What is the brute force solution for Shortest Path in a Grid with Obstacles Elimination?

The brute force approach for Shortest Path in a Grid with Obstacles Elimination is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves exploring all possible paths from the starting point to the destination, considering the number of obstacles we can eliminate. This method is straightforward but inefficient, as it may explore many unnecessary paths. The optimal Optimal Solution approach improves this to O(m * n * k).

What algorithm or data structure is used to solve Shortest Path in a Grid with Obstacles Elimination?

Shortest Path in a Grid with Obstacles Elimination uses the following concepts: Array, Breadth-First Search, Matrix. The recommended patterns to study are: Breadth-First Search, Dynamic Programming.

What are the interview tips for Shortest Path in a Grid with Obstacles Elimination?

Clearly explain your thought process and the trade-offs of different approaches. Practice BFS and DFS patterns, as they are commonly used in grid problems. Be prepared to discuss edge cases, such as grids with no obstacles or maximum obstacles.

What are common mistakes when solving Shortest Path in a Grid with Obstacles Elimination?

Failing to track the number of obstacles eliminated can lead to incorrect results. Not using a visited structure can cause infinite loops or excessive computations.

#1293

Shortest Path in a Grid with Obstacles Elimination

Hard

ArrayBreadth-First SearchMatrixBreadth-First SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(m * n * k)Space O(m * n * k)

The optimal solution uses Breadth-First Search (BFS) to explore the grid while keeping track of the number of obstacles eliminated. This approach ensures that we find the shortest path efficiently by exploring all possible paths level by level.

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue for BFS and a 3D visited array to track visited cells with the number of obstacles eliminated.
2Step 2: Start from the top-left corner (0, 0) and enqueue it with the initial steps and remaining k.
3Step 3: While the queue is not empty, dequeue an element and explore its neighbors (up, down, left, right). If a neighbor is an obstacle and we can eliminate it, enqueue it with updated steps and remaining k.

solution.py23 lines

1# Full working Python code
2from collections import deque
3
4def min_steps(grid, k):
5    m, n = len(grid), len(grid[0])
6    visited = [[[False] * (k + 1) for _ in range(n)] for _ in range(m)]
7    queue = deque([(0, 0, 0, k)])  # (x, y, steps, remaining_k)
8    directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
9
10    while queue:
11        x, y, steps, remaining_k = queue.popleft()
12        if x == m - 1 and y == n - 1:
13            return steps
14        for dx, dy in directions:
15            nx, ny = x + dx, y + dy
16            if 0 <= nx < m and 0 <= ny < n:
17                if grid[nx][ny] == 0 and not visited[nx][ny][remaining_k]:
18                    visited[nx][ny][remaining_k] = True
19                    queue.append((nx, ny, steps + 1, remaining_k))
20                elif grid[nx][ny] == 1 and remaining_k > 0 and not visited[nx][ny][remaining_k - 1]:
21                    visited[nx][ny][remaining_k - 1] = True
22                    queue.append((nx, ny, steps + 1, remaining_k - 1))
23    return -1

ℹ

Complexity note: The time complexity is O(m * n * k) because we may visit each cell with each possible number of obstacles eliminated. The space complexity is also O(m * n * k) due to the visited array.

1Using BFS allows us to explore all possible paths efficiently without revisiting cells unnecessarily.
2Tracking the number of obstacles eliminated helps in making optimal decisions at each step.

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