Minimum Obstacle Removal to Reach Corner
<p>You are given a <strong>0-indexed</strong> 2D integer array <code>grid</code> of size <code>m x n</code>. Each cell has one of two values:</p>

<ul>
	<li><code>0</code> represents an <strong>empty</strong> cell,</li>
	<li><code>1</code> represents an <strong>obstacle</strong> that may be removed.</li>
</ul>

<p>You can move up, down, left, or right from and to an empty cell.</p>

<p>Return <em>the <strong>minimum</strong> number of <strong>obstacles</strong> to <strong>remove</strong> so you can move from the upper left corner </em><code>(0, 0)</code><em> to the lower right corner </em><code>(m - 1, n - 1)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2290.Minimum%20Obstacle%20Removal%20to%20Reach%20Corner/images/example1drawio-1.png" style="width: 605px; height: 246px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1],[1,1,0],[1,1,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2290.Minimum%20Obstacle%20Removal%20to%20Reach%20Corner/images/example1drawio.png" style="width: 405px; height: 246px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,0,0,0],[0,1,0,1,0],[0,0,0,1,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> We can move from (0, 0) to (2, 4) without removing any obstacles, so we return 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>2 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> <strong>or</strong> <code>1</code>.</li>
	<li><code>grid[0][0] == grid[m - 1][n - 1] == 0</code></li>
</ul>
Breadth-First Search; Graph; Array; Matrix; Shortest Path; Heap (Priority Queue)
class Solution {
    public int minimumObstacles(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {0, 0, 0});
        int[] dirs = {-1, 0, 1, 0, -1};
        boolean[][] vis = new boolean[m][n];
        while (true) {
            var p = q.poll();
            int i = p[0], j = p[1], k = p[2];
            if (i == m - 1 && j == n - 1) {
                return k;
            }
            if (vis[i][j]) {
                continue;
            }
            vis[i][j] = true;
            for (int h = 0; h < 4; ++h) {
                int x = i + dirs[h], y = j + dirs[h + 1];
                if (x >= 0 && x < m && y >= 0 && y < n) {
                    if (grid[x][y] == 0) {
                        q.offerFirst(new int[] {x, y, k});
                    } else {
                        q.offerLast(new int[] {x, y, k + 1});
                    }
                }
            }
        }
    }
}