<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]]:
        m,n = len(mat), len(mat[0])
        dist = [[0] * n for _ in range(m)]

        visited = set()
        dr = [0,0,-1,1]
        dc = [-1,1,0,0]

        dq = deque()
        for i in range(m):
            for j in range(n):
                if mat[i][j] == 0:
                    dq.append((i,j,0))
                    visited.add((i,j))
        
        while dq:
            cur_r, cur_c, curdist = dq.popleft()
            for i in range(4):
                newr = cur_r + dr[i]
                newc = cur_c + dc[i]
                if newr < 0 or newr == m or newc < 0 or newc == n or (newr, newc) in visited:
                    continue
                print(newr, newc)
                visited.add((newr, newc))
                dq.append((newr, newc, curdist + 1))
                dist[newr][newc] = curdist + 1
        return dist

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