<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]]:
        visited = set()
        q = deque()
        res = [[0] * len(mat[0]) for _ in range(len(mat))]
        rows, cols = len(mat), len(mat[0])
        for r in range(rows):
            for c in range(cols):
                if mat[r][c] == 0:
                    q.append((r,c))
                    visited.add((r,c))
                else:
                    res[r][c] = float("inf")
        def bfs():
            while q:
                r, c = q.popleft()
                dirn = [[0,1], [0,-1], [1,0], [-1,0]]
                for dr,dc in dirn:
                    nr, nc = dr+r, dc+c
                    if nr in range(rows) and nc in range(cols) and (nr,nc) not in visited:
                        if res[nr][nc] > res[r][c] + 1:
                            res[nr][nc] = res[r][c] + 1
                            q.append((nr, nc))
                            visited.add((nr, nc))
        bfs()
        return res
                        

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