<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]]:
        
        width = len(mat)
        height = len(mat[0])

        visited = set()
        queue = deque()

        maxm = 10000

        # search for zeros
        for x in range(width):
            for y in range(height):
                if mat[x][y] == 0:
                    visited.add((x,y))
                    queue.append((x,y, 0))
                else:
                    mat[x][y] = maxm

        print(queue)

        directions = [(0,-1), (-1,0), (0, 1), (1,0) ]

        while queue:
            x, y, steps = queue.popleft()

            for dx, dy in directions:
                
                xn = x+dx
                yn = y+dy

                if 0 <= xn < width and 0 <= yn < height:

                    if mat[xn][yn] == maxm:
                        queue.append((xn, yn, steps+1))
                    
                    mat[xn][yn] = min(mat[xn][yn], steps+1)
                



        return mat
19500