<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>

//bfs state

void flip(vector<vector<int>>& mat, int i, int j){
    mat[i][j]=(mat[i][j]==0)?1:0;
    if(i!=0)mat[i-1][j]=(mat[i-1][j]==0)?1:0;
    if(i!=mat.size()-1)mat[i+1][j]=(mat[i+1][j]==0)?1:0;
    if(j!=0)mat[i][j-1]=(mat[i][j-1]==0)?1:0;
    if(j!=mat[0].size()-1)mat[i][j+1]=(mat[i][j+1]==0)?1:0;
}
void calcStatus(vector<vector<int>>& mat, string& next){
    int m = mat.size();
    int n = mat[0].size();
    int i,j,k;
    for(i=0;i<m;i++){
        for(j=0;j<n;j++){
            next.append(to_string(mat[i][j]));
        }
    }
    return;
}
int solve(vector<vector<int>>& mat){
    //origin = mat
    int m = mat.size();
    int n = mat[0].size();
    int i,j,k;
    string state;
    for(i=0;i<m;i++){
        for(j=0;j<n;j++)state.append(to_string(mat[i][j]));
    }
    map<string,int> level;
    level[state]=0;
    string dest;
    for(i=0;i<m;i++){
        for(j=0;j<n;j++)dest.append("0");
    }
    queue<string> q;
    q.push(state);
    while(!q.empty()){
        string curr = q.front();
        q.pop();
        for(i=0;i<m;i++){
            for(j=0;j<n;j++){
                mat[i][j]=curr[i*n+j]-48;
            }
        }
        for(i=0;i<m;i++){
            for(j=0;j<n;j++){
                if(level.find(dest)!=level.end())break;
                flip(mat,i,j);
                string next;
                calcStatus(mat,next);
                flip(mat,i,j);
                if(level.find(next)!=level.end())continue;
                level[next]=level[curr]+1;
                q.push(next);
            }
            if(level.find(dest)!=level.end())break;
        }
        if(level.find(dest)!=level.end())break;
    }
    if(level.find(dest)==level.end())return -1;
    return level[dest];
}
class Solution {
public:
    int minFlips(vector<vector<int>>& mat) {
        return solve(mat);
    }
};
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