<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:
    int minFlips(vector<vector<int>>& mat) {
        int m = mat.size();
        int n = mat[0].size();
        vector<vector<int>> t(m, vector<int> (n, 0));
        queue<vector<vector<int>>> q;
        set<vector<vector<int>>> st;
        q.push(mat);
        st.insert(mat);
        vector<vector<int>> move = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        int steps = 0;
        while (!q.empty())
        {
            int n1 = q.size();
            for (int i = 0;i<n1;i++)
            {
                vector<vector<int>> p = q.front();
                q.pop();
                if (p == t)
                {
                    return steps;
                }
                for (int j = 0;j<m;j++)
                {
                    for (int k = 0;k<n;k++)
                    {
                        vector<vector<int>> p2 = p;
                        p2[j][k] = 1 - p2[j][k];
                        for (int r = 0;r<4;r++)
                        {
                            int a = j + move[r][0];
                            int b = k + move[r][1];
                            if (a >= 0 and a < m and b >= 0 and b < n)
                            {
                                p2[a][b] = 1 - p2[a][b];
                            }
                        }
                        if (st.find(p2) == st.end())
                        {
                            st.insert(p2);
                            q.push(p2);
                        }
                    }
                }
            }
            steps++;
        }
        return -1;
    }
};
14000