<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>

class Solution:
    def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
        def valid(row, col):
            return 0 <= row < m and 0 <= col < n
        
        m = len(mat)
        n = len(mat[0])

        queue = collections.deque()
        seen = set()

        for row in range(m):
            for col in range(n):
                if mat[row][col] == 0:
                    queue.append((row, col, 0))
                    seen.add((row, col))
        
        ret = [row[:] for row in mat]
        while queue:
            row, col, step = queue.popleft()
            ret[row][col] = step
            for x, y in [(0,1), (0, -1), (1,0), (-1,0)]:
                n_row, n_col = row + x, col + y
                if (n_row, n_col) not in seen and valid(n_row, n_col):
                    seen.add((n_row, n_col))
                    queue.append((n_row, n_col, step + 1))
        return ret

20800