<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]]:
        if not mat:
            return []

        R, C = len(mat) ,len(mat[0])

        def in_bounds(row, col):
            return 0 <= row < R and 0 <= col < C

        res, seen, queue = [[0 for _ in range(C)] for _ in range(R)], [[False for _ in range(C)] for _ in range(R)], deque([])

        for row in range(R):
            for col in range(C):
                if mat[row][col] == 0:
                    queue.append((row, col, 0))
                    seen[row][col] = True

        while queue:
            row, col, pathlen = queue.popleft()
            for dr, dc in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
                if in_bounds(row + dr, col + dc) and not seen[row+dr][col+dc]:
                    seen[row+dr][col+dc], res[row+dr][col+dc] = True, pathlen + 1
                    queue.append((row+dr, col+dc, pathlen + 1))
        
        return res

20600