01 Matrix
<p>Given an <code>m x n</code> binary matrix <code>mat</code>, return <em>the distance of the nearest </em><code>0</code><em> for each cell</em>.</p>

<p>The distance between two cells sharing a common edge is <code>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/0500-0599/0542.01%20Matrix/images/01-1-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[0,0,0]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[0,0,0]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0542.01%20Matrix/images/01-2-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,0],[1,1,1]]
<strong>Output:</strong> [[0,0,0],[0,1,0],[1,2,1]]
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one <code>0</code> in <code>mat</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Note:</strong> This question is the same as 1765: <a href="https://leetcode.com/problems/map-of-highest-peak/description/" target="_blank">https://leetcode.com/problems/map-of-highest-peak/</a></p>
Breadth-First Search; Array; Dynamic Programming; Matrix
function updateMatrix(mat: number[][]): number[][] {
    const [m, n] = [mat.length, mat[0].length];
    const ans: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => -1));
    const q: [number, number][] = [];
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (mat[i][j] === 0) {
                q.push([i, j]);
                ans[i][j] = 0;
            }
        }
    }
    const dirs: number[] = [-1, 0, 1, 0, -1];
    for (const [i, j] of q) {
        for (let k = 0; k < 4; ++k) {
            const [x, y] = [i + dirs[k], j + dirs[k + 1]];
            if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] === -1) {
                ans[x][y] = ans[i][j] + 1;
                q.push([x, y]);
            }
        }
    }
    return ans;
}
