<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:
    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;
    }
    int minFlips(vector<vector<int>>& mat) {
        int n = mat.size();
        int m = mat[0].size();
        
        int ans = 0;
        
        queue<vector<vector<int>>> q;
        vector<vector<int>> temp, temp1;
        map<vector<vector<int>>, bool> mp;

        q.push(mat);
        mp[mat] = true;

        while(!q.empty()){
            int s = q.size();
            for(int i = 0; i < s; i++){
                temp = q.front();
                temp1 = temp;
                q.pop();

                if(isValid(temp)){
                    return ans;
                }

                for(int x = 0; x < n; x++){
                    for(int y = 0; y < m; y++){
                        // choose cells
                        temp1[x][y] ^= 1;

                        //uppercell
                        if(x-1 >= 0)temp1[x-1][y] ^= 1;
                        // belowcell
                        if(x+1 < n)temp1[x+1][y] ^=1;
                        // leftcell
                        if(y-1 >= 0) temp1[x][y-1] ^=1;
                        //rightcell
                        if(y+1 < m) temp1[x][y+1] ^=1;

                        if(mp.find(temp1)==mp.end()){
                            mp[temp1] = true;
                            q.push(temp1);
                        }
                        temp1 = temp;
                    }
                }
            }
        ans++;
        }
        return -1;
    }
};
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