<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 chk(vector<vector<int>> &mat)
    {
        for(auto &it : mat)
        {
            for(auto &i : it)
            {
                if(i)
                {
                    return false;
                }
            }
        }
        return true;
    }
    void flip(vector<vector<int>>& mat,int i,int j){
        int n = mat.size(), m = mat[0].size();
        mat[i][j]^=1;

        vector<int> dx = {0, 0, -1, 1};
        vector<int> dy = {-1, 1, 0, 0};

        for(int k=0; k < 4; k++)
        {
            int nx = i + dx[k];
            int ny = j + dy[k];
            if(nx >= 0 && nx < n && ny >= 0 && ny < m)
            {
                mat[nx][ny]^=1;
            }
        }
    }
    int rec(int i, int j, vector<vector<int>> &mat)
    {
        int n = mat.size(), m = mat[0].size();

        if(j == m)
        {
            j = 0;
            i++;
        }

        if(i == n)
        {
            if(chk(mat))
            {
                return 0;
            }
            else
            {
                return 1e9;
            }
        }


        int op1=0,op2=0;
        
        op1 = rec(i, j+1, mat);
        flip(mat,i,j);
        op2 = 1 + rec(i, j+1, mat);
        flip(mat,i,j);
        
        return min(op1,op2);

    }
    int minFlips(vector<vector<int>>& mat) {
        int n =  mat.size(), m = mat[0].size();
        int ans = rec(0, 0, mat);
        return ans == 1e9 ? -1 : ans;
    }
};
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