<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:

    vector<vector<int>> tmpl;
    unordered_map<string, int> dp;
    
    vector<vector<int>>& flip(vector<vector<int>>& mat, int i, int j, int n, int m) 
    {
        mat[i][j] ^= 1;

        if(i - 1 >= 0) mat[i - 1][j] ^= 1;
        if(i + 1 < n) mat[i + 1][j] ^= 1;
        if(j - 1 >= 0) mat[i][j - 1] ^= 1;
        if(j + 1 < m) mat[i][j + 1] ^= 1;
        
        return mat;
    }
    int solver(vector<vector<int>> & mat, int n, int m, int cntr) 
    {
        if(mat == tmpl) return 0;

        if(cntr > 8) return 1e9;

        string tmp;

        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                tmp += to_string(mat[i][j]) + "_"; 
            }
        }

        tmp += to_string(cntr);
        
        if(dp.find(tmp) != dp.end()) return dp[tmp];

        int res = 1e9;

        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {  
                flip(mat, i, j, n, m);

                res = min(res, 1 + solver(mat, n, m, cntr + 1));
                
                flip(mat, i, j, n, m);
            }
        }
        return dp[tmp] = res;
    }
    int minFlips(vector<vector<int>>& mat) 
    {
        int n = mat.size(), m = mat[0].size();

        tmpl.resize(n, vector<int> (m, 0));

        int res = solver(mat, n, m, 0);

        return res == 1e9 ? -1 : res;
    }
};
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