<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 isZero(vector<vector<int>>& mat){
        for(auto it:mat){
            for(auto at:it){
                if(at==1) return 0;
            }
        }
        return 1;
    }
    long long count(vector<vector<int>> mat, int c, vector<vector<int>>& act, unordered_map<string,int>& map){
        if(c && mat==act) return INT_MAX;
        string s;
        for(auto it:mat){
            for(auto at:it){
                if(at==1) s+='1';
                else s+='0';
            }
        }
        if(map.find(s)!=map.end()) return map[s];
        if(isZero(mat)) return 0;
        map[s]=INT_MAX;
        long long ans=INT_MAX;
        vector<pair<int,int>> v={{1,0},{0,1},{-1,0},{0,-1}};
        for(int i=0;i<mat.size();i++){
            for(int j=0;j<mat[0].size();j++){
                mat[i][j]=!mat[i][j];
                for(auto it:v){
                    int g=i+it.first;
                    int h=j+it.second;
                    if(g>=0 && h>=0 && g<mat.size() && h<mat[0].size()){
                        mat[g][h]=!mat[g][h];
                    }
                }
                ans=min(ans,1+count(mat,1,act,map));
                mat[i][j]=!mat[i][j];
                for(auto it:v){
                    int g=i+it.first;
                    int h=j+it.second;
                    if(g>=0 && h>=0 && g<mat.size() && h<mat[0].size()){
                        mat[g][h]=!mat[g][h];
                    }
                }
            }
        }
        return map[s]=ans;
    }
    int minFlips(vector<vector<int>> mat) {
        if(isZero(mat)) return 0;
        unordered_map<string,int> map;
        int ans=count(mat,0,mat,map);
        if(ans==INT_MAX) return -1;
        else return ans;
    }
};
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