<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:
    bool check(vector<vector<int>> &mat,int n,int m){
        for(int i=0;i<n;++i){
            for(int j=0;j<m;++j){
                if(mat[i][j]) return false;
            }
        }
        return true;
    }
    void flipCell(int i,vector<vector<int>> &mat,int n,int m){
        int row = i/m;
        int col = i%m;
        mat[row][col] = 1^mat[row][col];
        if(row > 0) mat[row-1][col] = 1^mat[row-1][col];
        if(col > 0) mat[row][col-1] = 1^mat[row][col-1];
        if(row + 1 < n) mat[row+1][col] = 1^mat[row+1][col];
        if(col + 1 < m) mat[row][col+1] = 1^mat[row][col+1];
    }
    int solve(int i,vector<vector<int>> &mat,int n,int m){
        if(i == m*n){
            return check(mat,n,m)? 0: 1e9;
        }

        // notflip
        int notFlip = solve(i+1,mat,n,m);

        //flip
        flipCell(i,mat,n,m);
        int flip = 1 + solve(i+1,mat,n,m);
        flipCell(i,mat,n,m);

        return min(flip,notFlip);
    }
    int minFlips(vector<vector<int>>& mat) {
        int n = mat.size();
        int m = mat[0].size();

        int ans = solve(0,mat,n,m);

        return ans == 1e9 ? -1 : ans;
    }
};
8000