<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 {
    int hasher(vector<vector<int>>&mat){
        int h=0;
        int n=mat.size(),m=mat[0].size();
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                h=(h<<1)+mat[i][j];
            }
        }
        return h;
    }
    int dp[1024];
    int solve(vector<vector<int>>&mat){
        int h=hasher(mat);
        if(h==0) return 0;
        int &ans=dp[h];
        if(ans!=-1) return ans;
        ans=1e9;
        auto flip=[&](int x,int y){
            mat[x][y]=1-mat[x][y];
            vector<int>dx={1,-1,0,0};
            vector<int>dy={0,0,1,-1};
            for(int d=0;d<4;d++){
                if(x+dx[d]<0 or y+dy[d]<0 or x+dx[d]>=mat.size() or y+dy[d]>=mat[0].size()) continue;
                mat[x+dx[d]][y+dy[d]]=1-mat[x+dx[d]][y+dy[d]];
            }
        };
        for(int i=0;i<mat.size();i++){
            for(int j=0;j<mat[0].size();j++){
                flip(i,j);
                ans=min(ans,1+solve(mat));
                flip(i,j);
            }
        }
        return ans;
    }
public:
    int minFlips(vector<vector<int>>& mat) {
        memset(dp,-1,sizeof dp);
        return solve(mat)==1e9?-1:solve(mat);
    }
};
14600