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
func maxDistance(grid [][]int) int {
	n := len(grid)
	q := [][2]int{}
	for i, row := range grid {
		for j, v := range row {
			if v == 1 {
				q = append(q, [2]int{i, j})
			}
		}
	}
	ans := -1
	if len(q) == 0 || len(q) == n*n {
		return ans
	}
	dirs := [5]int{-1, 0, 1, 0, -1}
	for len(q) > 0 {
		for i := len(q); i > 0; i-- {
			p := q[0]
			q = q[1:]
			for k := 0; k < 4; k++ {
				x, y := p[0]+dirs[k], p[1]+dirs[k+1]
				if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 0 {
					grid[x][y] = 1
					q = append(q, [2]int{x, y})
				}
			}
		}
		ans++
	}
	return ans
}