<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])
        result = [[float('inf')] * cols for _ in range(rows)]  # Initialize with infinity
        queue = deque()

        # Enqueue all cells with 0 and set their distance to 0
        for r in range(rows):
            for c in range(cols):
                if mat[r][c] == 0:
                    result[r][c] = 0
                    queue.append((r, c))

        # Directions for moving up, down, left, right
        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]

        # BFS from all 0's at once
        while queue:
            r, c = queue.popleft()

            # Check all 4 directions
            for dr, dc in directions:
                nr, nc = r + dr, c + dc

                # Check if the new position is within bounds and if the new distance is shorter
                if 0 <= nr < rows and 0 <= nc < cols and result[nr][nc] > result[r][c] + 1:
                    result[nr][nc] = result[r][c] + 1
                    queue.append((nr, nc))

        return result
19500