<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 getMask(int i, int j, int m, int n) const
    {
        int mask = 0;
        int deltas[5][2] = { { 1, 0 }, { -1, 0 }, { 0, 0 }, { 0, -1 }, { 0, 1 } };
        for (auto delta : deltas)
        {
            auto di = delta[0] + i;
            auto dj = delta[1] + j;
            if (  di >= 0
               && di < m
               && dj >= 0
               && dj < n)
            {
                auto index = di * n + dj;
                mask |= (1 << index);
            }
        }
        return mask;
    }

    int dfs(int matrixMask, int indexMask, int m, int n)
    {
        if (matrixMask == 0)
        {
            return 0;
        }

        if (m_memo[matrixMask][indexMask] != -1)
        {
            return m_memo[matrixMask][indexMask];
        }

        int minCount = INT_MAX;
        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                auto index = i * n + j;
                auto bit = (1 << index);
                if ((indexMask & bit) != bit)
                {
                    auto newMatrixMask = getMask(i, j, m, n);
                    auto count = dfs(matrixMask ^ newMatrixMask, indexMask | bit, m, n);
                    if (count != INT_MAX)
                    {
                        minCount = std::min(minCount, 1 + count);
                    }
                }
            }
        }
        return m_memo[matrixMask][indexMask] = minCount;
    }

    int minFlips(std::vector<std::vector<int>> &mat)
    {
        int m = mat.size();
        int matrixMask = 0;
        int n = mat[0].size();
        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                auto index = i * n + j;
                matrixMask |= (mat[i][j] << index);
            }
        }
        std::memset(m_memo, -1, 1024 * 1024 * sizeof(int));
        return [] (auto x) { return x == INT_MAX ? -1 : x; }(dfs(matrixMask, 0, m, n));
    }
    int m_memo[1024][1024];
};
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