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
func minFlips(mat [][]int) int {
	m, n := len(mat), len(mat[0])
	state := 0
	for i, row := range mat {
		for j, v := range row {
			if v == 1 {
				state |= 1 << (i*n + j)
			}
		}
	}
	q := []int{state}
	vis := map[int]bool{state: true}
	ans := 0
	dirs := []int{0, -1, 0, 1, 0, 0}
	for len(q) > 0 {
		for t := len(q); t > 0; t-- {
			state = q[0]
			if state == 0 {
				return ans
			}
			q = q[1:]
			for i := 0; i < m; i++ {
				for j := 0; j < n; j++ {
					nxt := state
					for k := 0; k < 5; k++ {
						x, y := i+dirs[k], 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[nxt] {
						vis[nxt] = true
						q = append(q, nxt)
					}
				}
			}
		}
		ans++
	}
	return -1
}