<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 to find shortest path unless we already know?
        """
        if not mat or not mat[0]:
            return []
        distances = [[0 for i in range(len(mat[0]))] for j in range(len(mat))]

        q = deque()
        for r in range(len(mat)):
            for c in range(len(mat[r])):
                if mat[r][c] == 0:
                    distances[r][c] = 0
                    q.append((r, c))
                else:
                    distances[r][c] = len(mat) * len(mat[0])


        row_lim = len(mat)
        col_lim = len(mat[0])
        while q:
            (r, c) = q.popleft()

            neighbors = [(r+1,c), (r-1,c), (r,c+1), (r,c-1)]

            for (row, col) in neighbors:
                
                if row >= 0 and row < row_lim and col >= 0 and col < col_lim and distances[r][c] + 1 < distances[row][col]:
                    q.append((row, col))
                    print(distances[row][col])
                    print((r,c))
                    distances[row][col] = distances[r][c] + 1
        
        return distances
20500