<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:
    void solve(int index, int temp, vector<bool>& flipped, vector<vector<int>>& mat, int n, int m, int& ans) {
        if (index >= n * m) {
            // Check if the matrix is all zeroes after flipping
            if (isAllZeroes(mat, flipped, n, m)) {
                if (ans == -1) ans = temp;
                else ans = min(ans, temp);
            }
            return;
        }

        // Try not flipping the current cell
        solve(index + 1, temp, flipped, mat, n, m, ans);

        // Flip the current cell and its adjacent cells
        flipCells(index, flipped, n, m);
        solve(index + 1, temp + 1, flipped, mat, n, m, ans);
        flipCells(index, flipped, n, m); // Backtrack
    }

    int minFlips(vector<vector<int>>& mat) {
        int n = mat.size();
        int m = mat[0].size();
        int ans = -1;
        vector<bool> flipped(n * m, false);
        solve(0, 0, flipped, mat, n, m, ans);
        return ans;
    }

private:
    void flipCells(int index, vector<bool>& flipped, int n, int m) {
        int row = index / m;
        int col = index % m;
        flipped[index] = !flipped[index];
        if (row > 0) flipped[(row - 1) * m + col] = !flipped[(row - 1) * m + col];
        if (col > 0) flipped[row * m + col - 1] = !flipped[row * m + col - 1];
        if (row < n - 1) flipped[(row + 1) * m + col] = !flipped[(row + 1) * m + col];
        if (col < m - 1) flipped[row * m + col + 1] = !flipped[row * m + col + 1];
    }

    bool isAllZeroes(vector<vector<int>>& mat, vector<bool>& flipped, int n, int m) {
        vector<vector<int>> tempMat = mat;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                int index = i * m + j;
                if (flipped[index]) {
                    tempMat[i][j] = 1 - tempMat[i][j];
                }
                if (tempMat[i][j] == 1) {
                    return false;
                }
            }
        }
        return true;
    }
};
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