2556. Disconnect Path in a Binary Matrix by at Most One Flip
Description
You are given a 0-indexed m x n
binary matrix grid
. You can move from a cell (row, col)
to any of the cells (row + 1, col)
or (row, col + 1)
that has the value 1
. The matrix is disconnected if there is no path from (0, 0)
to (m - 1, n - 1)
.
You can flip the value of at most one (possibly none) cell. You cannot flip the cells (0, 0)
and (m - 1, n - 1)
.
Return true
if it is possible to make the matrix disconnect or false
otherwise.
Note that flipping a cell changes its value from 0
to 1
or from 1
to 0
.
Example 1:
Input: grid = [[1,1,1],[1,0,0],[1,1,1]] Output: true Explanation: We can change the cell shown in the diagram above. There is no path from (0, 0) to (2, 2) in the resulting grid.
Example 2:
Input: grid = [[1,1,1],[1,0,1],[1,1,1]] Output: false Explanation: It is not possible to change at most one cell such that there is not path from (0, 0) to (2, 2).
Constraints:
m == grid.length
n == grid[i].length
1 <= m, n <= 1000
1 <= m * n <= 105
grid[i][j]
is either0
or1
.grid[0][0] == grid[m - 1][n - 1] == 1
Solutions
Solution 1: Two DFS Traversals
First, we perform a DFS traversal to determine whether there is a path from $(0, 0)$ to $(m - 1, n - 1)$, and we denote the result as $a$. During the DFS process, we set the value of the visited cells to $0$ to prevent revisiting.
Next, we set the values of $(0, 0)$ and $(m - 1, n - 1)$ to $1$, and perform another DFS traversal to determine whether there is a path from $(0, 0)$ to $(m - 1, n - 1)$, and we denote the result as $b$. During the DFS process, we set the value of the visited cells to $0$ to avoid revisiting.
Finally, if both $a$ and $b$ are true
, we return false
, otherwise, we return true
.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the matrix, respectively.
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