<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:
    int minFlips(vector<vector<int>>& mat) {
        int m = mat.size(), n = mat[0].size();
        queue<vector<vector<int>>> q;
        set<vector<vector<int>>> s;
        s.insert(mat);
        q.push(mat);
        int step = -1;
        while (!q.empty()) {
            step++;
            int size = q.size();
            for (int a=0; a<size; ++a) {
                auto x = q.front();
                int tmp = 0;
                for (int i=0; i<m; ++i) {
                    for (int j=0; j<n; ++j) {
                        tmp += x[i][j];
                        if (tmp > 0) break;
                    }
                    if (tmp > 0) break;
                }
                if (tmp == 0) return step;
                for (int i=0; i<m; ++i) {
                    for (int j=0; j<n; ++j) {
                        vector<vector<int>> t = x;
                        t[i][j] ^= 1;
                        if (i - 1 >= 0) {
                            t[i-1][j] ^= 1;                            
                        }
                        if (i + 1 < m) {
                            t[i+1][j] ^= 1;                            
                        }                        
                        if (j - 1 >= 0) {
                            t[i][j-1] ^= 1;                            
                        }                        
                        if (j + 1 < n) {
                            t[i][j+1] ^= 1;                            
                        }
                        if (s.find(t) == s.end()) {
                            s.insert(t);
                            q.push(t);
                        }
                        // print(t);
                    }
                }
                q.pop();        
            }
        }
        return -1;
    }
    void print(vector<vector<int>> t) {
        int m = t.size(), n = t[0].size();
        for (int i=0; i<m; ++i) {
            for (int j=0; j<n; ++j) {
                printf("%d ", t[i][j]);
            }
            puts("");
        }
        puts("");        
    }
};
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