Shortest Path in a Grid with Obstacles Elimination
Problem Description
You are given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step.
Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.
Example 1:
Input: grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1
Output: 6
Explanation:
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
Explanation: We need to eliminate at least two obstacles to find such a walk.
Constraints:
m == grid.lengthn == grid[i].length1 <= m, n <= 401 <= k <= m * ngrid[i][j]is either0or1.grid[0][0] == grid[m - 1][n - 1] == 0
Solve it on LeetCode
Approaches
1. BFS with State Tracking
Intuition:
The problem requires finding the shortest path in a grid while overcoming obstacles with a limited number of eliminations allowed for obstacles. A natural approach to explore all possible paths in an unweighted grid and account for the obstacle eliminations is to use Breadth-First Search (BFS).
In this solution:
- We treat the grid as a graph where each cell is a node.
- Edges exist between each node (cell) to its neighbor nodes (adjacent cells).
- We incorporate additional state variables in our search to account for the number of eliminations performed so far and only proceed if doing so does not exceed the maximum allowable eliminations.
- We use a 3D boolean array to track visited states, ensuring that we do not process the same state multiple times unnecessarily.
Code:
- Time Complexity: O(N * M * K)), where (N) is the number of rows, (M) is the number of columns, and (K) is the limit on obstacle eliminations.
- Space Complexity: O(n), due to the use of additional array.
2. Optimized BFS using A* Heuristic
Intuition:
This approach enhances the BFS by adding an A* heuristic to prioritize the exploration of paths that seem promising, based on a heuristic function that estimates the cost to reach the destination. The heuristic used can be the Manhattan distance to the bottom-right corner. By incorporating the A* heuristic, we guide the BFS to expand nodes that are likely part of the shortest path, potentially reducing the number of states we explore.
Code:
- Time Complexity: Typically better than plain BFS, potentially O(N * M * K) in worst-case but often reduced by choosing promising nodes earlier.
- Space Complexity: O(N * M * K) similar to BFS due to state storage.