Disconnect Path in a Binary Matrix by at Most One Flip

Problem 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.

Examples

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
• grid[0][0] == grid[m - 1][n - 1] == 1
• grid[i][j] is either 0 or 1

Approach

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.

Python3

class Solution:
    def isPossibleToCutPath(self, grid: List[List[int]]) -> bool:
        def dfs(i, j):
            if i >= m or j >= n or grid[i][j] == 0:
                return False
            grid[i][j] = 0
            if i == m - 1 and j == n - 1:
                return True
            return dfs(i + 1, j) or dfs(i, j + 1)

        m, n = len(grid), len(grid[0])
        a = dfs(0, 0)
        grid[0][0] = grid[-1][-1] = 1
        b = dfs(0, 0)
        return not (a and b)

Java

class Solution {
    private int[][] grid;
    private int m;
    private int n;

    public boolean isPossibleToCutPath(int[][] grid) {
        this.grid = grid;
        m = grid.length;
        n = grid[0].length;
        boolean a = dfs(0, 0);
        grid[0][0] = 1;
        grid[m - 1][n - 1] = 1;
        boolean b = dfs(0, 0);
        return !(a && b);
    }

    private boolean dfs(int i, int j) {
        if (i >= m || j >= n || grid[i][j] == 0) {
            return false;
        }
        if (i == m - 1 && j == n - 1) {
            return true;
        }
        grid[i][j] = 0;
        return dfs(i + 1, j) || dfs(i, j + 1);
    }
}

C++

class Solution {
public:
    bool isPossibleToCutPath(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        function<bool(int, int)> dfs = [&](int i, int j) -> bool {
            if (i >= m || j >= n || grid[i][j] == 0) {
                return false;
            }
            if (i == m - 1 && j == n - 1) {
                return true;
            }
            grid[i][j] = 0;
            return dfs(i + 1, j) || dfs(i, j + 1);
        };
        bool a = dfs(0, 0);
        grid[0][0] = grid[m - 1][n - 1] = 1;
        bool b = dfs(0, 0);
        return !(a && b);
    }
};