<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>

static const int _ = [](){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 0;
}();
class Solution {
public: 
    int dx[4]={0,0,-1,1};
    int dy[4]={1,-1,0,0};
    void flip(vector<vector<int>>& mat,int i,int j){
        mat[i][j]^=1;
        int n=mat.size();
        int m=mat[0].size();
        for(int k=0;k<4;k++){
            int ni=dx[k]+i;
            int nj=dy[k]+j;
            if(ni>=0 && nj>=0 && ni<n && nj<m){
                mat[ni][nj]^=1;
            }
        }
    };
    int minFlips(vector<vector<int>>& mat) {
        int ans=INT_MAX;
        int n=mat.size();
        int m=mat[0].size();
        vector<vector<int>> fin(n,vector<int>(m,0));
        vector<pair<int,int>> temp;
        int c=0;
        for(int i=0;i<n;i++)for(int j=0;j<m;j++)temp.push_back({i,j});
        map<vector<vector<int>>,int> vis;
        queue<pair<vector<vector<int>>,int>> q;
        q.push({mat,0});
        while(!q.empty()){
            auto tp=q.front();
            q.pop();
            auto g=tp.first;
            auto d=tp.second;
            if(g==fin){
                ans=min(ans,d);
                break;
            }
            vis[g]=1;
            for(int k=0;k<temp.size();k++){
                auto tg=g;
                flip(tg,temp[k].first,temp[k].second);
                if(vis.find(tg)==vis.end()){
                    q.push({tg,d+1});
                    vis[tg]++;
                }
            }
        }
        if(ans==INT_MAX)return -1;
        return ans;
    }
};
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