<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 n=mat.size();
        int m=mat[0].size();
        map<vector<vector<int>>,int> vis;
        queue<vector<vector<int>>> q;
        q.push(mat);
        vis[mat]=1;
        int steps=0;
        auto flip=[&](int i, int j, vector<vector<int>>& mat){
            int dx[]={-1,1,0,0};
            int dy[]={0,0,-1,1};
            mat[i][j]=1-mat[i][j];
            for(int k=0; k<4; k++){
                int ni=i+dx[k];
                int nj=j+dy[k];

                if(ni<0 || ni==n || nj<0 || nj==m) continue;
                mat[ni][nj]=1-mat[ni][nj];
            }
        };
        while(!q.empty()){
            int sz=q.size();
            while(sz--){
                auto top=q.front();
                q.pop();
                int sum=0;
                for(auto r:top){
                    for(auto c:r) sum+=c;
                }
                if(sum==0) return steps;
                for(int i=0; i<mat.size(); i++){
                    for(int j=0; j<mat[0].size(); j++){
                        flip(i,j,top);
                        if(!vis[top]){
                            q.push(top);
                            vis[top]=1;
                        }
                        flip(i,j,top);
                    }
                }
            }
            steps++;
        }

        return -1;
    }
};
11800