<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 auto init_io = []() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  return 0;
}();
using VEC = vector<int>;
class Solution {
public:
    int minFlips(vector<vector<int>>& mat) {
		int m = mat.size();
		int n = mat[0].size();
		if( m == 1 && n == 1 ) return mat[0][0] == 1;
		vector<VEC> modArr;
		int val = 0;
		modArr.resize(m*n);
		for(int i = 0; i < m; i++)
		{
			for(int j = 0; j < n; j++)	
			{
				//cout << "(" << i << ", " << j << ") can modify:" << endl;
				int idx = IDX(m, i, j);
				modArr[idx].push_back(idx);
				if(i-1 >= 0) modArr[idx].push_back(IDX(m, i-1, j  ));
				if(i+1  < m) modArr[idx].push_back(IDX(m, i+1, j  ));
				if(j-1 >= 0) modArr[idx].push_back(IDX(m, i  , j-1));
				if(j+1  < n) modArr[idx].push_back(IDX(m, i  , j+1));
				
				val += mat[i][j]* (1 << idx);
			}
		}

		vector<int> Q;		
		unordered_set<int> vis;
		Q.push_back(val);
		vis.insert(val);
		int step = 0;
		while( !Q.empty() )
		{
			vector<int> tempQ;
			for(auto&num : Q)
			{				
				if( num == 0 ) return step;
				for(int i = 0; i < m*n; i++)
				{
					int temp = num;
					for(auto&m : modArr[i])
					{
						int bit = ( 1 << m);
						temp ^= bit;
					}
					//cout << i << ", " << num << ", " << temp << endl;
					if(!vis.count(temp))
					{
						tempQ.push_back(temp);
						vis.insert(temp);
					}
				}
			}
			swap(Q, tempQ);
			step++;
		}
		return -1;
    }
	int IDX(int x, int i, int j)
	{
		//cout << "(" << i << ", " << j << ") can be modified." << endl;
		return j * x + i;
	}
};
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