<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 n, m;
    vector<vector<int>>& flip(vector<vector<int>>& mat, int y, int x) {
        mat[y][x] ^= 1;
        if(y - 1 >= 0) mat[y-1][x] ^= 1;
        if(y + 1 <  n) mat[y+1][x] ^= 1;
        if(x - 1 >= 0) mat[y][x-1] ^= 1;
        if(x + 1 <  m) mat[y][x+1] ^= 1;
        return mat;
    }

    bool isZeroMat(vector<vector<int>>& mat) {
        for(auto i = 0; i < n; ++i)
            for(auto j = 0; j < m; ++j)
                if(mat[i][j])
                    return false;
        return true;
    }

    int FlipOrNotFlip(vector<vector<int>> mat, int y, int x) {
        if(x == m) y++, x = 0;
        if(y == n) return isZeroMat(mat) ? 0:10000;

        auto ret1 = FlipOrNotFlip(mat, y, x+1);
        auto ret2 = FlipOrNotFlip(flip(mat, y, x), y, x+1) + 1;
        return min(ret1, ret2);
    }

    int minFlips(vector<vector<int>>& mat) {
        n = mat.size(), m = mat[0].size();
        auto ret = FlipOrNotFlip(mat, 0, 0);
        return (ret >= 10000 ? -1 : ret);
    }
};
14900