<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 {
    vector<int> dx={-1,1,0,0};
    vector<int> dy={0,0,-1,1};
public:
    int minFlips(vector<vector<int>>& mat) {
        int m=mat.size();
        int n=mat[0].size();

        int mn = m*n;
        int ans = mn+1;
        for(int i=0; i<(1<<mn); i++){
            ans=min(ans, check(mat, i));
        }

        return ans==mn+1 ? -1 : ans;
    }
    
    int check(auto mat, int mask){
        int m=mat.size();
        int n=mat[0].size();

        int count=0;

        for(int i=0; i<m*n; i++){
            if(mask&(1<<i)){
                int r = i/n;
                int c = i%n;

                mat[r][c]^=1;
                for(int d=0; d<4; d++){
                    int nr = r+dx[d];
                    int nc = c+dy[d];

                    if(nr>=0 && nr<m && nc>=0 &&nc<n){
                        mat[nr][nc]^=1;
                    }
                }

                count++;
            }
        }
        for(int i=0; i<m;i++){
            for(int j=0; j<n; j++){
                if(mat[i][j]!=0)return m*n+1;
            }
        }
        return count;
    }
};


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