<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:
    vector<vector<int>>dir={{0,1},{0,-1},{-1,0},{1,0}};
    int minFlips(vector<vector<int>>& mat) {
        int m=mat.size();
        int n=mat[0].size();
        map<vector<vector<int>>,bool>mp;
        int steps=0;
        queue<vector<vector<int>>>q;
        vector<vector<int>>temp;
        vector<vector<int>>temp1;
        q.push(mat);
        mp[mat]=true;
        while(!q.empty())
        {
            int size=q.size();
            for(int idx=0;idx<size;idx++)
            {
                temp=q.front();
                temp1=temp;
                q.pop();
                if(isvalid(temp))
                {
                    return steps;
                }
                for(int i=0;i<m;i++)
                {
                    for(int j=0;j<n;j++)
                    {
                        //choose this cell
                        temp1[i][j]^=1;
                        for(int d=0;d<4;d++)
                        {
                            int r=i+dir[d][0];
                            int c=j+dir[d][1];
                            if(r>=0 and c>=0 and r<m and c<n)
                            {
                                temp1[r][c]^=1;
                            }
                        }
                        if(mp.find(temp1)==mp.end())
                        {
                            mp[temp1]=true;
                            q.push(temp1);
                        }
                        temp1=temp;
                    }
                }
            }
            steps++;
        }
        return -1;
    }
    bool isvalid(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]!=0)
                    return false;
            }
        }
        return true;
    }
};
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