What is the time complexity of Disconnect Path in a Binary Matrix by at Most One Flip?

The optimal solution for Disconnect Path in a Binary Matrix by at Most One Flip runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the grid a limited number of times, making it linear in terms of the number of cells.

What is the brute force solution for Disconnect Path in a Binary Matrix by at Most One Flip?

The brute force approach for Disconnect Path in a Binary Matrix by at Most One Flip is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible cell in the matrix to see if flipping it would disconnect the path from (0, 0) to (m - 1, n - 1). This method is straightforward but inefficient, as it involves multiple checks for each cell. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Disconnect Path in a Binary Matrix by at Most One Flip?

Explain your thought process clearly while coding. Consider edge cases and discuss them with the interviewer. Practice BFS/DFS problems to become comfortable with grid traversal.

What are common mistakes when solving Disconnect Path in a Binary Matrix by at Most One Flip?

Not considering edge cases where the grid is already disconnected. Failing to reset the state of the grid after checking a flip.

#2556

Disconnect Path in a Binary Matrix by at Most One Flip

Medium

ArrayDynamic ProgrammingDepth-First SearchBreadth-First SearchMatrixGraph TraversalBreadth-First SearchDepth-First Search

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the idea of finding two non-intersecting paths from (0, 0) to (m - 1, n - 1). If such paths exist, flipping one cell won't disconnect the paths. This approach is efficient and avoids unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Perform a BFS or DFS to find the first path from (0, 0) to (m - 1, n - 1).
2Step 2: Mark all cells in the found path as visited.
3Step 3: Perform another BFS or DFS from (0, 0) to find a second path that does not intersect with the first path.
4Step 4: If a second path is found, return false. If not, return true.

solution.py31 lines

1# Full working Python code
2from collections import deque
3
4def canDisconnect(grid):
5    m, n = len(grid), len(grid[0])
6
7    def bfs(start, visited):
8        queue = deque([start])
9        path = []
10        while queue:
11            x, y = queue.popleft()
12            path.append((x, y))
13            if (x, y) == (m - 1, n - 1):
14                return path
15            for dx, dy in [(1, 0), (0, 1)]:
16                nx, ny = x + dx, y + dy
17                if 0 <= nx < m and 0 <= ny < n and grid[nx][ny] == 1 and (nx, ny) not in visited:
18                    visited.add((nx, ny))
19                    queue.append((nx, ny))
20        return []
21
22    visited = set()
23    path1 = bfs((0, 0), visited)
24    if not path1:
25        return True
26
27    for x, y in path1:
28        visited.add((x, y))
29
30    path2 = bfs((0, 0), visited)
31    return len(path2) == 0

ℹ

Complexity note: The time complexity is O(n) because we only traverse the grid a limited number of times, making it linear in terms of the number of cells.

1Finding two non-intersecting paths is crucial to determine if a disconnection is possible.
2BFS/DFS can efficiently explore paths in a grid-like structure.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.