#1293

Hard

auto_awesomeArray plus Breadth-First Search

LeetCode Problem Workspace

Shortest Path in a Grid with Obstacles Elimination

Find the shortest path in a grid with obstacles, allowing the elimination of up to k obstacles using BFS.

Problem Statement

You are given a m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You start at the top-left corner (0, 0) and need to reach the bottom-right corner (m - 1, n - 1), but you can only move up, down, left, or right to an empty cell. The challenge is to minimize the number of steps while being allowed to eliminate at most k obstacles. If it is impossible to find such a path, return -1.

The key difficulty lies in navigating the grid while accounting for obstacles and the ability to eliminate them, requiring an efficient BFS strategy. With grid constraints where m and n are up to 40, and k can range from 1 to m * n, the problem involves a careful balance of exploration and obstacle management to find the shortest possible path.

Examples

Example 1

Input: grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1

Output: 6

The shortest path without eliminating any obstacle is 10. The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).

Example 2

Input: grid = [[0,1,1],[1,1,1],[1,0,0]], k = 1

Output: -1

We need to eliminate at least two obstacles to find such a walk.

Constraints

m == grid.length
n == grid[i].length
1 <= m, n <= 40
1 <= k <= m * n
grid[i][j] is either 0 or 1.
grid[0][0] == grid[m - 1][n - 1] == 0

Solution Approach

Breadth-First Search (BFS)

The optimal approach for this problem is to use a breadth-first search (BFS) to explore the grid while considering the obstacles that can be eliminated. During BFS, each node should carry a state indicating the number of obstacles eliminated so far, which helps prioritize valid paths and minimize steps.

State Representation with a Queue

To implement BFS, represent the grid state with a queue where each element holds the current position (row, column) and the number of obstacles eliminated. This way, BFS can propagate through the grid dynamically, adjusting the obstacle elimination count at each step, ensuring optimal pathfinding.

Tracking Visited States

A 3D visited array should be used to track visited positions along with the number of obstacles eliminated at each position. This ensures that no path is revisited unnecessarily, saving computation time and allowing BFS to proceed efficiently.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on BFS's exploration of the grid, which is O(m * n * k), where m is the number of rows, n is the number of columns, and k is the maximum number of obstacles that can be eliminated. Space complexity is O(m * n * k) for storing visited states and the BFS queue.

What Interviewers Usually Probe

Can the candidate identify the correct algorithm pattern (BFS)?
Does the candidate understand the importance of managing obstacle eliminations during pathfinding?
Is the candidate able to optimize BFS by utilizing an efficient state-tracking mechanism?

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the problem with unnecessary algorithms or search techniques that don't account for obstacle eliminations efficiently.
Failing to properly track the state of eliminated obstacles, leading to redundant processing of paths.
Not leveraging BFS optimally, causing excessive computation or incorrect answers when paths with fewer eliminated obstacles are possible.

Follow-up variants

Allowing more than k obstacles to be eliminated.
Handling edge cases with very small grids (1x1 or 2x2).
Expanding the problem to work with more complex movement rules, such as diagonal movement.

FAQ

What is the core pattern of the Shortest Path in a Grid with Obstacles Elimination problem?

The core pattern is applying breadth-first search (BFS) with a state-tracking mechanism to manage obstacle eliminations while finding the shortest path.

How do I avoid redundant path explorations in this problem?

By using a 3D visited array to track each position's obstacle-elimination state, you ensure that no path is revisited unnecessarily, saving computation time.

What is the best way to handle grid traversal with obstacle eliminations?

BFS is the most efficient way to handle grid traversal, with the additional complexity of tracking how many obstacles have been eliminated so far.

How do I decide when to eliminate obstacles?

The decision to eliminate an obstacle is based on BFS exploring the grid and determining if eliminating the obstacle results in a shorter path to the destination.

How can GhostInterview assist with this problem?

GhostInterview provides structured guidance on using BFS effectively, managing state tracking for obstacle eliminations, and avoiding common pitfalls in grid traversal problems.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def shortestPath(self, grid: List[List[int]], k: int) -> int:
        m, n = len(grid), len(grid[0])
        if k >= m + n - 3:
            return m + n - 2
        q = deque([(0, 0, k)])
        vis = {(0, 0, k)}
        ans = 0
        while q:
            ans += 1
            for _ in range(len(q)):
                i, j, k = q.popleft()
                for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n:
                        if x == m - 1 and y == n - 1:
                            return ans
                        if grid[x][y] == 0 and (x, y, k) not in vis:
                            q.append((x, y, k))
                            vis.add((x, y, k))
                        if grid[x][y] == 1 and k > 0 and (x, y, k - 1) not in vis:
                            q.append((x, y, k - 1))
                            vis.add((x, y, k - 1))
        return -1

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