<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
from typing import List

class Solution:
    def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
        n = len(mat)
        m = len(mat[0])
        
        # Initialize distance matrix with high values (infinity)
        dist = [[float('inf') for _ in range(m)] for _ in range(n)]
        queue = deque()

        # Add all 0's to the queue and set their distance to 0
        for i in range(n):
            for j in range(m):
                if mat[i][j] == 0:
                    dist[i][j] = 0
                    queue.append((i, j))

        # Directions for up, down, left, and right
        directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]

        # Perform BFS
        while queue:
            x, y = queue.popleft()

            for dx, dy in directions:
                nx, ny = x + dx, y + dy
                
                # Check bounds
                if 0 <= nx < n and 0 <= ny < m:
                    # If we find a shorter path, update the distance
                    if dist[nx][ny] > dist[x][y] + 1:
                        dist[nx][ny] = dist[x][y] + 1
                        queue.append((nx, ny))

        return dist

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