Minimum Number of Flips to Convert Binary Matrix to Zero Matrix
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1284.Minimum%20Number%20of%20Flips%20to%20Convert%20Binary%20Matrix%20to%20Zero%20Matrix/images/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>
Bit Manipulation; Breadth-First Search; Array; Hash Table; Matrix
class Solution {
public:
    int minFlips(vector<vector<int>>& mat) {
        int m = mat.size(), n = mat[0].size();
        int state = 0;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (mat[i][j])
                    state |= (1 << (i * n + j));
        queue<int> q{{state}};
        unordered_set<int> vis{{state}};
        int ans = 0;
        vector<int> dirs = {0, -1, 0, 1, 0, 0};
        while (!q.empty()) {
            for (int t = q.size(); t; --t) {
                state = q.front();
                if (state == 0) return ans;
                q.pop();
                for (int i = 0; i < m; ++i) {
                    for (int j = 0; j < n; ++j) {
                        int nxt = state;
                        for (int k = 0; k < 5; ++k) {
                            int x = i + dirs[k], y = j + dirs[k + 1];
                            if (x < 0 || x >= m || y < 0 || y >= n) continue;
                            if ((nxt & (1 << (x * n + y))) != 0)
                                nxt -= 1 << (x * n + y);
                            else
                                nxt |= 1 << (x * n + y);
                        }
                        if (!vis.count(nxt)) {
                            vis.insert(nxt);
                            q.push(nxt);
                        }
                    }
                }
            }
            ++ans;
        }
        return -1;
    }
};