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

Problem Description

The problem presents us with an m x n binary matrix, where 0 and 1 represent blocked and open cells, respectively. From any cell, (row, col), we can move to an adjacent cell to the right (row, col + 1) or below (row + 1, col) providing it contains a 1.

The task is to determine whether we can make the matrix "disconnected" by flipping at most one cell's value, except for the cells at (0, 0) and (m - 1, n - 1). A disconnected matrix means there is no path from the start (0, 0) to the end (m - 1, n - 1).

Flipping a cell switches its value: if the cell was 1, it becomes 0, and if it was 0, it becomes 1. The challenge lies in deciding if it's possible to disrupt any existing path with a single flip, assuming such a path exists.

Intuition

The key intuition for this solution is to perform a Depth-First Search (DFS) from the starting cell (0, 0) two times.

Initially, we execute DFS to check if there exists a path from (0, 0) to (m - 1, n - 1) without flipping any cells. If such a path exists, then we proceed to the next step. Otherwise, if there is no such path, the grid is already disconnected, and we can return true early (which the code doesn't handle explicitly).

After conducting the initial DFS, we mark the beginning and ending cells as open (if they aren't already), hoping to test whether forcing these cells open and disconnecting another cell can lead to a disconnected path. Therefore, another DFS is executed with these conditions.

Here's the process:

1. Start by marking the current cell as visited (flipping from 1 to 0 to avoid revisiting).
2. If the end cell (m - 1, n - 1) is reached, we've found a path.
3. If we cannot find a path to the end cell, mark it as not possible to disconnect by only flipping one cell.
4. Perform the first DFS to check for the existing path.
5. If a path is found in step 4, flip the start and end cells to 1 (to ensure they are considered in the next DFS), then conduct the second DFS, enforcing the use of at least one flip elsewhere.
6. Lastly, assess the outcomes from both DFS searches. If both searches find a path, we conclude that it is not possible to disconnect the grid with a single flip. If the second DFS does not find a path, it shows that flipping at most one cell (besides the start and end) can disconnect the grid. We return true if the grid can be disconnected, false otherwise.

This approach works because flipping the cells to 1 after the initial search guarantees that if there is another path aside from the one found during the first traversal, the second traversal will reveal whether blocking one additional cell leads to disconnection.

Learn more about Depth-First Search, Breadth-First Search and Dynamic Programming patterns.

Solution Approach

The solution leverages the Depth-First Search (DFS) algorithm, which is a common technique for exploring all the paths in a grid-like structure or graph. It works by moving from the starting node and venturing as far as possible along each branch before backtracking.

In the Python code provided, a nested function dfs(i, j) is defined for the DFS traversal. This function aims to determine whether a path exists from the current cell (i, j) to the bottom-right cell (m - 1, n - 1).

Here's a breakdown of the DFS implementation:

1. Base Case: If the current cell indices are out of bounds (greater than or equal to m or n), or if the current cell value is 0 (blocked), DFS returns False, indicating no path can be made through this cell.

2. Visiting Cells: When a cell with a value 1 is visited, the function first marks it as visited by flipping it to 0. This prevents the DFS from revisiting the same cell and getting stuck in a loop.

3. Path Completion Check: If the end cell (m - 1, n - 1) is reached, DFS returns True, confirming a path has been found.

4. Exploring Adjacent Cells: The DFS continues by making recursive calls to the cell directly below (i + 1, j) and to the cell directly to the right (i, j + 1). If any of those recursive calls return True, the current function also returns True.

5. Determining Disconnectability:

   • Conduct the first DFS from the start cell (0, 0) using a = dfs(0, 0) to check for an existing path.
   • After the initial DFS, enforce the start and end cells to be open by setting grid[0][0] and grid[-1][-1] to 1.
   • Perform a second DFS with the modified grid using b = dfs(0, 0).

6. Result Evaluation: The piece of code not (a and b) returns True only if the first DFS found a path (a is True), and the second DFS did not find a path (b is False). Thus, the expression evaluates to True if we can disconnect the matrix, otherwise, it returns False.

The effectiveness of the solution is due to the algorithm's ability to explore all possible paths from the start to end cell, and the logical use of DFS to determine if disrupting any cell's connectivity (by changing a single 1 to 0) will disconnect the grid without testing every single cell explicitly. The solution cleverly checks the disconnectability of the matrix by attempting to find two distinct paths before and after assuming at least one flip elsewhere in the grid.

Example Walkthrough

Let's consider a simple 3 x 3 binary matrix example to illustrate the solution approach:

1Initial Grid:
21 1 1
30 1 0
41 0 1

Following the solution steps:

1. We attempt to perform DFS from the start cell (0, 0). We check for the path existence without any flip and mark the visited cells.
2. Since we start at (0, 0), we mark it as 0, to not revisit and move right to (0, 1) and then to (0, 2).
   • Now in the top right corner, (0, 2), we go down to (1, 2) but find that it's blocked (0).
   • We backtrack and from (0, 2) and move down to (1, 2) and find it is also blocked.
   • Backtracking again, we move down from (0, 1) to (1, 1) and since it has a 1, we continue by going right to (1, 2) which is blocked.
   • We go down from (1, 1) to (2, 1) but it is blocked.
   • Backtracking again, we go down from (0, 0) straight to (1, 0) which is blocked.
   • There is no more path to explore and we couldn't reach (2, 2), the end cell.

1Grid after first DFS attempt:
20 0 1
30 1 0
41 0 1

3. Since the first DFS didn't find any path, we know that the grid is already disconnected.
4. There is no need to proceed with the second DFS in this case since the grid is disconnected after the first DFS. We can directly return true.

Given the grid is already disconnected, this example demonstrates the scenario where the initial DFS itself concludes that no path exists from start to end without flipping any cells, thus no need for a second DFS or flipping any additional cells to conclude that the grid can be disconnected by at most one flip.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given algorithm is primarily dependent on the depth-first search (DFS) implementation that explores all possible paths from the top-left to the bottom-right corner of the grid, if such a path exists. In the worst case, every cell in the grid might be visited during the DFS.

For a grid with m rows and n columns, the worst-case time complexity is O(m*n) since each cell is visited at most once during the search.

Space Complexity

The space complexity is primarily driven by the call stack used by the recursive DFS function. In the worst case, the function may recurse to a depth equal to the sum of the number of rows and columns of the grid, because at most, the path can be as long as moving to the rightmost column and then downwards to the bottommost row (or vice versa).

Therefore, the space complexity, in the worst case, is O(m+n).

Additionally, it should be noted that the input grid is modified during the search process, which means the algorithm uses O(1) extra space aside from the stack space used by the recursion.

Learn more about how to find time and space complexity quickly using problem constraints.