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
use std::collections::VecDeque;

impl Solution {
    pub fn update_matrix(mat: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let m = mat.len();
        let n = mat[0].len();
        let mut ans = vec![vec![-1; n]; m];
        let mut q = VecDeque::new();

        for i in 0..m {
            for j in 0..n {
                if mat[i][j] == 0 {
                    q.push_back((i, j));
                    ans[i][j] = 0;
                }
            }
        }

        let dirs = [-1, 0, 1, 0, -1];
        while let Some((i, j)) = q.pop_front() {
            for k in 0..4 {
                let x = i as isize + dirs[k];
                let y = j as isize + dirs[k + 1];
                if x >= 0 && x < m as isize && y >= 0 && y < n as isize {
                    let x = x as usize;
                    let y = y as usize;
                    if ans[x][y] == -1 {
                        ans[x][y] = ans[i][j] + 1;
                        q.push_back((x, y));
                    }
                }
            }
        }

        ans
    }
}
