<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>

from typing import List

class Solution:
    def isPossibleToCutPath(self, grid: List[List[int]]) -> bool:
        m, n = len(grid), len(grid[0])
        
        # Helper function to perform DFS and find a path
        def dfs(x, y):
            if x == m - 1 and y == n - 1:
                return True
            grid[x][y] = 0  # Mark the cell as visited
            # Check if we can move down or right
            if x + 1 < m and grid[x + 1][y] == 1 and dfs(x + 1, y):
                return True
            if y + 1 < n and grid[x][y + 1] == 1 and dfs(x, y + 1):
                return True
            return False
        
        # Check if there's an initial path from (0, 0) to (m - 1, n - 1)
        if not dfs(0, 0):
            return True  # Already disconnected
        
        # Restore the grid for a second DFS
        grid[0][0] = 1
        
        # Try to find a second distinct path
        if not dfs(0, 0):
            return True  # If we can't find a second path, we can disconnect by flipping one cell
        
        return False  # If there are two paths, we can't disconnect by flipping just one cell

# Example usage:
# sol = Solution()
# print(sol.isPossibleToCutPath([[1,1,1],[1,0,0],[1,1,1]]))  # Output: True
# print(sol.isPossibleToCutPath([[1,1,1],[1,0,1],[1,1,1]]))  # Output: False
20400