<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) {
        const int m = mat.size(), n = mat[0].size();
        const int limit = 1 << (m * n);
        auto flip = [&](int mask, int r, int c) -> int {
            static const vector<pair<int, int>> dirs = {{0, 0}, {0, 1}, {1, 0}, 
                                                        {0, -1}, {-1, 0}};
            for (const auto& [dr, dc] : dirs) {
                int nr = r + dr, nc = c + dc;
                if (nr >= 0 && nr < m && nc >= 0 && nc < n) {
                    mask ^= (1 << (nr * n + nc));
                }
            }
            return mask;
        };
        auto encode = [&](const vector<vector<int>>& mat) -> int {
            int mask = 0;
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (mat[i][j]) {
                        mask |= 1 << (i * n + j);
                    }
                }
            }
            return mask;
        };
        auto isZero = [](int mask) -> bool {
            return mask == 0;
        };
        int target = encode(mat);
        if (isZero(target)) {
            return 0;
        }
        queue<pair<int, int>> q;
        q.emplace(target, 0);
        unordered_set<int> seen = {target};
        while (!q.empty()) {
            auto [mask, step] = q.front(); q.pop();
            for (int r = 0; r < m; ++r) {
                for (int c = 0; c < n; ++c) {
                    int newMask = flip(mask, r, c);
                    if (isZero(newMask)) {
                        return step + 1;
                    }
                    if (!seen.count(newMask)) {
                        seen.insert(newMask);
                        q.emplace(newMask, step + 1);
                    }
                }
            }
        }
        return -1;
    }
};
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