<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 {
    map<vector<vector<int>>, int > visted;
public:
    int minFlips(vector<vector<int>>& mat) {
     if(mat.empty()) return -1;     
        int steps = 0;
        queue<vector<vector<int> > > q;
        q.push(mat);
        visted[mat] = 1;
        while(!q.empty()) {
            int len = q.size();
            cout<<len<<" ";
            for(int i = 0; i < len; i++) {
                vector<vector<int> > v = q.front();
                q.pop();
                if(!check(v)) {
                    return steps;
                }
                for(int i = 0; i < v.size(); i++) {
                    for(int j = 0;j<v[0].size();j++) {
                        vector<vector<int> > temp = v;
                        changeMat(temp,i,j);
                        if(!visted[temp]) {
                            q.push(temp);
                            visted[temp] = 1;
                        }
                    }
                }
            }
            steps++;
        }
        
        return -1;
    }
    
    
    bool check(vector<vector<int>> mat) {
        int f = 0;
        for(int i = 0; i< mat.size();i++) {
            for(int j=0;j<mat[0].size();j++) {
                if(mat[i][j]) {
                    f=1;
                    break;
                }
            }
        }
        return f;
    }
    
    void changeMat(vector<vector<int> > &v, int i,int j) {
        int n,m;
        n=v.size();
        m=v[0].size();
        
        v[i][j] = !v[i][j];
        
        if(i+1<n) {
            v[i+1][j] = !v[i+1][j];
        }
        
        if(j+1<m) {
             v[i][j+1] = !v[i][j+1];
        }
        
        if(i-1>=0) {
             v[i-1][j] = !v[i-1][j];
        }
        
        if(j-1>=0) {
             v[i][j-1] = !v[i][j-1];
        }
    }
 };
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