<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>

from collections import deque
class Solution:
    def minFlips(self, mat: List[List[int]]) -> int:
        def isSolved(mat):
            for row in mat:
                if any(cell == 1 for cell in row):
                    return False
            return True
        def performMove(mat, row, col):
            directions = [(0, 0), (1, 0), (-1, 0), (0, 1), (0, -1)]
            new_mat = [r[:] for r in mat]
            for dx, dy in directions:
                if 0 <= row + dx < len(mat) and 0 <= col + dy < len(mat[0]):
                    new_mat[row + dx][col + dy] = 1 if new_mat[row + dx][col + dy] == 0 else 0
            return new_mat

        ROWS = len(mat)
        COLS = len(mat[0])

        if isSolved(mat):
            return 0

        # bfs step
        initial_state = tuple(tuple(row) for row in mat)     
        visited = set()
        visited.add(initial_state)

        q = deque([(mat, 0)])
        while q:
            curr_mat, curr_moves = q.popleft()
            for i in range(ROWS):
                for j in range(COLS):
                    next_mat = performMove(curr_mat, i, j)
                    tuple_next_mat = tuple(tuple(row) for row in next_mat)
                    if isSolved(next_mat):
                        return curr_moves + 1
                    if tuple_next_mat not in visited:
                        visited.add(tuple_next_mat)
                        q.append((next_mat, curr_moves + 1))
        
        return -1
        
16800