<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 check(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] == 1) {
                    return false;
                }
            }
        }
        return true;
    }
    
    int minFlips(vector<vector<int>>& mat) {
        if(check(mat)) return 0;
        
        set<vector<vector<int>>> vis;
        int delRow[] = {-1, 0, 1, 0};
        int delCol[] = {0, 1, 0, -1};
        int m = mat.size();
        int n = mat[0].size();
        
        queue<pair<int, vector<vector<int>>>> q;
        q.push({0, mat});
        vis.insert(mat);
        
        while(!q.empty()) {
            auto [step, curMat] = q.front();
            q.pop();
            
            if(check(curMat)) return step;
            
            for(int i = 0; i < m; i++) {
                for(int j = 0; j < n; j++) {
                    vector<vector<int>> newMat = curMat;
                    newMat[i][j] = !newMat[i][j];
                    
                    for(int k = 0; k < 4; k++) {
                        int newRow = i + delRow[k];
                        int newCol = j + delCol[k];
                        
                        if (newRow >= 0 && newRow < m && newCol >= 0 && newCol < n) {
                            newMat[newRow][newCol] = !newMat[newRow][newCol];
                        }
                    }
                    
                    if(vis.find(newMat) == vis.end()) {
                        vis.insert(newMat);
                        q.push({step + 1, newMat});
                    }
                }
            }
        }
        
        return -1;
    }
};

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