<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]]:
        m, n = len(mat), len(mat[0])

        res = [[-1 for _ in range(n)] for _ in range(m)]
        
        def valid(i, j):
            nonlocal n, m
            if i >= 0 and i < m and j >= 0 and j < n:
                return True
        
        visited = set()
        queue = deque()
        
        for row in range(m):
            for col in range(n):
                if mat[row][col] == 0:
                    queue.append((row, col, 0))
                    visited.add((row, col))

        while queue:
            i, j, depth = queue.popleft()
            
            res[i][j] = depth
            ds = [(1, 0), (-1, 0), (0, 1), (0, -1)]
            for d in ds:
                x, y = i + d[0], j + d[1]
                if valid(x, y) and (x, y) not in visited:
                    queue.append((x, y, depth + 1))
                    visited.add((x, y))

        return res



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