<p>You are given a <strong>0-indexed</strong> <code>m x n</code> <strong>binary</strong> matrix <code>grid</code>. You can move from a cell <code>(row, col)</code> to any of the cells <code>(row + 1, col)</code> or <code>(row, col + 1)</code> that has the value <code>1</code>.&nbsp;The matrix is <strong>disconnected</strong> if there is no path from <code>(0, 0)</code> to <code>(m - 1, n - 1)</code>.</p>

<p>You can flip the value of <strong>at most one</strong> (possibly none) cell. You <strong>cannot flip</strong> the cells <code>(0, 0)</code> and <code>(m - 1, n - 1)</code>.</p>

<p>Return <code>true</code> <em>if it is possible to make the matrix disconnect or </em><code>false</code><em> otherwise</em>.</p>

<p><strong>Note</strong> that flipping a cell changes its value from <code>0</code> to <code>1</code> or from <code>1</code> to <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/12/07/yetgrid2drawio.png" style="width: 441px; height: 151px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,1],[1,0,0],[1,1,1]]
<strong>Output:</strong> true
<strong>Explanation:</strong> We can change the cell shown in the diagram above. There is no path from (0, 0) to (2, 2) in the resulting grid.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/12/07/yetgrid3drawio.png" />
<pre>
<strong>Input:</strong> grid = [[1,1,1],[1,0,1],[1,1,1]]
<strong>Output:</strong> false
<strong>Explanation:</strong> It is not possible to change at most one cell such that there is not path from (0, 0) to (2, 2).
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 1000</code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li><code>grid[0][0] == grid[m - 1][n - 1] == 1</code></li>
</ul>

class Solution:
    def isPossibleToCutPath(self, grid: List[List[int]]) -> bool:
        rows = len(grid)
        cols = len(grid[0])

        if rows == 1 and cols == 1:
            return grid[0][0] == 0

        # BFS from bottom right to top left
        # and set the visited positions to 2
        visited = set()
        q = collections.deque([(rows - 1, cols - 1)])
        grid[-1][-1]= 2
        reach_source = False
        while q:
            next_q = collections.deque()
            reach_source = False
            while q:
                node = q.popleft()
                for dr, dc in (-1, 0), (0, -1):
                    new_row, new_col = node[0] + dr, node[1] + dc
                    if (new_row, new_col) in visited or new_row < 0 or new_col < 0 or grid[new_row][new_col] == 0:
                        continue
                    visited.add((new_row, new_col))
                    next_q.append((new_row, new_col))
                    grid[new_row][new_col] = 2
                    if new_row == 0 and new_col == 0:
                        reach_source = True
            if not reach_source and len(next_q) < 2:
                return True
            if reach_source:
                break
            q = next_q

        if not reach_source:
            return True

        visited = set()
        q = collections.deque([(0, 0)])
        while q:
            next_q = collections.deque()
            reach_dest = False
            while q:
                node = q.popleft()
                for dr, dc in (0, 1), (1, 0):
                    new_row, new_col = node[0] + dr, node[1] + dc
                    if (new_row, new_col) in visited or new_row >= rows or new_col >= cols or grid[new_row][new_col] != 2:
                        continue
                    visited.add((new_row, new_col))
                    next_q.append((new_row, new_col))
                    if new_row == rows - 1 and new_col == cols - 1:
                        reach_dest = True
            if not reach_dest and len(next_q) < 2:
                return True
            if reach_dest:
                return False
            q = next_q

        return False
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