LeetCode-in-Java

2556. Disconnect Path in a Binary Matrix by at Most One Flip

Medium

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 either 0 or 1.
• grid[0][0] == grid[m - 1][n - 1] == 1

Solution

public class Solution {
    private int n;
    private int m;

    public boolean isPossibleToCutPath(int[][] g) {
        n = g.length;
        m = g[0].length;
        if (!dfs(0, 0, g)) {
            return true;
        }
        g[0][0] = 1;
        return !dfs(0, 0, g);
    }

    private boolean dfs(int r, int c, int[][] g) {
        if (r == n - 1 && c == m - 1) {
            return true;
        }
        if (r == n || c == m || g[r][c] == 0) {
            return false;
        }
        g[r][c] = 0;
        return dfs(r, c + 1, g) || dfs(r + 1, c, g);
    }
}

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