<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]]:
        # bfs, use mat as the distance matrix
        # strategy: visit the 0's, not the non-zeros
        # update dist of the 1's from a 0. update via the shortest path by comparing, and update when checking adjacents

        queue = deque()
        
        for i in range(len(mat)):
            for j in range(len(mat[0])):
                if mat[i][j] == 0:
                    queue.append((i, j))

                else:
                    mat[i][j] = float('inf')

        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]

        while queue:
            currR, currC = queue.popleft()

            for dirR, dirC in directions:
                nextR, nextC = currR + dirR, currC + dirC

                if (nextR >= 0 and nextC >= 0 and nextR < len(mat) and nextC < len(mat[0])) and mat[nextR][nextC] > mat[currR][currC] + 1:
                    mat[nextR][nextC] = mat[currR][currC] + 1
                    queue.append((nextR, nextC))

        return mat;
19200