            #print('tyt ret')
            return 2
17700