<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) {
        unordered_set<long long> visited;

        queue<vector<vector<int>>> q;
        q.push(mat);
        
        int steps = 0;
        bool found = false;
        while(!q.empty()){
            int size = q.size();
            for(int i = 0; i < size; i++){
                auto m = q.front();
                q.pop();
                if(isZero(m)){
                    return steps;
                } else {
                     for(int i = 0; i < m.size(); i++){
                        for(int j = 0; j < m[0].size();j++){
                            flip(m, i, j);
                            auto code = encode(m);
                            if(!visited.contains(code)){
                                visited.insert(code);
                                q.push(m);
                            }
                            flip(m, i, j);
                        }
                    }
                }
            }
            steps++;
        }

        return -1;
    }

    void flip(vector<vector<int>>& m, int i, int j){
        m[i][j] = !m[i][j];
        
        if(i > 0){
            m[i - 1][j] = !m[i - 1][j];
        }

        if(j > 0){
            m[i][j - 1] = !m[i][j - 1];
        }

        if(i < m.size() - 1){
            m[i + 1][j] = !m[i + 1][j];
        }

        if(j < m[0].size() - 1){
            m[i][j + 1] = !m[i][j + 1];
        }
    }

    long long encode(vector<vector<int>>& mat){
        long long num = 0;
        for(int i = 0; i < mat.size(); i++){
            for(int j = 0; j < mat[0].size();j++){
                num *= 10;
                num += mat[i][j];
            }
        }

        return num;
    }

    bool isZero(vector<vector<int>>& mat){
        bool found = false;
        for(int i = 0; i < mat.size(); i++){
            for(int j = 0; j < mat[0].size();j++){
                if(mat[i][j] == 1){
                    found = true;
                    break;
                }
            }
        }

        return !found;
    }
};
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