<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:
    map<vector<vector<int>>, vector<vector<int>>> parent;
    map<vector<vector<int>>, int> dist;
    int minFlips(vector<vector<int>>& mat) {
        queue<vector<vector<int>>> q;
        q.push(mat);

        parent.insert({mat, mat});
        dist.insert({mat, 0});

        bool done = false;

        int dx[5] = {00, 01, 00, -1, 00};
        int dy[5] = {00, 00, -1, 00, 01};
        while (q.size() > 0) {
            vector<vector<int>> cur = q.front();
            q.pop();
            for (int i = 0; i < cur.size(); i++) {
                for (int j = 0; j < cur[i].size(); j++) {
                    vector<vector<int>> next(cur);
                    for (int d = 0; d < 5; d++) {
                        if (i + dx[d] < 0 || i + dx[d] >= next.size()) continue;
                        if (j + dy[d] < 0 || j + dy[d] >= next[i + dx[d]].size()) continue;
                        next[i + dx[d]][j + dy[d]] = 1 - next[i + dx[d]][j + dy[d]];
                    }

                    if (parent.find(next) != parent.end()) continue;
                    parent.insert({next, cur});
                    dist.insert({next, dist.at(cur) + 1});
                    q.push(next);
                    int sum = 0;
                    for (auto v : next) {
                        for (auto vv : v) sum += vv;
                    }
                    if (sum == 0) done = true;
                    if (done) break;
                }
                if (done) break;
            }
            if (done) break;
        }

        vector<vector<int>> desired(mat.size(), vector<int>(mat[0].size(), 0));

        if (dist.find(desired) == dist.end()) return -1;
        return dist.at(desired);
    }
};
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