<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:
    void helper(int times,vector<vector<int>> &mat,vector<vector<bool>>& c,int counter,bool& checker,int ni)
    {
        if(checker) return;
        if(times==counter)
                {
                    if(check(mat)) {checker=true;return;}
                    else return;
                }
        for(int i=ni/mat[0].size();i<mat.size();i++)
        {
            for(int j=ni%mat[0].size();j<mat[i].size();j++)
            {
                
                if(!c[i][j])
                {
                    counter++;
                    c[i][j]=true;
                    flip(mat,i,j);
                    helper(times,mat,c,counter,checker,i*mat[0].size());
                    flip(mat,i,j);
                    c[i][j]=false;
                    counter--;
                }
            }
        }



    }
    void flip(vector<vector<int>> & mat,int i,int j)
    {
        mat[i][j]^=1;
        if(i!=0){mat[i-1][j]^=1;}
      
        if(i!=mat.size()-1){mat[i+1][j]^=1;}
    
        if(j!=0){mat[i][j-1]^=1;}
      
        if(j!=mat[0].size()-1){mat[i][j+1]^=1;}
       
    }
    bool check(vector<vector<int>> &mat)
    {
        for(int i=0;i<mat.size();i++)
        {
            for(int j=0;j<mat[i].size();j++)
            {
                if(mat[i][j]!=0) return false;
            }
        }
        return true;
    }
    int minFlips(vector<vector<int>>& mat) {
         if(check(mat)) return 0;
         if(mat.size()==1&&mat[0].size()==1) return 1;
         vector<vector<bool>> c;
         vector<bool> v(mat[0].size(),false);
         bool checker=false;
             for(int j=0;j<mat.size();j++)
             {
                 c.push_back(v);
             }
         
         for(int i=1;i<mat.size()*mat[0].size()+1;i++)
         {
             helper(i,mat,c,0,checker,0);
             if(checker) return i;
         }
         return -1;
          
    }
};
9200