<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 adjacent cells is <code>1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/01-1-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://assets.leetcode.com/uploads/2021/04/24/01-2-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>

from collections import deque

class Solution:
    def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
        rows, cols = len(mat), len(mat[0])
        res = [
            [-1 for col in range(cols)]
            for row in range(rows)
        ]
        steps = [[0, 1], [1, 0], [-1, 0], [0, -1]]
        queue = deque()
        for row in range(rows):
            for col in range(cols):
                if mat[row][col] == 0:
                    queue.append((row, col, 0))
        while queue:
            r, c, depth = queue.popleft()
            if r < 0 or c < 0 or r == rows or c == cols:
                continue
            if res[r][c] != -1:
                continue
            res[r][c] = depth
            for r_, c_ in steps:
                new_row, new_col = r + r_, c + c_
                queue.append((new_row, new_col, depth + 1))
        return res
20900