<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 minFlips(vector<vector<int>>& mat) {
         int row=mat.size();
         int col=mat[0].size();
         map<vector<vector<int>>,int>vis;
         map<vector<vector<int>>,int>dist;
		 queue<vector<vector<int>>>q;
		 q.push(mat);
		 vis[mat]=1;
		 dist[mat]=0;
		 int x[]={-1,1,0,0,0};
		 int y[]={0,0,-1,1,0};
		 while(q.size()>0){
			 vector<vector<int>>tp=q.front();
			 q.pop();
			 bool check=true;
			 for(int i=0;i<tp.size();i++){
				 for(int j=0;j<tp[i].size();j++){
					  if(tp[i][j]==1){
						  check=false;
					  }
				 }
			 }
			 if(check) return dist[tp];
			 for(int i=0;i<tp.size();i++){
				 for(int j=0;j<tp[i].size();j++){
					 vector<vector<int>>new_matrix=tp;
					 for(int k=0;k<5;k++){
						 int x_axis=x[k]+i;
						 int y_axis=y[k]+j;
						 if(x_axis>=0 and x_axis<row and y_axis>=0 and y_axis<col){
							 new_matrix[x_axis][y_axis]=1-new_matrix[x_axis][y_axis];
						 }
					 }
					 if(!vis[new_matrix]){
						 q.push(new_matrix);
                         vis[new_matrix]=1;
						 dist[new_matrix]=dist[tp]+1;
					 }
				 }
			 }
			 
		 }
		 return -1;
    }
};
16200