Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1284.Minimum%20Number%20of%20Flips%20to%20Convert%20Binary%20Matrix%20to%20Zero%20Matrix/images/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>
Bit Manipulation; Breadth-First Search; Array; Hash Table; Matrix
class Solution:
    def minFlips(self, mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        state = sum(1 << (i * n + j) for i in range(m) for j in range(n) if mat[i][j])
        q = deque([state])
        vis = {state}
        ans = 0
        dirs = [0, -1, 0, 1, 0, 0]
        while q:
            for _ in range(len(q)):
                state = q.popleft()
                if state == 0:
                    return ans
                for i in range(m):
                    for j in range(n):
                        nxt = state
                        for k in range(5):
                            x, y = i + dirs[k], j + dirs[k + 1]
                            if not 0 <= x < m or not 0 <= y < n:
                                continue
                            if nxt & (1 << (x * n + y)):
                                nxt -= 1 << (x * n + y)
                            else:
                                nxt |= 1 << (x * n + y)
                        if nxt not in vis:
                            vis.add(nxt)
                            q.append(nxt)
            ans += 1
        return -1
