<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])

        vis = [[0 for _ in range(m)]for _ in range(n)]

        dis = [[0 for _ in range(m)]for _ in range(n)]
        
        q = deque()

        for i in range(n):
            for j in range(m):

                if mat[i][j]==0:
                    q.append((i,j,0))
                    vis[i][j]=1
        delr = [-1,0,1,0]
        delc = [0,1,0,-1]

        while q:
            row,col,steps = q.popleft()
            dis[row][col]=steps

            for i in range(4):
                nrow=row+delr[i]
                ncol = col +delc[i]
                if nrow >= 0 and nrow < n and ncol >= 0 and ncol < m and vis[nrow][ncol] == 0 :

                    vis[nrow][ncol] = 1 
                    q.append((nrow, ncol, steps+1))
	            
        return dis
20400