<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>

static const int fastIO = [] {
	std::ios_base::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);
	return 0;
}();

class Solution {
public:
	int minFlips(vector<vector<int>>& mat) {
		const int n = (int)mat.size(), m = (int)mat[0].size();
		if (n == 1 && m == 1) return mat[0][0];
		int dest = 0;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				dest = (dest << 1) + mat[i][j];
		vector<int> dist(1 << (n * m), 1e9);
		queue<int> q;
		dist[0] = 0;
		q.push(0);
		while (!q.empty()) {
			int u = q.front();
			q.pop();
			for (int i = 0; i < n * m; i++) {
				int v = u ^ (1 << i);
				if (i >= m) v ^= (1 << (i - m));
				if (i < (n - 1) * m) v ^= (1 << (i + m));
				if (i % m != 0) v ^= (1 << (i - 1));
				if (i % m != m - 1) v ^= (1 << (i + 1));
				if (dist[v] == 1e9) {
					dist[v] = dist[u] + 1;
					if (v == dest) return dist[v];
					q.push(v);
				}
			}
		}
		return -1;
	}
};
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