<p>Given a <code>m x n</code> binary matrix <code>mat</code>. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing <code>1</code> to <code>0</code> and <code>0</code> to <code>1</code>). A pair of cells are called neighbors if they share one edge.</p>

<p>Return the <em>minimum number of steps</em> required to convert <code>mat</code> to a zero matrix or <code>-1</code> if you cannot.</p>

<p>A <strong>binary matrix</strong> is a matrix with all cells equal to <code>0</code> or <code>1</code> only.</p>

<p>A <strong>zero matrix</strong> is a matrix with all cells equal to <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/11/28/matrix.png" style="width: 409px; height: 86px;" />
<pre>
<strong>Input:</strong> mat = [[0,0],[0,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> mat = [[0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Given matrix is a zero matrix. We do not need to change it.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,0,0],[1,0,0]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> Given matrix cannot be a zero matrix.
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 3</code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution:

    def all_zero(self, flatten_m):
        for i in range(len(flatten_m)):
            if flatten_m[i] != 0:
                return False
        return True

    def minFlips(self, mat: List[List[int]]) -> int:
        # order of flips doesn't matter
        # size of the matrix <= 3*3, at most 2^9 combintations
        flatten_mat = []
        for i in range(len(mat)):
            for j in range(len(mat[i])):
                flatten_mat.append(mat[i][j])
        q = collections.deque()
        q.append((0, flatten_mat))
        visited = set()
        visited.add(tuple(flatten_mat))
        while len(q) >0:
            ct, m = q.popleft()
            if self.all_zero(m) == True:
                return ct
            for i in range(len(mat)):
                for j in range(len(mat[0])):
                    new_m = m[:]
                     # flip center and neighbors
                    for di, dj in [(1,0),(0,1),(-1,0),(0,-1),(0,0)]:
                        if i+di>=0 and i+di<len(mat) and j+dj>=0 and j+dj<len(mat[0]):
                            new_m[(i+di)*len(mat[0])+j+dj] = 0 if new_m[(i+di)*len(mat[0])+j+dj]==1 else 1
                    if tuple(new_m) not in visited:
                        visited.add(tuple(new_m))
                        q.append((ct+1, new_m))
        return -1
            

16800