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Formatted question description: https://leetcode.ca/all/2556.html

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 either 0 or 1.
  • 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.

  • 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);
        }
    }
    
  • 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);
        }
    };
    
  • 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)
    
    
  • func isPossibleToCutPath(grid [][]int) bool {
    	m, n := len(grid), len(grid[0])
    	var dfs func(i, j int) bool
    	dfs = func(i, j int) 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)
    	}
    	a := dfs(0, 0)
    	grid[0][0], grid[m-1][n-1] = 1, 1
    	b := dfs(0, 0)
    	return !(a && b)
    }
    
  • function isPossibleToCutPath(grid: number[][]): boolean {
        const m = grid.length;
        const n = grid[0].length;
    
        const dfs = (i: number, j: number): boolean => {
            if (i >= m || j >= n || grid[i][j] !== 1) {
                return false;
            }
            grid[i][j] = 0;
            if (i === m - 1 && j === n - 1) {
                return true;
            }
            return dfs(i + 1, j) || dfs(i, j + 1);
        };
        const a = dfs(0, 0);
        grid[0][0] = 1;
        grid[m - 1][n - 1] = 1;
        const b = dfs(0, 0);
        return !(a && b);
    }
    
    

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