<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:
        if grid == [[1,1]] or grid == [[1],[1]] or grid == [[1]]:
            return False
        rows=len(grid)
        cols=len(grid[0])
        if rows<2 or cols<2:
            return True

        if grid[0][1] == 0 or grid[1][0] ==0:
            return True

        if grid[rows-1][cols-2] == 0 or grid[rows-2][cols-1] == 0:
            return True
        for row in grid:
            if sum(row) <= 1:
                return True
        def first(trail):
            if trail[-1] == (rows-1,cols-1):
                return trail
            r,c = trail[-1]
            if c < cols - 1 and grid[r][c+1] == 1:
                trail2 = first(trail + [(r,c+1)])
                if trail2[-1] == (rows-1,cols-1):
                    return trail2
            if r < rows - 1 and grid[r+1][c] == 1:
                trail2 = first(trail + [(r+1,c)])
                if trail2[-1]==(rows-1,cols-1):
                    return trail2
            return trail

        def second(cell):
            if cell == (rows-1,cols-1):
                return True
            r,c = cell
            if r < rows - 1 and grid[r+1][c] == 1 and (r+1,c) not in path:
                cell2 = second((r+1,c))
                if cell2 is True:
                    return True
            if c < cols - 1 and grid[r][c+1] == 1 and (r,c+1) not in path:
                cell2 = second((r,c+1))
                if cell2 is True:
                    return True
            return False
        
        tr = first([(0,1)])
        if tr[-1] != (rows-1,cols-1):
            return True
        path = set(tr[:-1])
        return not second((1,0))
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