<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):
        m,n = len(grid),len(grid[0])

        stack, visited, res1 = [(0,0,[[0,0]])], set(), []

        visited.add((0,0))

        while stack:
            x,y,path = stack.pop()

            if x == m-1 and y == n-1:
                res1 = path

            for (nx,ny) in [(x+1,y),(x,y+1)]:
                if 0 <= nx < m and 0 <= ny < n and grid[nx][ny] == 1 and (nx,ny) not in visited:
                    stack.append((nx,ny,path+[[nx,ny]]))
                    visited.add((nx,ny))

        if not res1:
            return True 

        for i,j in res1:
            if [i,j] != [0,0] and [i,j] != [m-1,n-1]:
                grid[i][j] = 0 

        stack, visited = [(0,0)], set()

        visited.add((0,0))

        while stack:
            x,y = stack.pop()

            if x == m-1 and y == n-1:
                return False 

            for (nx,ny) in [(x+1,y),(x,y+1)]:
                if 0 <= nx < m and 0 <= ny < n and grid[nx][ny] == 1 and (nx,ny) not in visited:
                    stack.append((nx,ny))
                    visited.add((nx,ny))

        return True
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