<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 {
public:
    static constexpr uint32_t masks[3][3] = {
        { 0b110'100'000, 0b111'010'000, 0b011'001'000 },
        { 0b100'110'100, 0b010'111'010, 0b001'011'001 },
        { 0b000'100'110, 0b000'010'111, 0b000'001'011 }
    };

    pair<int, int> parse(vector<vector<int>>& mat) {
        int first = 0, mask = 0;
        for (auto& r : mat) {
            first >>= 3, mask >>= 3;
            for (int i = 0; i < r.size(); ++i) {
                mask |= 1 << (8 - i);
                first |= (r[i] == 1) << (8 - i);
            }
        }
        return {first, mask};
    }

    int minFlips(vector<vector<int>>& mat) {
        int m = mat.size(), n = mat[0].size();
        bool visited[1 << 9] = {};
        queue<uint32_t> q;
        auto [first, mask] = parse(mat);
        q.push(first);
    
        int res = 0;
        while (!q.empty()) {
            auto size = q.size();
            while (size--) {
                uint32_t b = q.front(); q.pop();
                if ((b & mask) == 0) return res;
                visited[b & mask] = true;

                for (int i = 0; i < m; ++i)
                    for (int j = 0; j < n; ++j)
                        if (!visited[(b ^ masks[i][j]) & mask])
                            q.push(b ^ masks[i][j]);
            }
            ++res;
        }

        return -1;
    }
};
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