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
function maxDistance(grid: number[][]): number {
    const n = grid.length;
    const q: [number, number][] = [];
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] === 1) {
                q.push([i, j]);
            }
        }
    }
    let ans = -1;
    if (q.length === 0 || q.length === n * n) {
        return ans;
    }
    const dirs: number[] = [-1, 0, 1, 0, -1];
    while (q.length > 0) {
        for (let m = q.length; m; --m) {
            const [i, j] = q.shift()!;
            for (let k = 0; k < 4; ++k) {
                const x = i + dirs[k];
                const y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] === 0) {
                    grid[x][y] = 1;
                    q.push([x, y]);
                }
            }
        }
        ++ans;
    }
    return ans;
}
