<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 {
    private:
    bool check(vector<vector<int>>& mat){
        for(int i=0;i<mat.size();i++){
            for(int j=0;j<mat[0].size();j++){
                if(mat[i][j]==1) return false;
            }
        }
        return true;
    }
    void sol(vector<vector<int>>& mat,int i,int j){
        int a1[4]={0,0,1,-1};
        int a2[4]={1,-1,0,0};
        mat[i][j]=1-mat[i][j];
        for(int ii=0;ii<4;ii++){
            int ni=i+a1[ii],nj=j+a2[ii];
            if(ni>=0 and ni<mat.size() and nj>=0 and nj<mat[0].size()){
                mat[ni][nj]=1-mat[ni][nj];
            }
        }
    }
public:
    int minFlips(vector<vector<int>>& mat) {
        map<vector<vector<int>>,int> vis;
        queue<pair<vector<vector<int>>,int>>q;
        q.push({mat,0});
        vis[mat]=1;
        while(!q.empty()){
            vector<vector<int>>tem=q.front().first;
            int cst=q.front().second;
            q.pop();
            if(check(tem)) return cst;
            for(int i=0;i<tem.size();i++){
                for(int j=0;j<tem[0].size();j++){
                    vector<vector<int>>v=tem;
                    sol(v,i,j);
                    if(vis[v]==0){
                        vis[v]=1;
                        q.push({v,cst+1});
                    }
                }
            }
        }
        return -1;
    }
};
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