Leetcode 2556: Disconnect Path in a Binary Matrix by at Most One Flip
You are given a binary matrix
grid where you can move from any cell (row, col) to adjacent cells (row + 1, col) or (row, col + 1) only if they have the value 1. The grid is disconnected if there is no path from the top-left corner (0, 0) to the bottom-right corner (m - 1, n - 1). You are allowed to flip at most one cell (changing a 1 to 0 or vice versa), but you cannot flip the cells (0, 0) or (m - 1, n - 1). Return true if it is possible to disconnect the grid by flipping at most one cell, otherwise return false.Problem
Approach
Steps
Complexity
Input: You are given an `m x n` binary matrix `grid` where each cell is either 0 or 1.
Example: grid = [[1, 1, 1], [1, 0, 0], [1, 1, 1]]
Constraints:
• 1 <= m, n <= 1000
• 1 <= m * n <= 10^5
• grid[i][j] is either 0 or 1
• grid[0][0] == grid[m - 1][n - 1] == 1
Output: Return true if it is possible to disconnect the grid by flipping at most one cell, otherwise return false.
Example: true
Constraints:
Goal: Check if it is possible to disconnect the grid by flipping at most one cell.
Steps:
• 1. Check if the grid is already disconnected. If there is no path from (0, 0) to (m-1, n-1), return true.
• 2. Try flipping each cell (except (0, 0) and (m-1, n-1)) and check if the grid becomes disconnected.
• 3. If flipping any cell makes the grid disconnected, return true. Otherwise, return false.
Goal: The solution should efficiently handle the grid of up to 1000x1000 in size.
Steps:
• The grid will always contain at least one valid path from (0, 0) to (m-1, n-1) initially.
Assumptions:
• The grid is initially connected and there is a valid path from (0, 0) to (m-1, n-1).
• Input: grid = [[1, 1, 1], [1, 0, 0], [1, 1, 1]]
• Explanation: Flipping the cell at (1, 1) disconnects the path between (0, 0) and (2, 2), making the grid disconnected.
• Input: grid = [[1, 1, 1], [1, 0, 1], [1, 1, 1]]
• Explanation: No single flip can disconnect the grid, so the result is false.
Approach: The task is to check whether flipping one cell can disconnect the grid. We will check the connectivity of the grid both before and after flipping each cell.
Observations:
• To solve this, we need to first verify if the grid is initially disconnected and if any flip can disconnect it.
• We can use a depth-first search (DFS) or breadth-first search (BFS) to verify if the grid is connected or disconnected before and after flipping any cell.
Steps:
• 1. Implement a DFS or BFS to check if there is a valid path from (0, 0) to (m-1, n-1).
• 2. Flip each cell (except (0, 0) and (m-1, n-1)) and check if the grid becomes disconnected after the flip.
• 3. If any flip causes the grid to disconnect, return true. Otherwise, return false.
Empty Inputs:
• The grid will always have a valid initial path as per the constraints.
Large Inputs:
• Ensure that the solution works efficiently with grids up to the maximum size of 1000x1000.
Special Values:
• Handle cases where there are few cells (e.g., 2x2 grids) or all cells are already 1s or 0s.
Constraints:
• The solution should work for grids with sizes up to 1000x1000.
vector<vector<int>> grid, memo;
int m, n;
bool isPossibleToCutPath(vector<vector<int>>& grid) {
m = grid.size();
n = grid[0].size();
this->grid = grid;
if(isconnected(0, 0) == false) return true;
this->grid[0][0] = 1;
return !isconnected(0, 0);
}
bool isconnected(int i, int j) {
if(i == m - 1 && j == n - 1) return true;
if(i >= m || j >= n || grid[i][j] == 0) return false;
grid[i][j] = 0;
return isconnected(i + 1, j) || isconnected(i, j + 1);
}
1 : Variable Initialization
vector<vector<int>> grid, memo;
Two 2D vectors 'grid' and 'memo' are declared. 'grid' represents the input grid, and 'memo' can be used for storing intermediate results (though it's not utilized in this function).
2 : Variable Initialization
int m, n;
Two integers, 'm' and 'n', are declared to store the dimensions of the grid.
3 : Function Definition
bool isPossibleToCutPath(vector<vector<int>>& grid) {
The 'isPossibleToCutPath' function is defined, which takes the grid as input and returns a boolean indicating whether it is possible to cut the path.
4 : Grid Dimensions
m = grid.size();
The variable 'm' is assigned the number of rows in the grid by getting the size of the 'grid' vector.
5 : Grid Dimensions
n = grid[0].size();
The variable 'n' is assigned the number of columns in the grid by getting the size of the first row ('grid[0]').
6 : Grid Assignment
this->grid = grid;
The input 'grid' is assigned to the class-level 'grid' variable.
7 : Initial Check
if(isconnected(0, 0) == false) return true;
This line checks if there is no valid path from the top-left corner (0, 0) to the bottom-right corner using the helper function 'isconnected'. If the path doesn't exist, it returns 'true', indicating the path can be cut.
8 : Modify Grid
this->grid[0][0] = 1;
The top-left corner of the grid is marked as 1, indicating that the starting point has been 'cut' for the next step.
9 : Final Check
return !isconnected(0, 0);
This final check calls 'isconnected' again to verify if there is still a valid path from (0, 0). If the path no longer exists, it returns 'true', meaning the path was successfully cut.
10 : Helper Function Definition
bool isconnected(int i, int j) {
The 'isconnected' function is defined, which checks if there is a valid path from the current position (i, j) to the bottom-right corner.
11 : Base Case Check
if(i == m - 1 && j == n - 1) return true;
This is the base case: if the current position (i, j) is the bottom-right corner, the function returns 'true', indicating a valid path exists.
12 : Out-of-Bounds and Blocked Check
if(i >= m || j >= n || grid[i][j] == 0) return false;
This line checks if the current position (i, j) is out of bounds or blocked (grid[i][j] == 0). If so, it returns 'false'.
13 : Mark Visited
grid[i][j] = 0;
This line marks the current cell as visited by setting grid[i][j] to 0.
14 : Recursive Call
return isconnected(i + 1, j) || isconnected(i, j + 1);
This line makes recursive calls to check the right (i, j + 1) and down (i + 1, j) cells to see if there is a valid path.
Best Case: O(m * n), where m is the number of rows and n is the number of columns in the grid.
Average Case: O(m * n), as we may need to check all cells to determine connectivity.
Worst Case: O(m * n), as we need to check connectivity multiple times for various flips.
Description:
Best Case: O(m * n), as the space complexity depends on the size of the grid.
Worst Case: O(m * n), for the space required to track visited cells during DFS/BFS.
Description: Space complexity is mainly dominated by the storage used for the visited cells during the traversal.
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