<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:

    bool isValid(int x, int y, int m, int n)
    {
        return x >= 0 && y >= 0 && x < m && y < n;
    }

    int minFlips(vector<vector<int>>& mat) {
        
        int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, -1, 1};
        int m = mat.size(), n = mat[0].size();
        set<vector<vector<int>>> vis;

        priority_queue<pair<int, vector<vector<int>>>, vector<pair<int, vector<vector<int>>>>, greater<pair<int, vector<vector<int>>>>> pq;
        pq.emplace(0, mat);

        while(!pq.empty())
        {
            auto it = pq.top();
            pq.pop();
            int steps = it.first;
            vector<vector<int>> curr = it.second;

            if (vis.count(curr))
                continue;

            vis.emplace(curr);

            int sum = 0;

            for (auto &row : curr)
            {
                for (auto it : row)
                {
                    sum += it;

                    if (sum)
                        break;
                }
            }

            if (!sum)
                return steps;

            for (int i = 0; i < m; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    vector<vector<int>> next = curr;
                    next[i][j] ^= 1;

                    for (int k = 0; k < 4; k++)
                    {
                        int newi = i + dx[k], newj = j + dy[k];

                        if (isValid(newi, newj, m, n))
                        {
                            next[newi][newj] ^= 1;
                        }
                    }

                    if (!vis.count(next))
                        pq.emplace(steps + 1, next);
                }
            }   
        } 

        return -1;
    }
};
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