<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 matrix_to_bitmask(vector<vector<int>>& mat){
        int num=0;
        int it=0;
        for(int i=0;i<mat.size();i++){
            for(int j=0;j<mat[0].size();j++){
                if(mat[i][j]) num=(num | (1ll<<it));

                it++;
            }
        }

        return num;
    }

    vector<vector<int>> bitmask_to_matrix(int num,int n,int  m){
        vector<vector<int>>ans(n,vector<int>(m));

        int it=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if((num>>it)&1) ans[i][j]=1;

                it++;
            }
        }

        return ans;
    }

    int minFlips(vector<vector<int>>& mat) {
        
        set<int>st;
        int n=mat.size(),m=mat[0].size();

        queue<int>q;
        q.push(matrix_to_bitmask(mat));
        
        int step=0;
        while(q.size()>0 ){
            int sz=q.size();
            while(sz--){

                int cur=q.front();
                q.pop();
                if(cur==0) return step;
                
                mat=bitmask_to_matrix(cur,n,m);

                for(int i=0;i<n;i++){
                    for(int j=0;j<m;j++){
                        vector<vector<int>>v=mat;

                        v[i][j] = !v[i][j];
                        if(i+1 < n) v[i+1][j] = !v[i+1][j];
                        if(i-1 >=0) v[i-1][j] = !v[i-1][j];
                        if(j+1 < m) v[i][j+1] = !v[i][j+1];
                        if(j-1 >=0) v[i][j-1] = !v[i][j-1];

                        int num=matrix_to_bitmask(v);
                        if(st.count(num)) continue;
                        st.insert(num);
                        q.push(num);
                    }
                }
            }

            step++;
        }


        return -1;
    }
};
13000