<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();
        vector<vector<int>> directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}, {0, 0}};
        
        auto flip = [&](int x, int y) {
            for (auto& dir : directions) {
                int nx = x + dir[0], ny = y + dir[1];
                if (nx >= 0 && nx < m && ny >= 0 && ny < n) {
                    mat[nx][ny] ^= 1;
                }
            }
        };
        
        int totalSteps = m * n;
        int result = INT_MAX;

        for (int mask = 0; mask < (1 << totalSteps); ++mask) {
            vector<vector<int>> backup = mat;
            int steps = 0;
            for (int i = 0; i < totalSteps; ++i) {
                if (mask & (1 << i)) {
                    int x = i / n;
                    int y = i % n;
                    flip(x, y);
                    ++steps;
                }
            }

            bool allZero = true;
            for (auto& row : mat) {
                for (auto& cell : row) {
                    if (cell != 0) {
                        allZero = false;
                        break;
                    }
                }
                if (!allZero) break;
            }

            if (allZero) {
                result = min(result, steps);
            }
            mat = backup;
        }
        return result == INT_MAX ? -1 : result;
    }
};
9000