As Far from Land as Possible
<p>Given an <code>n x n</code> <code>grid</code>&nbsp;containing only values <code>0</code> and <code>1</code>, where&nbsp;<code>0</code> represents water&nbsp;and <code>1</code> represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance.&nbsp;If no land or water exists in the grid, return <code>-1</code>.</p>

<p>The distance used in this problem is the Manhattan distance:&nbsp;the distance between two cells <code>(x0, y0)</code> and <code>(x1, y1)</code> is <code>|x0 - x1| + |y0 - y1|</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/1100-1199/1162.As%20Far%20from%20Land%20as%20Possible/images/1336_ex1.jpg" style="width: 185px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,1],[0,0,0],[1,0,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The cell (1, 1) is as far as possible from all the land with distance 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1162.As%20Far%20from%20Land%20as%20Possible/images/1336_ex2.jpg" style="width: 184px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The cell (2, 2) is as far as possible from all the land with distance 4.
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n&nbsp;&lt;= 100</code></li>
	<li><code>grid[i][j]</code>&nbsp;is <code>0</code> or <code>1</code></li>
</ul>
Breadth-First Search; Array; Dynamic Programming; Matrix
class Solution {
    public int maxDistance(int[][] grid) {
        int n = grid.length;
        Deque<int[]> q = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    q.offer(new int[] {i, j});
                }
            }
        }
        int ans = -1;
        if (q.isEmpty() || q.size() == n * n) {
            return ans;
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            for (int i = q.size(); i > 0; --i) {
                int[] p = q.poll();
                for (int k = 0; k < 4; ++k) {
                    int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
                    if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 0) {
                        grid[x][y] = 1;
                        q.offer(new int[] {x, y});
                    }
                }
            }
            ++ans;
        }
        return ans;
    }
}