How do you solve Minimum Obstacle Removal to Reach Corner on LeetCode?

Minimum Obstacle Removal to Reach Corner (LeetCode #2290) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n log(m * n)) time complexity and O(m * n) space complexity. Key insight: Model the grid as a graph where each cell is a node and edges represent possible moves.

What is the brute force solution for Minimum Obstacle Removal to Reach Corner?

The brute force approach for Minimum Obstacle Removal to Reach Corner is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves exploring all possible paths from the start to the end cell, counting the number of obstacles encountered along the way. This method is straightforward but inefficient for larger grids. The optimal Optimal Solution approach improves this to O(m * n log(m * n)).

What are the interview tips for Minimum Obstacle Removal to Reach Corner?

Always clarify the problem and constraints before diving into coding. Discuss your thought process and potential approaches with the interviewer. Be prepared to optimize your initial solution and explain your reasoning.

What are common mistakes when solving Minimum Obstacle Removal to Reach Corner?

Not considering all possible paths and getting stuck in a single direction. Failing to account for already visited cells, leading to infinite loops.

#2290

Minimum Obstacle Removal to Reach Corner

Hard

ArrayBreadth-First SearchGraph TheoryHeap (Priority Queue)MatrixShortest PathGraph TraversalDijkstra's AlgorithmPriority Queue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

Using a modified Dijkstra's algorithm allows us to efficiently find the minimum number of obstacles to remove by treating the grid as a graph where moving into an obstacle has a cost of 1.

⚙️

Algorithm

3 steps

1Step 1: Initialize a priority queue to explore cells based on the number of obstacles removed.
2Step 2: Start from the top-left corner and push it into the queue with a cost of 0.
3Step 3: While the queue is not empty, pop the cell with the lowest cost, and explore its neighbors, updating their costs if a better path is found.

solution.py20 lines

1# Full working Python code
2import heapq
3from typing import List
4
5def min_obstacles(grid: List[List[int]]) -> int:
6    m, n = len(grid), len(grid[0])
7    pq = [(0, 0, 0)]  # (cost, x, y)
8    visited = set()
9    while pq:
10        cost, x, y = heapq.heappop(pq)
11        if (x, y) in visited:
12            continue
13        visited.add((x, y))
14        if x == m - 1 and y == n - 1:
15            return cost
16        for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
17            nx, ny = x + dx, y + dy
18            if 0 <= nx < m and 0 <= ny < n and (nx, ny) not in visited:
19                heapq.heappush(pq, (cost + grid[nx][ny], nx, ny))
20    return -1

ℹ

Complexity note: The complexity arises from the priority queue operations, where we may need to process each cell multiple times, leading to a logarithmic factor due to the heap.

1Model the grid as a graph where each cell is a node and edges represent possible moves.
2Using a priority queue helps efficiently find the minimum cost path in terms of obstacles removed.

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