<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]]:
        rows,cols = len(mat), len(mat[0])
        res = [[0 for _ in range(cols)] for _ in range(rows)]
        q = deque([])
        visited=set()

        for i in range(rows):
            for j in range(cols):
                if mat[i][j]==0:
                    q.append((i,j))
                    visited.add((i,j))

        directions = [(-1,0), (0,1),(1,0), (0,-1)]
        while q:
            for i in range(len(q)):
                r,c = q.popleft()
                for dr, dc in directions:
                    rr, cc = r+dr, c+dc
                    # only works for non zero elements. 0s are already seen and added to the queue before entrering the while
                    if 0<=rr<rows and 0<=cc<cols and (rr,cc) not in visited: 
                        mat[rr][cc] = mat[r][c]+1
                        visited.add((rr,cc))
                        q.append((rr,cc))

        return mat
21200