# Solving Leetcode Interviews in Seconds with AI: Minimum Obstacle Removal to Reach Corner


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2290" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a 0-indexed 2D integer array grid of size m x n. Each cell has one of two values:  0 represents an empty cell, 1 represents an obstacle that may be removed.  You can move up, down, left, or right from and to an empty cell. Return the minimum number of obstacles to remove so you can move from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1).   Example 1:   Input: grid = [[0,1,1],[1,1,0],[1,1,0]] Output: 2 Explanation: We can remove the obstacles at (0, 1) and (0, 2) to create a path from (0, 0) to (2, 2). It can be shown that we need to remove at least 2 obstacles, so we return 2. Note that there may be other ways to remove 2 obstacles to create a path.  Example 2:   Input: grid = [[0,1,0,0,0],[0,1,0,1,0],[0,0,0,1,0]] Output: 0 Explanation: We can move from (0, 0) to (2, 4) without removing any obstacles, so we return 0.    Constraints:  m == grid.length n == grid[i].length 1 <= m, n <= 105 2 <= m * n <= 105 grid[i][j] is either 0 or 1. grid[0][0] == grid[m - 1][n - 1] == 0  

	# Explanation
	Here's an efficient solution to find the minimum obstacles to remove, along with an explanation:

*   **Core Idea:** Use Dijkstra's algorithm (or 0-1 BFS) to find the shortest path from the top-left to the bottom-right corner, where the "cost" of moving to a cell is the number of obstacles encountered (0 for an empty cell, 1 for an obstacle).

*   **0-1 BFS Optimization:** Since the weights are only 0 or 1, we can use a deque (double-ended queue) to implement a 0-1 BFS, which is a specialized version of Dijkstra's algorithm that's more efficient for this specific scenario.  When we move to a neighbor with weight 0, we push it to the *front* of the queue, and when we move to a neighbor with weight 1, we push it to the *back* of the queue. This ensures that we explore paths with fewer obstacles first.

*   **Distance Tracking:** Keep track of the minimum number of obstacles needed to reach each cell.

*   **Runtime & Storage Complexity:** O(m*n) for both runtime and space, where m is the number of rows and n is the number of columns in the grid.

	
	# Code
	```python
	from collections import deque

def minimumObstacles(grid):
    m, n = len(grid), len(grid[0])
    dist = [[float('inf')] * n for _ in range(m)]
    dist[0][0] = 0
    q = deque([(0, 0)])
    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    while q:
        x, y = q.popleft()

        for dx, dy in directions:
            nx, ny = x + dx, y + dy
            if 0 <= nx < m and 0 <= ny < n:
                weight = grid[nx][ny]
                if dist[x][y] + weight < dist[nx][ny]:
                    dist[nx][ny] = dist[x][y] + weight
                    if weight == 0:
                        q.appendleft((nx, ny))
                    else:
                        q.append((nx, ny))

    return dist[m - 1][n - 1]
	```