<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:
        
        m = len(grid)
        n = len(grid[0])

        nz = numerozonas(grid, m, n)
        dias = 0
        if nz != 1:
            return (dias)

        unos = contar(grid, m,n,1)
        if unos < 3:
            return unos
        
        aux = [[10 for _ in range(n)] for _ in range(m)]

        for i in range(m):
            for j in range(n):
                if (grid[i][j] == 1):
                    aux[i][j] = conexion(grid, m, n, i, j)
        
        if min(min(fila) for fila in aux) == 1: return 1

        for i in range(m):
            for j in range(n):
                if (grid[i][j] == 1):
                    grid[i][j] = 0
                    nz = numerozonas(grid, m, n)
                    if nz != 1: return 1
                    grid[i][j] = 1
        
        return 2

                          

def zona(matriz, f,c,n,tam1, tam2):
    if matriz[f][c] == n:
        if f>0:
            if matriz[f-1][c] == -1:
                matriz[f-1][c] = n
                matriz = zona(matriz, f-1,c,n,tam1,tam2)
        if f<tam1-1:
            if matriz[f+1][c] == -1:
                matriz[f+1][c] = n
                matriz = zona(matriz, f+1,c,n,tam1,tam2)
        if c>0:
            if matriz[f][c-1] == -1:
                matriz[f][c-1] = n
                matriz = zona(matriz, f,c-1,n,tam1,tam2)
        if c<tam2-1:
            if matriz[f][c+1] == -1:
                matriz[f][c+1] = n
                matriz = zona(matriz, f,c+1,n,tam1,tam2)
    return matriz

def numerozonas(grid, m,n):
    aux = [[0 for _ in range(n)] for _ in range(m)]

    for i in range(m):
            for j in range(n):
                if grid[i][j] == 1: aux[i][j] = -1

    contador = 0;        
    for i in range(m): 
        for j in range(n):
            if (aux[i][j] == -1):
                contador += 1
                aux[i][j] = contador
                aux = zona(aux,i,j,contador,m,n)

    return(contador)

def contar(mat,m,n,nc):
    numero = 0
    for i in range(m):
        for j in range(n):
            if mat[i][j] == nc: numero +=1
            if numero >= 3: return numero
    return numero

def conexion(original, m,n, fil, col):
    lados = 0
    hor = 0
    ver = 0
    if fil>0:
        if original[fil-1][col] == 1: lados +=1; ver +=1
    if fil<m-1:
        if original[fil+1][col] == 1: lados +=1; ver +=1
    if col>0:
        if original[fil][col-1] == 1: lados +=1; hor +=1
    if col<n-1:
        if original[fil][col+1] == 1: lados +=1; hor +=1
    if lados == 2 and (hor == 2 or ver ==2): lados = 100
    return lados
      
350