Solution Explanation

This problem asks for the minimum number of steps to reach the bottom-right corner of a grid from the top-left corner, given that you can eliminate at most k obstacles. The solution uses a Breadth-First Search (BFS) algorithm.

Approach:

1. Base Case: If k is large enough to remove all possible obstacles along the shortest path (without considering obstacles), the shortest path is simply the sum of rows and columns minus 1.

2. BFS Implementation: The core of the solution is a BFS traversal of the grid. Each node in the BFS queue represents a state (row, column, remaining_obstacles).

3. State Representation: The state (row, column, remaining_obstacles) is crucial. It keeps track of not only the current cell's coordinates but also how many obstacles can still be removed.

4. Exploration: At each step, we explore the four adjacent cells (up, down, left, right). For each neighbor:

   • If the cell is empty (0), we add the state (neighbor_row, neighbor_column, remaining_obstacles) to the queue if it hasn't been visited before.
   • If the cell is an obstacle (1) and we still have obstacles to remove (remaining_obstacles > 0), we add the state (neighbor_row, neighbor_column, remaining_obstacles - 1) to the queue if it hasn't been visited before.

5. Visited States: We use a 3D boolean array vis to keep track of visited states. This prevents cycles and ensures we find the shortest path. The three dimensions correspond to row, column, and remaining obstacles.

6. Termination: The BFS terminates when:

   • We reach the bottom-right corner (m-1, n-1). The number of steps taken is returned.
   • The queue becomes empty, meaning no path exists within the given constraints. -1 is returned.

Time Complexity: O(M * N * K), where M and N are the dimensions of the grid and K is the maximum number of obstacles allowed to be removed. In the worst case, we might visit each cell with each possible value of remaining obstacles.

Space Complexity: O(M * N * K), dominated by the vis array used to track visited states.

Code Explanation (Python3)

from collections import deque
 
class Solution:
    def shortestPath(self, grid: List[List[int]], k: int) -> int:
        m, n = len(grid), len(grid[0])
        # Base case: Sufficient k to remove all obstacles along the shortest path
        if k >= m + n - 3:
            return m + n - 2
 
        q = deque([(0, 0, k)])  # Queue of (row, col, remaining_obstacles)
        vis = {(0, 0, k)}       # Set to track visited states
        ans = 0                # Steps taken
 
        while q:
            ans += 1
            for _ in range(len(q)): #Process all nodes at current distance
                i, j, cur_k = q.popleft()
 
                # Check for target
                if i == m - 1 and j == n - 1:
                    return ans
 
                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 grid[x][y] == 0 and (x, y, cur_k) not in vis:
                            q.append((x, y, cur_k))
                            vis.add((x, y, cur_k))
                        elif grid[x][y] == 1 and cur_k > 0 and (x, y, cur_k - 1) not in vis:
                            q.append((x, y, cur_k - 1))
                            vis.add((x, y, cur_k - 1))
 
        return -1  # No path found

The other language versions follow a similar structure, adapting the syntax and data structures to the specific language. The core algorithm remains the same BFS with state tracking.