<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 minFlips(self, mat: List[List[int]]) -> int:

        neighbors = [[0, 1], [0, -1], [1, 0], [-1, 0]]
        M, N = len(mat), len(mat[0])
        moves = 0
        def convert(d1):
            res = 0
            for i in range(M):
                for j in range(N):
                    res <<= 1
                    res |= d1[i][j]
            return res
        def construct(d2):
            res = [[0 for _ in range(N)] for _ in range(M)]
            for i in range(M-1, -1, -1):
                for j in range(N-1, -1, -1):
                    res[i][j] = d2 & 0b1
                    d2 >>= 1
            return res
        first = convert(mat) << 4 | 0b1111
        q = deque([first])
        visited = set([first])
        while q:
            k = len(q)
            for _ in range(k):
                cur = q.popleft()
                state, x, y = cur >> 4, (cur >> 2) & 0b11, cur & 0b11
                #print("moves = ", moves, ", state = ", f'{state:04b}', ", x = ", x, ", y = ", y)
                if state == 0:
                    return moves
                data = construct(state)
                #print(" data = ", data)
                for i in range(M):
                    for j in range(N):
                        new_data = copy.deepcopy(data)
                        #print("    i = ", i, ", j = ", j, ", data = ", new_data)
                        #print("        before change i = ", i, ", j = ", j, ", val = ", new_data[i][j])
                        new_data[i][j] ^= 1
                        #print("        after change i = ", i, ", j = ", j, ", val = ", new_data[i][j])
                        for di, dj in neighbors:
                            ni, nj = i + di, j + dj
                            if ni < 0 or ni >= M or nj < 0 or nj >= N:
                                continue
                            new_data[ni][nj] ^= 1
                        new_state = convert(new_data)
                        #print("        new data = ", new_data, ", i = ", i, ", j = ", j)
                        next = (new_state << 4) | (i << 2) | j
                        if next not in visited:
                            visited.add(next)
                            q.append(next)
            moves += 1

        return -1

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