<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]]:
        # BFS, O(mn), O(mn), minimum steps problem, mark 1s as unprocessed and start with 0s
        # the first time a one-cell is met, its minimum distance to a zero is determined
        # m, n = len(mat), len(mat[0])
        # q = deque([])
        # for i in range(m):
        #     for j in range(n):
        #         if mat[i][j] == 1:
        #             mat[i][j] = -1
        #         else:
        #             q.append((i, j))
        # while q:
        #     i, j = q.popleft()
        #     for ni, nj in (i+1,j), (i-1,j), (i,j+1), (i,j-1):
        #         # It's fine to skip neighboring 0s because it won't be the shortest path
        #         if not 0<=ni<m or not 0<=nj<n or mat[ni][nj] != -1:
        #             continue
        #         mat[ni][nj] = mat[i][j] + 1
        #         q.append((ni, nj))
        # return mat

        # DP, O(mn), O(1)
        m, n = len(mat), len(mat[0])
        for r in range(m):
            for c in range(n):
                if mat[r][c] > 0:
                    top = mat[r-1][c] if r > 0 else float('inf')
                    left = mat[r][c-1] if c > 0 else float('inf')
                    mat[r][c] = min(top, left) + 1
        for r in range(m-1, -1, -1):
            for c in range(n-1, -1, -1):
                if mat[r][c] > 0:
                    bottom = mat[r+1][c] if r < m-1 else float('inf')
                    right = mat[r][c+1] if c < n-1 else float('inf')
                    mat[r][c] = min(mat[r][c], bottom+1, right+1)
        return mat
18700