<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>

struct VectorHash {
    std::size_t operator()(const std::vector<std::vector<int>>& vec) const {
        std::size_t hash = 0;
        for (const auto& row : vec) {
            for (int val : row) {
                hash ^= std::hash<int>{}(val) + 0x9e3779b9 + (hash << 6) +
                        (hash >> 2);
            }
        }
        return hash;
    }
};

struct VectorEqual {
    bool operator()(const std::vector<std::vector<int>>& lhs,
                    const std::vector<std::vector<int>>& rhs) const {
        return lhs == rhs;
    }
};

vector<vector<int>> dxdy = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

class Solution {
public:
    void flip(vector<vector<int>>& matrix, int i, int j, int n, int m) {
        matrix[i][j] ^= 1;
        for (auto& vec : dxdy) {
            int x = i + vec[0];
            int y = j + vec[1];
            if (x < 0 or y < 0 or x >= n or y >= m) {
                continue;
            }
            matrix[x][y] ^= 1;
        }
    }

    int minFlips(vector<vector<int>>& mat) {
        int n = mat.size();
        int m = mat[0].size();
        vector<vector<int>> end(n, vector<int>(m, 0));
        queue<vector<vector<int>>> q;
        set<vector<vector<int>>> visited;
        visited.insert(mat);
        int d = 0;
        q.push(mat);
        while (!q.empty()) {
            int qsize = q.size();
            while (qsize--) {
                vector<vector<int>> node = q.front();
                if (node == end) {
                    return d;
                }
                q.pop();
                for (int i = 0; i < node.size(); i++) {
                    for (int j = 0; j < node[i].size(); j++) {
                        flip(node, i, j, n, m);
                        if (visited.find(node) == visited.end()) {
                            visited.insert(node);
                            q.push(node);
                        }
                        flip(node, i, j, n, m);
                    }
                }
            }
            d++;
        }
        return -1;
    }
};
10400