<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:
    static constexpr pair<int, int> dirs[] = {{-1, 0}, {0 , 1}, {1, 0}, {0, -1}}; // Using constexpr for compile-time initialization

    bool testBit(int mask, int i) {
        return (mask >> i) & 1;
    }

    // Pass the mat vector by reference to modify it in place
    void flips(int i, int j, vector<vector<int>>& mat) {
        mat[i][j] ^= 1;
        for (auto [dx, dy] : dirs) {
            int i2 = i + dx, j2 = j + dy;
            if (i2 >= 0 && i2 < mat.size() && j2 >= 0 && j2 < mat[i2].size()) {
                mat[i2][j2] ^= 1;
            }
        }
    }

    int check(int mask, int m, int n, vector<vector<int>> mat) {
        int cnt = 0;
        // Apply the flips based on the mask for the first row
        for (int j = 0; j < n; ++j) {
            if (testBit(mask, j)) {
                flips(0, j, mat);
                cnt++;
            }
        }
        // Check and flip for subsequent rows
        for (int i = 1; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (mat[i - 1][j] == 1) { 
                    flips(i, j, mat);
                    cnt++;
                }
            }
        }
        // If there's any 1 left in the last row, return a large number (indicating failure)
        for (int j = 0; j < n; ++j) {
            if (mat[m - 1][j] == 1) return 1e9;
        }
        return cnt;
    }

    int minFlips(vector<vector<int>>& mat) {
        int res = 1e9;
        int m = mat.size(), n = mat[0].size();
        // Try all possible combinations for the first row
        for (int mask = 0; mask < (1 << n); ++mask) {
            int cntFlips = check(mask, m, n, mat);
            if (cntFlips < res) res = cntFlips;
        }
        return res == 1e9 ? -1 : res; // If no valid solution, return -1
    }
};

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