<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) 
        matrix = [row[:] for row in mat]
        m = len(matrix)
        n = len(matrix[0])
        queue = deque()
        seen = set()

        for row in range(m):
            for col in range(n):
                if matrix[row][col] == 0:
                    queue.append((row, col, 0))
                    seen.add((row, col))

        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        while queue:
            row, col, steps = queue.popleft()
            for dx, dy in directions:
                next_row, next_col = row + dy, col + dx
                if (next_row, next_col) not in seen and valid(next_row, next_col):
                    seen.add((next_row, next_col))
                    queue.append((next_row, next_col, steps + 1))
                    matrix[next_row][next_col] = steps + 1
        
        return matrix
20700