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.

Solution 2: Two Explicit Stack Passes

Thinking

Moves are only right or down. We may flip at most one non-endpoint cell to disconnect $(0,0)$ from the exit. Trying every flip is cubic. Two internally vertex-disjoint paths cannot be cut by a single flip. Search down-first, zero the visited cells, restore the two ends, and search again. There can be $1000$ rows, and the first move is always down, so a recursive search has depth $m$. An explicit stack keeps that order: push the right neighbor, then the cell below.

The search uses an explicit stack, clears a cell when it is entered, and handles the cell below before the cell to the right.

First, we search once with an explicit stack to determine whether there is a path from $(0, 0)$ to $(m - 1, n - 1)$, and we denote the result as $a$. During the search, 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 search once more with an explicit stack to determine whether there is a path from $(0, 0)$ to $(m - 1, n - 1)$, and we denote the result as $b$. During the search, 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);
    }
}


// Solution 2
class Solution {
    public boolean isPossibleToCutPath(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        boolean a = dfs(grid, m, n);
        grid[0][0] = 1;
        grid[m - 1][n - 1] = 1;
        boolean b = dfs(grid, m, n);
        return !(a && b);
    }

    private boolean dfs(int[][] grid, int m, int n) {
        int[] stk = new int[m * n * 2];
        int top = 0;
        stk[top++] = 0;
        while (top > 0) {
            int cur = stk[--top];
            int i = cur / n;
            int j = cur % n;
            if (i >= m || j >= n || grid[i][j] == 0) {
                continue;
            }
            grid[i][j] = 0;
            if (i == m - 1 && j == n - 1) {
                return true;
            }
            if (j + 1 < n) {
                stk[top++] = i * n + j + 1;
            }
            if (i + 1 < m) {
                stk[top++] = (i + 1) * n + j;
            }
        }
        return false;
    }
}

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);
    }
};


// Solution 2
class Solution {
public:
    bool isPossibleToCutPath(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        auto dfs = [&]() {
            vector<pair<int, int>> stk{{0, 0}};
            while (!stk.empty()) {
                auto [i, j] = stk.back();
                stk.pop_back();
                if (i >= m || j >= n || grid[i][j] == 0) {
                    continue;
                }
                grid[i][j] = 0;
                if (i == m - 1 && j == n - 1) {
                    return true;
                }
                stk.emplace_back(i, j + 1);
                stk.emplace_back(i + 1, j);
            }
            return false;
        };
        bool a = dfs();
        grid[0][0] = grid[m - 1][n - 1] = 1;
        bool b = dfs();
        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)



# Solution 2
class Solution:
    def isPossibleToCutPath(self, grid: List[List[int]]) -> bool:
        def dfs() -> bool:
            stk = [(0, 0)]
            while stk:
                i, j = stk.pop()
                if i >= m or j >= n or grid[i][j] == 0:
                    continue
                grid[i][j] = 0
                if i == m - 1 and j == n - 1:
                    return True
                stk.append((i, j + 1))
                stk.append((i + 1, j))
            return False

        m, n = len(grid), len(grid[0])
        a = dfs()
        grid[0][0] = grid[-1][-1] = 1
        b = dfs()
        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)
}


// Solution 2
func isPossibleToCutPath(grid [][]int) bool {
	m, n := len(grid), len(grid[0])
	dfs := func() bool {
		stk := [][2]int{{0, 0}}
		for len(stk) > 0 {
			i, j := stk[len(stk)-1][0], stk[len(stk)-1][1]
			stk = stk[:len(stk)-1]
			if i >= m || j >= n || grid[i][j] == 0 {
				continue
			}
			grid[i][j] = 0
			if i == m-1 && j == n-1 {
				return true
			}
			stk = append(stk, [2]int{i, j + 1}, [2]int{i + 1, j})
		}
		return false
	}
	a := dfs()
	grid[0][0], grid[m-1][n-1] = 1, 1
	b := dfs()
	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);
}



// Solution 2
function isPossibleToCutPath(grid: number[][]): boolean {
    const m = grid.length;
    const n = grid[0].length;
    const dfs = (): boolean => {
        const stk: number[][] = [[0, 0]];
        while (stk.length) {
            const [i, j] = stk.pop()!;
            if (i >= m || j >= n || grid[i][j] !== 1) {
                continue;
            }
            grid[i][j] = 0;
            if (i === m - 1 && j === n - 1) {
                return true;
            }
            stk.push([i, j + 1], [i + 1, j]);
        }
        return false;
    };
    const a = dfs();
    grid[0][0] = 1;
    grid[m - 1][n - 1] = 1;
    const b = dfs();
    return !(a && b);
}

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

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

Description

Solutions

Solution 2: Two Explicit Stack Passes

All Problems

All Solutions

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

Description

Solutions

Solution 2: Two Explicit Stack Passes

All Problems

All Solutions